{"id":540584,"date":"2021-09-25T07:05:54","date_gmt":"2021-09-25T07:05:54","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=540584"},"modified":"2022-12-26T09:25:32","modified_gmt":"2022-12-26T09:25:32","slug":"rd-sharma-solutions-for-class-12-maths-chapter-6-determinants","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-12-maths-chapter-6-determinants\/","title":{"rendered":"RD Sharma Solutions for Class 12 Maths Chapter 6\u2013Determinants"},"content":{"rendered":"\n

Class 12: Maths Chapter 6 solutions. Complete Class 12 Maths Chapter 6 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 6\u2013Determinants<\/h2>\n\n\n\n

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions<\/p>\n\n\n\n

Exercise 6.1 Page No: 6.10<\/p>\n\n\n\n

1. Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Let Mij<\/sub> and Cij<\/sub> represents the minor and co\u2013factor of an element, where i and j represent the row and column.The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.<\/p>\n\n\n\n

Also, Cij<\/sub> = (\u20131)i+j<\/sup> \u00d7 Mij<\/sub><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

From the given matrix we have,<\/p>\n\n\n\n

M11<\/sub> = \u20131<\/p>\n\n\n\n

M21<\/sub> = 20<\/p>\n\n\n\n

C11<\/sub> = (\u20131)1+1<\/sup> \u00d7 M11<\/sub><\/p>\n\n\n\n

= 1 \u00d7 \u20131<\/p>\n\n\n\n

= \u20131<\/p>\n\n\n\n

C21<\/sub> = (\u20131)2+1<\/sup> \u00d7 M21<\/sub><\/p>\n\n\n\n

= 20 \u00d7 \u20131<\/p>\n\n\n\n

= \u201320<\/p>\n\n\n\n

Now expanding along the first column we get<\/p>\n\n\n\n

|A| = a11<\/sub> \u00d7 C11<\/sub> + a21<\/sub>\u00d7 C21<\/sub><\/p>\n\n\n\n

= 5\u00d7 (\u20131) + 0 \u00d7 (\u201320)<\/p>\n\n\n\n

= \u20135<\/p>\n\n\n\n

(ii) Let Mij<\/sub> and Cij<\/sub> represents the minor and co\u2013factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.<\/p>\n\n\n\n

Also, Cij<\/sub> = (\u20131)i+j<\/sup> \u00d7 Mij<\/sub><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

From the above matrix we have<\/p>\n\n\n\n

M11<\/sub> = 3<\/p>\n\n\n\n

M21<\/sub> = 4<\/p>\n\n\n\n

C11<\/sub> = (\u20131)1+1<\/sup> \u00d7 M11<\/sub><\/p>\n\n\n\n

= 1 \u00d7 3<\/p>\n\n\n\n

= 3<\/p>\n\n\n\n

C21<\/sub> = (\u20131)2+1<\/sup> \u00d7 4<\/p>\n\n\n\n

= \u20131 \u00d7 4<\/p>\n\n\n\n

= \u20134<\/p>\n\n\n\n

Now expanding along the first column we get<\/p>\n\n\n\n

|A| = a11<\/sub> \u00d7 C11<\/sub> + a21<\/sub>\u00d7 C21<\/sub><\/p>\n\n\n\n

= \u20131\u00d7 3 + 2 \u00d7 (\u20134)<\/p>\n\n\n\n

= \u201311<\/p>\n\n\n\n

(iii) Let Mij<\/sub> and Cij<\/sub> represents the minor and co\u2013factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.<\/p>\n\n\n\n

Also, Cij<\/sub> = (\u20131)i+j<\/sup> \u00d7 Mij<\/sub><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

M31<\/sub> = \u20133 \u00d7 2 \u2013 (\u20131) \u00d7 2<\/p>\n\n\n\n

M31<\/sub> = \u20134<\/p>\n\n\n\n

C11<\/sub> = (\u20131)1+1<\/sup> \u00d7 M11<\/sub><\/p>\n\n\n\n

= 1 \u00d7 \u201312<\/p>\n\n\n\n

= \u201312<\/p>\n\n\n\n

C21<\/sub> = (\u20131)2+1<\/sup> \u00d7 M21<\/sub><\/p>\n\n\n\n

= \u20131 \u00d7 \u201316<\/p>\n\n\n\n

= 16<\/p>\n\n\n\n

C31<\/sub> = (\u20131)3+1<\/sup> \u00d7 M31<\/sub><\/p>\n\n\n\n

= 1 \u00d7 \u20134<\/p>\n\n\n\n

= \u20134<\/p>\n\n\n\n

Now expanding along the first column we get<\/p>\n\n\n\n

|A| = a11<\/sub> \u00d7 C11<\/sub> + a21<\/sub>\u00d7 C21<\/sub>+ a31<\/sub>\u00d7 C31<\/sub><\/p>\n\n\n\n

= 1\u00d7 (\u201312) + 4 \u00d7 16 + 3\u00d7 (\u20134)<\/p>\n\n\n\n

= \u201312 + 64 \u201312<\/p>\n\n\n\n

= 40<\/p>\n\n\n\n

(iv) Let Mij<\/sub> and Cij<\/sub> represents the minor and co\u2013factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.<\/p>\n\n\n\n

Also, Cij<\/sub> = (\u20131)i+j<\/sup> \u00d7 Mij<\/sub><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

M31<\/sub> = a \u00d7 c a \u2013 b \u00d7 bc<\/p>\n\n\n\n

M31<\/sub> = a2<\/sup>c \u2013 b2<\/sup>c<\/p>\n\n\n\n

C11<\/sub> = (\u20131)1+1<\/sup> \u00d7 M11<\/sub><\/p>\n\n\n\n

= 1 \u00d7 (ab2<\/sup> \u2013 ac2<\/sup>)<\/p>\n\n\n\n

= ab2<\/sup> \u2013 ac2<\/sup><\/p>\n\n\n\n

C21<\/sub> = (\u20131)2+1<\/sup> \u00d7 M21<\/sub><\/p>\n\n\n\n

= \u20131 \u00d7 (a2<\/sup>b \u2013 c2<\/sup>b)<\/p>\n\n\n\n

= c2<\/sup>b \u2013 a2<\/sup>b<\/p>\n\n\n\n

C31<\/sub> = (\u20131)3+1<\/sup> \u00d7 M31<\/sub><\/p>\n\n\n\n

= 1 \u00d7 (a2<\/sup>c \u2013 b2<\/sup>c)<\/p>\n\n\n\n

= a2<\/sup>c \u2013 b2<\/sup>c<\/p>\n\n\n\n

Now expanding along the first column we get<\/p>\n\n\n\n

|A| = a11<\/sub> \u00d7 C11<\/sub> + a21<\/sub>\u00d7 C21<\/sub>+ a31<\/sub>\u00d7 C31<\/sub><\/p>\n\n\n\n

= 1\u00d7 (ab2<\/sup> \u2013 ac2<\/sup>) + 1 \u00d7 (c2<\/sup>b \u2013 a2<\/sup>b) + 1\u00d7 (a2<\/sup>c \u2013 b2<\/sup>c)<\/p>\n\n\n\n

= ab2<\/sup> \u2013 ac2<\/sup> + c2<\/sup>b \u2013 a2<\/sup>b + a2<\/sup>c \u2013 b2<\/sup>c<\/p>\n\n\n\n

(v) Let Mij<\/sub> and Cij<\/sub> represents the minor and co\u2013factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.<\/p>\n\n\n\n

Also, Cij<\/sub> = (\u20131)i+j<\/sup> \u00d7 Mij<\/sub><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

M31<\/sub> = 2\u00d70 \u2013 5\u00d76<\/p>\n\n\n\n

M31<\/sub> = \u201330<\/p>\n\n\n\n

C11<\/sub> = (\u20131)1+1<\/sup> \u00d7 M11<\/sub><\/p>\n\n\n\n

= 1 \u00d7 5<\/p>\n\n\n\n

= 5<\/p>\n\n\n\n

C21<\/sub> = (\u20131)2+1<\/sup> \u00d7 M21<\/sub><\/p>\n\n\n\n

= \u20131 \u00d7 \u201340<\/p>\n\n\n\n

= 40<\/p>\n\n\n\n

C31<\/sub> = (\u20131)3+1<\/sup> \u00d7 M31<\/sub><\/p>\n\n\n\n

= 1 \u00d7 \u201330<\/p>\n\n\n\n

= \u201330<\/p>\n\n\n\n

Now expanding along the first column we get<\/p>\n\n\n\n

|A| = a11<\/sub> \u00d7 C11<\/sub> + a21<\/sub>\u00d7 C21<\/sub>+ a31<\/sub>\u00d7 C31<\/sub><\/p>\n\n\n\n

= 0\u00d7 5 + 1 \u00d7 40 + 3\u00d7 (\u201330)<\/p>\n\n\n\n

= 0 + 40 \u2013 90<\/p>\n\n\n\n

= 50<\/p>\n\n\n\n

(vi) Let Mij<\/sub> and Cij<\/sub> represents the minor and co\u2013factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.<\/p>\n\n\n\n

Also, Cij<\/sub> = (\u20131)i+j<\/sup> \u00d7 Mij<\/sub><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

M31<\/sub> = h \u00d7 f \u2013 b \u00d7 g<\/p>\n\n\n\n

M31<\/sub> = hf \u2013 bg<\/p>\n\n\n\n

C11<\/sub> = (\u20131)1+1<\/sup> \u00d7 M11<\/sub><\/p>\n\n\n\n

= 1 \u00d7 (bc\u2013 f2<\/sup>)<\/p>\n\n\n\n

= bc\u2013 f2<\/sup><\/p>\n\n\n\n

C21<\/sub> = (\u20131)2+1<\/sup> \u00d7 M21<\/sub><\/p>\n\n\n\n

= \u20131 \u00d7 (hc \u2013 fg)<\/p>\n\n\n\n

= fg \u2013 hc<\/p>\n\n\n\n

C31<\/sub> = (\u20131)3+1<\/sup> \u00d7 M31<\/sub><\/p>\n\n\n\n

= 1 \u00d7 (hf \u2013 bg)<\/p>\n\n\n\n

= hf \u2013 bg<\/p>\n\n\n\n

Now expanding along the first column we get<\/p>\n\n\n\n

|A| = a11<\/sub> \u00d7 C11<\/sub> + a21<\/sub>\u00d7 C21<\/sub>+ a31<\/sub>\u00d7 C31<\/sub><\/p>\n\n\n\n

= a\u00d7 (bc\u2013 f2<\/sup>) + h\u00d7 (fg \u2013 hc) + g\u00d7 (hf \u2013 bg)<\/p>\n\n\n\n

= abc\u2013 af2<\/sup> + hgf \u2013 h2<\/sup>c +ghf \u2013 bg2<\/sup><\/p>\n\n\n\n

(vii) Let Mij<\/sub> and Cij<\/sub> represents the minor and co\u2013factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.<\/p>\n\n\n\n

Also, Cij<\/sub> = (\u20131)i+j<\/sup> \u00d7 Mij<\/sub><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

M31<\/sub> = \u20131(1 \u00d7 0 \u2013 5 \u00d7 (\u20132)) \u2013 0(0 \u00d7 0 \u2013 (\u20131) \u00d7 (\u20132)) + 1(0 \u00d7 5 \u2013 (\u20131) \u00d7 1)<\/p>\n\n\n\n

M31<\/sub> = \u20139<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

M41<\/sub> = \u20131(1\u00d71 \u2013 (\u20131) \u00d7 (\u20132)) \u2013 0(0 \u00d7 1 \u2013 1 \u00d7 (\u20132)) + 1(0 \u00d7 (\u20131) \u2013 1 \u00d7 1)<\/p>\n\n\n\n

M41<\/sub> = 0<\/p>\n\n\n\n

C11<\/sub> = (\u20131)1+1<\/sup> \u00d7 M11<\/sub><\/p>\n\n\n\n

= 1 \u00d7 (\u20139)<\/p>\n\n\n\n

= \u20139<\/p>\n\n\n\n

C21<\/sub> = (\u20131)2+1<\/sup> \u00d7 M21<\/sub><\/p>\n\n\n\n

= \u20131 \u00d7 9<\/p>\n\n\n\n

= \u20139<\/p>\n\n\n\n

C31<\/sub> = (\u20131)3+1<\/sup> \u00d7 M31<\/sub><\/p>\n\n\n\n

= 1 \u00d7 \u20139<\/p>\n\n\n\n

= \u20139<\/p>\n\n\n\n

C41<\/sub> = (\u20131)4+1<\/sup> \u00d7 M41<\/sub><\/p>\n\n\n\n

= \u20131 \u00d7 0<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Now expanding along the first column we get<\/p>\n\n\n\n

|A| = a11<\/sub> \u00d7 C11<\/sub> + a21<\/sub>\u00d7 C21<\/sub>+ a31<\/sub>\u00d7 C31<\/sub> + a41<\/sub>\u00d7 C41<\/sub><\/p>\n\n\n\n

= 2 \u00d7 (\u20139) + (\u20133) \u00d7 \u20139 + 1 \u00d7 (\u20139) + 2 \u00d7 0<\/p>\n\n\n\n

= \u2013 18 + 27 \u20139<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

2. Evaluate the following determinants:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 |A| = x (5x + 1) \u2013 (\u20137) x<\/p>\n\n\n\n

|A| = 5x2<\/sup> + 8x<\/p>\n\n\n\n

(ii) Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 |A| = cos \u03b8 \u00d7 cos \u03b8 \u2013 (\u2013sin \u03b8) x sin \u03b8<\/p>\n\n\n\n

|A| = cos2<\/sup>\u03b8 + sin2<\/sup>\u03b8<\/p>\n\n\n\n

We know that cos2<\/sup>\u03b8 + sin2<\/sup>\u03b8 = 1<\/p>\n\n\n\n

|A| = 1<\/p>\n\n\n\n

(iii) Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 |A| = cos15\u00b0 \u00d7 cos75\u00b0 + sin15\u00b0 x sin75\u00b0<\/p>\n\n\n\n

We know that cos (A \u2013 B) = cos A cos B + Sin A sin B<\/p>\n\n\n\n

By substituting this we get, |A| = cos (75 \u2013 15)\u00b0<\/p>\n\n\n\n

|A| = cos60\u00b0<\/p>\n\n\n\n

|A| = 0.5<\/p>\n\n\n\n

(iv) Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 |A| = (a + ib) (a \u2013 ib) \u2013 (c + id) (\u2013c + id)<\/p>\n\n\n\n

= (a + ib) (a \u2013 ib) + (c + id) (c \u2013 id)<\/p>\n\n\n\n

= a2<\/sup> \u2013 i2<\/sup> b2<\/sup> + c2<\/sup> \u2013 i2<\/sup> d2<\/sup><\/p>\n\n\n\n

We know that i2<\/sup> = -1<\/p>\n\n\n\n

= a2<\/sup> \u2013 (\u20131) b2<\/sup> + c2<\/sup> \u2013 (\u20131) d2<\/sup><\/p>\n\n\n\n

= a2<\/sup> + b2<\/sup> + c2<\/sup> + d2<\/sup><\/p>\n\n\n\n

3. Evaluate:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Since |AB|= |A||B|<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 2(17 \u00d7 12 \u2013 5 \u00d7 20) \u2013 3(13 \u00d7 12 \u2013 5 \u00d7 15) + 7(13 \u00d7 20 \u2013 15 \u00d7 17)<\/p>\n\n\n\n

= 2 (204 \u2013 100) \u2013 3 (156 \u2013 75) + 7 (260 \u2013 255)<\/p>\n\n\n\n

= 2\u00d7104 \u2013 3\u00d781 + 7\u00d75<\/p>\n\n\n\n

= 208 \u2013 243 +35<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Now |A|2<\/sup> = |A|\u00d7|A|<\/p>\n\n\n\n

|A|2<\/sup>= 0<\/p>\n\n\n\n

4. Show that<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Let the given determinant as A<\/p>\n\n\n\n

Using sin (A+B) = sin A \u00d7 cos B + cos A \u00d7 sin B<\/p>\n\n\n\n

\u21d2 |A| = sin 10\u00b0 \u00d7 cos 80\u00b0 + cos 10\u00b0 x sin 80\u00b0<\/p>\n\n\n\n

|A| = sin (10 + 80)\u00b0<\/p>\n\n\n\n

|A| = sin90\u00b0<\/p>\n\n\n\n

|A| = 1<\/p>\n\n\n\n

Hence Proved<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 2(1 \u00d7 1 \u2013 4 \u00d7 (\u20132)) \u2013 3(7 \u00d7 1 \u2013 (\u20132) \u00d7 (\u20133)) \u2013 5(7 \u00d7 4 \u2013 1 \u00d7 (\u20133))<\/p>\n\n\n\n

= 2(1 + 8) \u2013 3(7 \u2013 6) \u2013 5(28 + 3)<\/p>\n\n\n\n

= 2 \u00d7 9 \u2013 3 \u00d7 1 \u2013 5 \u00d7 31<\/p>\n\n\n\n

= 18 \u2013 3 \u2013 155<\/p>\n\n\n\n

= \u2013140<\/p>\n\n\n\n

Now by expanding along the second column<\/p>\n\n\n\n

\"RD<\/figure>\n\n\n\n

= 2(1 \u00d7 1 \u2013 4 \u00d7 (\u20132)) \u2013 7(3 \u00d7 1 \u2013 4 \u00d7 (\u20135)) \u2013 3(3 \u00d7 (\u20132) \u2013 1 \u00d7 (\u20135))<\/p>\n\n\n\n

= 2 (1 + 8) \u2013 7 (3 + 20) \u2013 3 (\u20136 + 5)<\/p>\n\n\n\n

= 2 \u00d7 9 \u2013 7 \u00d7 23 \u2013 3 \u00d7 (\u20131)<\/p>\n\n\n\n

= 18 \u2013 161 +3<\/p>\n\n\n\n

= \u2013140<\/p>\n\n\n\n

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 |A| = 0 (0 \u2013 sin\u03b2 (\u2013sin\u03b2)) \u2013sin\u03b1 (\u2013sin\u03b1 \u00d7 0 \u2013 sin\u03b2 cos\u03b1) \u2013 cos\u03b1 ((\u2013sin\u03b1) (\u2013sin\u03b2) \u2013 0 \u00d7 cos\u03b1)<\/p>\n\n\n\n

|A| = 0 + sin\u03b1 sin\u03b2 cos\u03b1 \u2013 cos\u03b1 sin\u03b1 sin\u03b2<\/p>\n\n\n\n

|A| = 0<\/p>\n\n\n\n

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions<\/p>\n\n\n\n

Exercise 6.2 Page No: 6.57<\/p>\n\n\n\n

1. Evaluate the following determinant:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(ii) Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 1[(109) (12) \u2013 (119) (11)]<\/p>\n\n\n\n

= 1308 \u2013 1309<\/p>\n\n\n\n

= \u2013 1<\/p>\n\n\n\n

So, \u0394 = \u2013 1<\/p>\n\n\n\n

(iii) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= a (bc \u2013 f2<\/sup>) \u2013 h (hc \u2013 fg) + g (hf \u2013 bg)<\/p>\n\n\n\n

= abc \u2013 af2<\/sup> \u2013 ch2<\/sup> + fgh + fgh \u2013 bg2<\/sup><\/p>\n\n\n\n

= abc + 2fgh \u2013 af2<\/sup> \u2013 bg2<\/sup> \u2013 ch2<\/sup><\/p>\n\n\n\n

So, \u0394 = abc + 2fgh \u2013 af2<\/sup> \u2013 bg2<\/sup> \u2013 ch2<\/sup><\/p>\n\n\n\n

(iv) Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 2[1(24 \u2013 4)] = 40<\/p>\n\n\n\n

So, \u0394 = 40<\/p>\n\n\n\n

(v) Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

= 1[(\u2013 7) (\u2013 36) \u2013 (\u2013 20) (\u2013 13)]<\/p>\n\n\n\n

= 252 \u2013 260<\/p>\n\n\n\n

= \u2013 8<\/p>\n\n\n\n

So, \u0394 = \u2013 8<\/p>\n\n\n\n

(vi) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(vii) Given<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(viii) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions<\/p>\n\n\n\n

2. Without expanding, show that the value of each of the following determinants is zero:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(ii) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(iii) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iv) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(v) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(vi) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(vii) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(viii) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(ix) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

As, C1<\/sub> = C2<\/sub>, hence determinant is zero<\/p>\n\n\n\n

(x) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(xi) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(xii) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(xiii) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(xiv) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(xv) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(xvi) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(xvii) Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions<\/p>\n\n\n\n

Evaluate the following (3 \u2013 9):<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= (a + b + c) (b \u2013 a) (c \u2013 a) (b \u2013 c)<\/p>\n\n\n\n

So, \u0394 = (a + b + c) (b \u2013 a) (c \u2013 a) (b \u2013 c)<\/p>\n\n\n\n

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

= a [a (a + x + y) + az] + 0 + 0<\/p>\n\n\n\n

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions<\/p>\n\n\n\n

= a2 <\/sup>(a + x + y + z)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

So, \u0394 = a2 <\/sup>(a + x + y + z)<\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Prove the following identities (11 \u2013 45):<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

= \u2013 (a + b + c) [(b \u2013 c) (a + b \u2013 2c) \u2013 (c \u2013 a) (c + a \u2013 2b)]<\/p>\n\n\n\n

= 3abc \u2013 a3<\/sup> \u2013 b3<\/sup> \u2013 c3<\/sup><\/p>\n\n\n\n

Therefore, L.H.S = R.H.S,<\/p>\n\n\n\n

Hence the proof.<\/p>\n\n\n\n

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

L.H.S =
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now by applying, R1<\/sub>\u2192R1<\/sub> + R2<\/sub> + R3<\/sub>, we get,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Hence, the proof.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= \u2013 xyz(x \u2013 y) (z \u2013 y) [z2<\/sup> + y2<\/sup> + zy \u2013 x2<\/sup> \u2013 y2<\/sup> \u2013 xy]<\/p>\n\n\n\n

= \u2013 xyz(x \u2013 y) (z \u2013 y) [(z \u2013 x) (z + x0 + y (z \u2013 x)]<\/p>\n\n\n\n

= \u2013 xyz(x \u2013 y) (z \u2013 y) (z \u2013 x) (x + y + z)<\/p>\n\n\n\n

= R.H.S<\/p>\n\n\n\n

Hence, the proof.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

= (a2<\/sup> + b2<\/sup> + c2<\/sup>) (b \u2013 a) (c \u2013 a) [(b + a) (\u2013 b) \u2013 (\u2013 c) (c + a)]<\/p>\n\n\n\n

= (a2<\/sup> + b2<\/sup> + c2<\/sup>) (a \u2013 b) (c \u2013 a) (b \u2013 c) (a + b + c)<\/p>\n\n\n\n

= R.H.S<\/p>\n\n\n\n

Hence, the proof.<\/p>\n\n\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= [(2a + 4) (1) \u2013 (1) (2a + 6)]<\/p>\n\n\n\n

= \u2013 2<\/p>\n\n\n\n

= R.H.S<\/p>\n\n\n\n

Hence, the proof.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

= \u2013 (a2<\/sup> + b2<\/sup> + c2<\/sup>) (a \u2013 b) (c \u2013 a) [(\u2013 (b + a)) (\u2013 b) \u2013 (c) (c + a)]<\/p>\n\n\n\n

= (a \u2013 b) (b \u2013 c) (c \u2013 a) (a + b + c) (a2<\/sup> + b2<\/sup> + c2<\/sup>)<\/p>\n\n\n\n

= R.H.S<\/p>\n\n\n\n

Hence, the proof.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= R.H.S<\/p>\n\n\n\n

Hence, the proof.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Expanding the determinant along R1<\/sub>, we have<\/p>\n\n\n\n

\u0394 = 1[(1) (7) \u2013 (3) (2)] \u2013 0 + 0<\/p>\n\n\n\n

\u2234 \u0394 = 7 \u2013 6 = 1<\/p>\n\n\n\n

Thus,
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Hence the proof.<\/p>\n\n\n\n

Exercise 6.3 Page No: 6.71<\/p>\n\n\n\n

1. Find the area of the triangle with vertices at the points:<\/strong><\/p>\n\n\n\n

(i) (3, 8), (-4, 2) and (5, -1)<\/strong><\/p>\n\n\n\n

(ii) (2, 7), (1, 1) and (10, 8)<\/strong><\/p>\n\n\n\n

(iii) (-1, -8), (-2, -3) and (3, 2)<\/strong><\/p>\n\n\n\n

(iv) (0, 0), (6, 0) and (4, 3)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given (3, 8), (-4, 2) and (5, -1) are the vertices of the triangle.<\/p>\n\n\n\n

We know that, if vertices of a triangle are (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>) and (x3<\/sub>, y3<\/sub>), then the area of the triangle is given by:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(ii) Given (2, 7), (1, 1) and (10, 8) are the vertices of the triangle.<\/p>\n\n\n\n

We know that if vertices of a triangle are (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>) and (x3<\/sub>, y3<\/sub>), then the area of the triangle is given by:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iii) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle.<\/p>\n\n\n\n

We know that if vertices of a triangle are (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>) and (x3<\/sub>, y3<\/sub>), then the area of the triangle is given by:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

As we know area cannot be negative. Therefore, 15 square unit is the area<\/p>\n\n\n\n

Thus area of triangle is 15 square units<\/p>\n\n\n\n

(iv) Given (-1, -8), (-2, -3) and (3, 2) are the vertices of the triangle.<\/p>\n\n\n\n

We know that if vertices of a triangle are (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>) and (x3<\/sub>, y3<\/sub>), then the area of the triangle is given by:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

2. Using the determinants show that the following points are collinear:<\/strong><\/p>\n\n\n\n

(i) (5, 5), (-5, 1) and (10, 7)<\/strong><\/p>\n\n\n\n

(ii) (1, -1), (2, 1) and (10, 8)<\/strong><\/p>\n\n\n\n

(iii) (3, -2), (8, 8) and (5, 2)<\/strong><\/p>\n\n\n\n

(iv) (2, 3), (-1, -2) and (5, 8)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given (5, 5), (-5, 1) and (10, 7)<\/p>\n\n\n\n

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>) and (x3<\/sub>, y3<\/sub>), then the area of the triangle is given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(ii) Given (1, -1), (2, 1) and (10, 8)<\/p>\n\n\n\n

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>) and (x3<\/sub>, y3<\/sub>), then the area of the triangle is given by,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iii) Given (3, -2), (8, 8) and (5, 2)<\/p>\n\n\n\n

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>) and (x3<\/sub>, y3<\/sub>), then the area of the triangle is given by,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now, by substituting given value in above formula<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Since, Area of triangle is zero<\/p>\n\n\n\n

Hence, points are collinear.<\/p>\n\n\n\n

(iv) Given (2, 3), (-1, -2) and (5, 8)<\/p>\n\n\n\n

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>) and (x3<\/sub>, y3<\/sub>), then the area of the triangle is given by,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

3. If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given (a, 0), (0, b) and (1, 1) are collinear<\/p>\n\n\n\n

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>) and (x3<\/sub>, y3<\/sub>), then the area of the triangle is given by,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 a + b = ab<\/p>\n\n\n\n

Hence Proved<\/p>\n\n\n\n

4. Using the determinants prove that the points (a, b), (a\u2019, b\u2019) and (a \u2013 a\u2019, b \u2013 b) are collinear if a b\u2019 = a\u2019 b.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given (a, b), (a\u2019, b\u2019) and (a \u2013 a\u2019, b \u2013 b) are collinear<\/p>\n\n\n\n

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>) and (x3<\/sub>, y3<\/sub>), then the area of the triangle is given by,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 a b\u2019 = a\u2019 b<\/p>\n\n\n\n

Hence, the proof.<\/p>\n\n\n\n

5. Find the value of \u03bb so that the points (1, -5), (-4, 5) and (\u03bb, 7) are collinear.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given (1, -5), (-4, 5) and (\u03bb, 7) are collinear<\/p>\n\n\n\n

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>) and (x3<\/sub>, y3<\/sub>), then the area of the triangle is given by,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 \u2013 50 \u2013 10\u03bb = 0<\/p>\n\n\n\n

\u21d2 \u03bb = \u2013 5<\/p>\n\n\n\n

6. Find the value of x if the area of \u2206 is 35 square cms with vertices (x, 4), (2, -6) and (5, 4).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given (x, 4), (2, -6) and (5, 4) are the vertices of a triangle.<\/p>\n\n\n\n

We have the condition that three points to be collinear, the area of the triangle formed by these points will be zero. Now, we know that, vertices of a triangle are (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>) and (x3<\/sub>, y3<\/sub>), then the area of the triangle is given by,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

\u21d2 [x (\u2013 10) \u2013 4(\u2013 3) + 1(8 \u2013 30)] = \u00b1 70<\/p>\n\n\n\n

\u21d2 [\u2013 10x + 12 + 38] = \u00b1 70<\/p>\n\n\n\n

\u21d2 \u00b170 = \u2013 10x + 50<\/p>\n\n\n\n

Taking positive sign, we get<\/p>\n\n\n\n

\u21d2 + 70 = \u2013 10x + 50<\/p>\n\n\n\n

\u21d2 10x = \u2013 20<\/p>\n\n\n\n

\u21d2 x = \u2013 2<\/p>\n\n\n\n

Taking \u2013negative sign, we get<\/p>\n\n\n\n

\u21d2 \u2013 70 = \u2013 10x + 50<\/p>\n\n\n\n

\u21d2 10x = 120<\/p>\n\n\n\n

\u21d2 x = 12<\/p>\n\n\n\n

Thus x = \u2013 2, 12<\/p>\n\n\n\n

Exercise 6.4 Page No: 6.84<\/p>\n\n\n\n

Solve the following system of linear equations by Cramer\u2019s rule:<\/strong><\/p>\n\n\n\n

1. x \u2013 2y = 4<\/strong><\/p>\n\n\n\n

-3x + 5y = -7<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given x \u2013 2y = 4<\/p>\n\n\n\n

-3x + 5y = -7<\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D = 5(1) \u2013 (\u2013 3) (\u2013 2)<\/p>\n\n\n\n

\u21d2 D = 5 \u2013 6<\/p>\n\n\n\n

\u21d2 D = \u2013 1<\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 5(4) \u2013 (\u2013 7) (\u2013 2)<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 20 \u2013 14<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 6<\/p>\n\n\n\n

And<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 1(\u2013 7) \u2013 (\u2013 3) (4)<\/p>\n\n\n\n

\u21d2 D2<\/sub> = \u2013 7 + 12<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 5<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

2. 2x \u2013 y = 1<\/strong><\/p>\n\n\n\n

7x \u2013 2y = -7<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given 2x \u2013 y = 1 and<\/p>\n\n\n\n

7x \u2013 2y = -7<\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 1(\u2013 2) \u2013 (\u2013 7) (\u2013 1)<\/p>\n\n\n\n

\u21d2 D1<\/sub> = \u2013 2 \u2013 7<\/p>\n\n\n\n

\u21d2 D1<\/sub> = \u2013 9<\/p>\n\n\n\n

And<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 2(\u2013 7) \u2013 (7) (1)<\/p>\n\n\n\n

\u21d2 D2<\/sub> = \u2013 14 \u2013 7<\/p>\n\n\n\n

\u21d2 D2<\/sub> = \u2013 21<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

3. 2x \u2013 y = 17<\/strong><\/p>\n\n\n\n

3x + 5y = 6<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given 2x \u2013 y = 17 and<\/p>\n\n\n\n

3x + 5y = 6<\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 17(5) \u2013 (6) (\u2013 1)<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 85 + 6<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 91<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 2(6) \u2013 (17) (3)<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 12 \u2013 51<\/p>\n\n\n\n

\u21d2 D2<\/sub> = \u2013 39<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

4. 3x + y = 19<\/strong><\/p>\n\n\n\n

3x \u2013 y = 23<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D = 3(\u2013 1) \u2013 (3) (1)<\/p>\n\n\n\n

\u21d2 D = \u2013 3 \u2013 3<\/p>\n\n\n\n

\u21d2 D = \u2013 6<\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 19(\u2013 1) \u2013 (23) (1)<\/p>\n\n\n\n

\u21d2 D1<\/sub> = \u2013 19 \u2013 23<\/p>\n\n\n\n

\u21d2 D1<\/sub> = \u2013 42<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 3(23) \u2013 (19) (3)<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 69 \u2013 57<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 12<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

5. 2x \u2013 y = -2<\/strong><\/p>\n\n\n\n

3x + 4y = 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given 2x \u2013 y = -2 and<\/p>\n\n\n\n

3x + 4y = 3<\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 3(2) \u2013 (\u2013 2) (3)<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 6 + 6<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 12<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

6. 3x + ay = 4<\/strong><\/p>\n\n\n\n

2x + ay = 2, a \u2260 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given 3x + ay = 4 and<\/p>\n\n\n\n

2x + ay = 2, a \u2260 0<\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

3x + ay = 4<\/p>\n\n\n\n

2x + ay = 2, a\u22600<\/p>\n\n\n\n

So by comparing with the theorem, let\u2019s find D, D1<\/sub> and D2<\/sub><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D = 3(a) \u2013 (2) (a)<\/p>\n\n\n\n

\u21d2 D = 3a \u2013 2a<\/p>\n\n\n\n

\u21d2 D = a<\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 4(a) \u2013 (2) (a)<\/p>\n\n\n\n

\u21d2 D = 4a \u2013 2a<\/p>\n\n\n\n

\u21d2 D = 2a<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 3(2) \u2013 (2) (4)<\/p>\n\n\n\n

\u21d2 D = 6 \u2013 8<\/p>\n\n\n\n

\u21d2 D = \u2013 2<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

7. 2x + 3y = 10<\/strong><\/p>\n\n\n\n

x + 6y = 4<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D = 2 (6) \u2013 (3) (1)<\/p>\n\n\n\n

\u21d2 D = 12 \u2013 3<\/p>\n\n\n\n

\u21d2 D = 9<\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 10 (6) \u2013 (3) (4)<\/p>\n\n\n\n

\u21d2 D = 60 \u2013 12<\/p>\n\n\n\n

\u21d2 D = 48<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 2 (4) \u2013 (10) (1)<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 8 \u2013 10<\/p>\n\n\n\n

\u21d2 D2<\/sub> = \u2013 2<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

8. 5x + 7y = -2<\/strong><\/p>\n\n\n\n

4x + 6y = -3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Now, here we have<\/p>\n\n\n\n

5x + 7y = \u2013 2<\/p>\n\n\n\n

4x + 6y = \u2013 3<\/p>\n\n\n\n

So by comparing with the theorem, let\u2019s find D, D1<\/sub> and D2<\/sub><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D = 5(6) \u2013 (7) (4)<\/p>\n\n\n\n

\u21d2 D = 30 \u2013 28<\/p>\n\n\n\n

\u21d2 D = 2<\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = \u2013 2(6) \u2013 (7) (\u2013 3)<\/p>\n\n\n\n

\u21d2 D1<\/sub> = \u2013 12 + 21<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 9<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D2<\/sub> = \u2013 3(5) \u2013 (\u2013 2) (4)<\/p>\n\n\n\n

\u21d2 D2<\/sub> = \u2013 15 + 8<\/p>\n\n\n\n

\u21d2 D2<\/sub> = \u2013 7<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

9. 9x + 5y = 10<\/strong><\/p>\n\n\n\n

3y \u2013 2x = 8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D = 3(9) \u2013 (5) (\u2013 2)<\/p>\n\n\n\n

\u21d2 D = 27 + 10<\/p>\n\n\n\n

\u21d2 D = 37<\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 10(3) \u2013 (8) (5)<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 30 \u2013 40<\/p>\n\n\n\n

\u21d2 D1<\/sub> = \u2013 10<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 9(8) \u2013 (10) (\u2013 2)<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 72 + 20<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 92<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

10. x + 2y = 1<\/strong><\/p>\n\n\n\n

3x + y = 4<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D = 1(1) \u2013 (3) (2)<\/p>\n\n\n\n

\u21d2 D = 1 \u2013 6<\/p>\n\n\n\n

\u21d2 D = \u2013 5<\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 1(1) \u2013 (2) (4)<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 1 \u2013 8<\/p>\n\n\n\n

\u21d2 D1<\/sub> = \u2013 7<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 1(4) \u2013 (1) (3)<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 4 \u2013 3<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 1<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solve the following system of linear equations by Cramer\u2019s rule:<\/strong><\/p>\n\n\n\n

11. 3x + y + z = 2<\/strong><\/p>\n\n\n\n

2x \u2013 4y + 3z = -1<\/strong><\/p>\n\n\n\n

4x + y \u2013 3z = -11<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Now, here we have<\/p>\n\n\n\n

3x + y + z = 2<\/p>\n\n\n\n

2x \u2013 4y + 3z = \u2013 1<\/p>\n\n\n\n

4x + y \u2013 3z = \u2013 11<\/p>\n\n\n\n

So by comparing with the theorem, let\u2019s find D, D1<\/sub>, D2<\/sub> and D3<\/sub><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D = 3[(\u2013 4) (\u2013 3) \u2013 (3) (1)] \u2013 1[(2) (\u2013 3) \u2013 12] + 1[2 \u2013 4(\u2013 4)]<\/p>\n\n\n\n

\u21d2 D = 3[12 \u2013 3] \u2013 [\u2013 6 \u2013 12] + [2 + 16]<\/p>\n\n\n\n

\u21d2 D = 27 + 18 + 18<\/p>\n\n\n\n

\u21d2 D = 63<\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 2[( \u2013 4)( \u2013 3) \u2013 (3)(1)] \u2013 1[( \u2013 1)( \u2013 3) \u2013 ( \u2013 11)(3)] + 1[( \u2013 1) \u2013 ( \u2013 4)( \u2013 11)]<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 2[12 \u2013 3] \u2013 1[3 + 33] + 1[\u2013 1 \u2013 44]<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 2[9] \u2013 36 \u2013 45<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 18 \u2013 36 \u2013 45<\/p>\n\n\n\n

\u21d2 D1<\/sub> = \u2013 63<\/p>\n\n\n\n

Again<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 3[3 + 33] \u2013 2[\u2013 6 \u2013 12] + 1[\u2013 22 + 4]<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 3[36] \u2013 2(\u2013 18) \u2013 18<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 126<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 3[44 + 1] \u2013 1[\u2013 22 + 4] + 2[2 + 16]<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 3[45] \u2013 1(\u2013 18) + 2(18)<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 135 + 18 + 36<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 189<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

12. x \u2013 4y \u2013 z = 11<\/strong><\/p>\n\n\n\n

2x \u2013 5y + 2z = 39<\/strong><\/p>\n\n\n\n

-3x + 2y + z = 1<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

x \u2013 4y \u2013 z = 11<\/p>\n\n\n\n

2x \u2013 5y + 2z = 39<\/p>\n\n\n\n

-3x + 2y + z = 1<\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now, here we have<\/p>\n\n\n\n

x \u2013 4y \u2013 z = 11<\/p>\n\n\n\n

2x \u2013 5y + 2z = 39<\/p>\n\n\n\n

\u2013 3x + 2y + z = 1<\/p>\n\n\n\n

So by comparing with theorem, now we have to find D, D1<\/sub> and D2<\/sub><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D = 1[(\u2013 5) (1) \u2013 (2) (2)] + 4[(2) (1) + 6] \u2013 1[4 + 5(\u2013 3)]<\/p>\n\n\n\n

\u21d2 D = 1[\u2013 5 \u2013 4] + 4[8] \u2013 [\u2013 11]<\/p>\n\n\n\n

\u21d2 D = \u2013 9 + 32 + 11<\/p>\n\n\n\n

\u21d2 D = 34<\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 11[(\u2013 5) (1) \u2013 (2) (2)] + 4[(39) (1) \u2013 (2) (1)] \u2013 1[2 (39) \u2013 (\u2013 5) (1)]<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 11[\u2013 5 \u2013 4] + 4[39 \u2013 2] \u2013 1[78 + 5]<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 11[\u2013 9] + 4(37) \u2013 83<\/p>\n\n\n\n

\u21d2 D1<\/sub> = \u2013 99 \u2013 148 \u2013 45<\/p>\n\n\n\n

\u21d2 D1<\/sub> = \u2013 34<\/p>\n\n\n\n

Again<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 1[39 \u2013 2] \u2013 11[2 + 6] \u2013 1[2 + 117]<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 1[37] \u2013 11(8) \u2013 119<\/p>\n\n\n\n

\u21d2 D2<\/sub> = \u2013 170<\/p>\n\n\n\n

And,<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n\n\n

Solving determinant, expanding along 1st<\/sup> row<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 1[\u2013 5 \u2013 (39) (2)] \u2013 (\u2013 4) [2 \u2013 (39) (\u2013 3)] + 11[4 \u2013 (\u2013 5)(\u2013 3)]<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 1 [\u2013 5 \u2013 78] + 4 (2 + 117) + 11 (4 \u2013 15)<\/p>\n\n\n\n

\u21d2 D3<\/sub> = \u2013 83 + 4(119) + 11(\u2013 11)<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 272<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

13. 6x + y \u2013 3z = 5<\/strong><\/p>\n\n\n\n

x + 3y \u2013 2z = 5<\/strong><\/p>\n\n\n\n

2x + y + 4z = 8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

6x + y \u2013 3z = 5<\/p>\n\n\n\n

x + 3y \u2013 2z = 5<\/p>\n\n\n\n

2x + y + 4z = 8<\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now, here we have<\/p>\n\n\n\n

6x + y \u2013 3z = 5<\/p>\n\n\n\n

x + 3y \u2013 2z = 5<\/p>\n\n\n\n

2x + y + 4z = 8<\/p>\n\n\n\n

So by comparing with theorem, now we have to find D , D1<\/sub> and D2<\/sub><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> Row<\/p>\n\n\n\n

\u21d2 D = 6[(4) (3) \u2013 (1) (\u2013 2)] \u2013 1[(4) (1) + 4] \u2013 3[1 \u2013 3(2)]<\/p>\n\n\n\n

\u21d2 D = 6[12 + 2] \u2013 [8] \u2013 3[\u2013 5]<\/p>\n\n\n\n

\u21d2 D = 84 \u2013 8 + 15<\/p>\n\n\n\n

\u21d2 D = 91<\/p>\n\n\n\n

Again, Solve D1<\/sub> formed by replacing 1st<\/sup> column by B matrices<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> Row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 5[(4) (3) \u2013 (\u2013 2) (1)] \u2013 1[(5) (4) \u2013 (\u2013 2) (8)] \u2013 3[(5) \u2013 (3) (8)]<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 5[12 + 2] \u2013 1[20 + 16] \u2013 3[5 \u2013 24]<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 5[14] \u2013 36 \u2013 3(\u2013 19)<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 70 \u2013 36 + 57<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 91<\/p>\n\n\n\n

Again, Solve D2<\/sub> formed by replacing 1st<\/sup> column by B matrices<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 6[20 + 16] \u2013 5[4 \u2013 2(\u2013 2)] + (\u2013 3)[8 \u2013 10]<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 6[36] \u2013 5(8) + (\u2013 3) (\u2013 2)<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 182<\/p>\n\n\n\n

And, Solve D3<\/sub> formed by replacing 1st<\/sup> column by B matrices<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> Row<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 6[24 \u2013 5] \u2013 1[8 \u2013 10] + 5[1 \u2013 6]<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 6[19] \u2013 1(\u2013 2) + 5(\u2013 5)<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 114 + 2 \u2013 25<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 91<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

14. x + y = 5<\/strong><\/p>\n\n\n\n

y + z = 3<\/strong><\/p>\n\n\n\n

x + z = 4<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given x + y = 5<\/p>\n\n\n\n

y + z = 3<\/p>\n\n\n\n

x + z = 4<\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Let Dj<\/sub> be the determinant obtained from D after replacing the jth<\/sup> column by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now, here we have<\/p>\n\n\n\n

x + y = 5<\/p>\n\n\n\n

y + z = 3<\/p>\n\n\n\n

x + z = 4<\/p>\n\n\n\n

So by comparing with theorem, now we have to find D, D1<\/sub> and D2<\/sub><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> Row<\/p>\n\n\n\n

\u21d2 D = 1[1] \u2013 1[\u2013 1] + 0[\u2013 1]<\/p>\n\n\n\n

\u21d2 D = 1 + 1 + 0<\/p>\n\n\n\n

\u21d2 D = 2<\/p>\n\n\n\n

Again, Solve D1<\/sub> formed by replacing 1st<\/sup> column by B matrices<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> Row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 5[1] \u2013 1[(3) (1) \u2013 (4) (1)] + 0[0 \u2013 (4) (1)]<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 5 \u2013 1[3 \u2013 4] + 0[\u2013 4]<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 5 \u2013 1[\u2013 1] + 0<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 5 + 1 + 0<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 6<\/p>\n\n\n\n

Again, Solve D2<\/sub> formed by replacing 1st<\/sup> column by B matrices<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 1[3 \u2013 4] \u2013 5[\u2013 1] + 0[0 \u2013 3]<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 1[\u2013 1] + 5 + 0<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 4<\/p>\n\n\n\n

And, Solve D3<\/sub> formed by replacing 1st<\/sup> column by B matrices<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> Row<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 1[4 \u2013 0] \u2013 1[0 \u2013 3] + 5[0 \u2013 1]<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 1[4] \u2013 1(\u2013 3) + 5(\u2013 1)<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 4 + 3 \u2013 5<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 2<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

15. 2y \u2013 3z = 0<\/strong><\/p>\n\n\n\n

x + 3y = -4<\/strong><\/p>\n\n\n\n

3x + 4y = 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

2y \u2013 3z = 0<\/p>\n\n\n\n

x + 3y = -4<\/p>\n\n\n\n

3x + 4y = 3<\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now, here we have<\/p>\n\n\n\n

2y \u2013 3z = 0<\/p>\n\n\n\n

x + 3y = \u2013 4<\/p>\n\n\n\n

3x + 4y = 3<\/p>\n\n\n\n

So by comparing with theorem, now we have to find D, D1<\/sub> and D2<\/sub><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> Row<\/p>\n\n\n\n

\u21d2 D = 0[0] \u2013 2[(0) (1) \u2013 0] \u2013 3[1 (4) \u2013 3 (3)]<\/p>\n\n\n\n

\u21d2 D = 0 \u2013 0 \u2013 3[4 \u2013 9]<\/p>\n\n\n\n

\u21d2 D = 0 \u2013 0 + 15<\/p>\n\n\n\n

\u21d2 D = 15<\/p>\n\n\n\n

Again, Solve D1<\/sub> formed by replacing 1st<\/sup> column by B matrices<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> Row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 0[0] \u2013 2[(0) (\u2013 4) \u2013 0] \u2013 3[4 (\u2013 4) \u2013 3(3)]<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 0 \u2013 0 \u2013 3[\u2013 16 \u2013 9]<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 0 \u2013 0 \u2013 3(\u2013 25)<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 0 \u2013 0 + 75<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 75<\/p>\n\n\n\n

Again, Solve D2<\/sub> formed by replacing 2nd<\/sup> column by B matrices<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 0[0] \u2013 0[(0) (1) \u2013 0] \u2013 3[1 (3) \u2013 3(\u2013 4)]<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 0 \u2013 0 + (\u2013 3) (3 + 12)<\/p>\n\n\n\n

\u21d2 D2<\/sub> = \u2013 45<\/p>\n\n\n\n

And, Solve D3<\/sub> formed by replacing 3rd<\/sup> column by B matrices<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> Row<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 0[9 \u2013 (\u2013 4) 4] \u2013 2[(3) (1) \u2013 (\u2013 4) (3)] + 0[1 (4) \u2013 3 (3)]<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 0[25] \u2013 2(3 + 12) + 0(4 \u2013 9)<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 0 \u2013 30 + 0<\/p>\n\n\n\n

\u21d2 D3<\/sub> = \u2013 30<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

16. 5x \u2013 7y + z = 11<\/strong><\/p>\n\n\n\n

6x \u2013 8y \u2013 z = 15<\/strong><\/p>\n\n\n\n

3x + 2y \u2013 6z = 7<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

5x \u2013 7y + z = 11<\/p>\n\n\n\n

6x \u2013 8y \u2013 z = 15<\/p>\n\n\n\n

3x + 2y \u2013 6z = 7<\/p>\n\n\n\n

Let there be a system of n simultaneous linear equations and with n unknown given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Now, here we have<\/p>\n\n\n\n

5x \u2013 7y + z = 11<\/p>\n\n\n\n

6x \u2013 8y \u2013 z = 15<\/p>\n\n\n\n

3x + 2y \u2013 6z = 7<\/p>\n\n\n\n

So by comparing with theorem, now we have to find D, D1<\/sub> and D2<\/sub><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> Row<\/p>\n\n\n\n

\u21d2 D = 5[(\u2013 8) (\u2013 6) \u2013 (\u2013 1) (2)] \u2013 7[(\u2013 6) (6) \u2013 3(\u2013 1)] + 1[2(6) \u2013 3(\u2013 8)]<\/p>\n\n\n\n

\u21d2 D = 5[48 + 2] \u2013 7[\u2013 36 + 3] + 1[12 + 24]<\/p>\n\n\n\n

\u21d2 D = 250 \u2013 231 + 36<\/p>\n\n\n\n

\u21d2 D = 55<\/p>\n\n\n\n

Again, Solve D1<\/sub> formed by replacing 1st<\/sup> column by B matrices<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> Row<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 11[(\u2013 8) (\u2013 6) \u2013 (2) (\u2013 1)] \u2013 (\u2013 7) [(15) (\u2013 6) \u2013 (\u2013 1) (7)] + 1[(15)2 \u2013 (7) (\u2013 8)]<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 11[48 + 2] + 7[\u2013 90 + 7] + 1[30 + 56]<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 11[50] + 7[\u2013 83] + 86<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 550 \u2013 581 + 86<\/p>\n\n\n\n

\u21d2 D1<\/sub> = 55<\/p>\n\n\n\n

Again, Solve D2<\/sub> formed by replacing 2nd<\/sup> column by B matrices<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> Row<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 5[(15) (\u2013 6) \u2013 (7) (\u2013 1)] \u2013 11 [(6) (\u2013 6) \u2013 (\u2013 1) (3)] + 1[(6)7 \u2013 (15) (3)]<\/p>\n\n\n\n

\u21d2 D2 <\/sub>= 5[\u2013 90 + 7] \u2013 11[\u2013 36 + 3] + 1[42 \u2013 45]<\/p>\n\n\n\n

\u21d2 D2<\/sub> = 5[\u2013 83] \u2013 11(\u2013 33) \u2013 3<\/p>\n\n\n\n

\u21d2 D2<\/sub> = \u2013 415 + 363 \u2013 3<\/p>\n\n\n\n

\u21d2 D2<\/sub> = \u2013 55<\/p>\n\n\n\n

And, Solve D3<\/sub> formed by replacing 3rd<\/sup> column by B matrices<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solving determinant, expanding along 1st<\/sup> Row<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 5[(\u2013 8) (7) \u2013 (15) (2)] \u2013 (\u2013 7) [(6) (7) \u2013 (15) (3)] + 11[(6)2 \u2013 (\u2013 8) (3)]<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 5[\u2013 56 \u2013 30] \u2013 (\u2013 7) [42 \u2013 45] + 11[12 + 24]<\/p>\n\n\n\n

\u21d2 D3<\/sub> = 5[\u2013 86] + 7[\u2013 3] + 11[36]<\/p>\n\n\n\n

\u21d2 D3<\/sub> = \u2013 430 \u2013 21 + 396<\/p>\n\n\n\n

\u21d2 D3<\/sub> = \u2013 55<\/p>\n\n\n\n

Thus by Cramer\u2019s Rule, we have<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Exercise 6.5 Page No: 6.89<\/p>\n\n\n\n

Solve each of the following system of homogeneous linear equations:<\/strong><\/p>\n\n\n\n

1. x + y \u2013 2z = 0<\/strong><\/p>\n\n\n\n

2x + y \u2013 3z =0<\/strong><\/p>\n\n\n\n

5x + 4y \u2013 9z = 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given x + y \u2013 2z = 0<\/p>\n\n\n\n

2x + y \u2013 3z =0<\/p>\n\n\n\n

5x + 4y \u2013 9z = 0<\/p>\n\n\n\n

Any system of equation can be written in matrix form as AX = B<\/p>\n\n\n\n

Now finding the Determinant of these set of equations,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 1(1 \u00d7 (\u2013 9) \u2013 4 \u00d7 (\u2013 3)) \u2013 1(2 \u00d7 (\u2013 9) \u2013 5 \u00d7 (\u2013 3)) \u2013 2(4 \u00d7 2 \u2013 5 \u00d7 1)<\/p>\n\n\n\n

= 1(\u2013 9 + 12) \u2013 1(\u2013 18 + 15) \u2013 2(8 \u2013 5)<\/p>\n\n\n\n

= 1 \u00d7 3 \u20131 \u00d7 (\u2013 3) \u2013 2 \u00d7 3<\/p>\n\n\n\n

= 3 + 3 \u2013 6<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Since D = 0, so the system of equation has infinite solution.<\/p>\n\n\n\n

Now let z = k<\/p>\n\n\n\n

\u21d2 x + y = 2k<\/p>\n\n\n\n

And 2x + y = 3k<\/p>\n\n\n\n

Now using the Cramer\u2019s rule<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

2. 2x + 3y + 4z = 0<\/strong><\/p>\n\n\n\n

x + y + z = 0<\/strong><\/p>\n\n\n\n

2x + 5y \u2013 2z = 0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

2x + 3y + 4z = 0<\/p>\n\n\n\n

x + y + z = 0<\/p>\n\n\n\n

2x + 5y \u2013 2z = 0<\/p>\n\n\n\n

Any system of equation can be written in matrix form as AX = B<\/p>\n\n\n\n

Now finding the Determinant of these set of equations,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"RD<\/figure>\n\n\n\n

= 2(1 \u00d7 (\u2013 2) \u2013 1 \u00d7 5) \u2013 3(1 \u00d7 (\u2013 2) \u2013 2 \u00d7 1) + 4(1 \u00d7 5 \u2013 2 \u00d7 1)<\/p>\n\n\n\n

= 2(\u2013 2 \u2013 5) \u2013 3(\u2013 2 \u2013 2) + 4(5 \u2013 2)<\/p>\n\n\n\n

= 1 \u00d7 (\u2013 7) \u2013 3 \u00d7 (\u2013 4) + 4 \u00d7 3<\/p>\n\n\n\n

= \u2013 7 + 12 + 12<\/p>\n\n\n\n

= 17<\/p>\n\n\n\n

Since D \u2260 0, so the system of equation has infinite solution.<\/p>\n\n\n\n

Therefore the system of equation has only solution as x = y = z = 0.<\/p>\n\n\n\n

RD Sharma 12th Maths Chapter 6, Class 12 Maths Chapter 6 solutions<\/p>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 6\u2013Determinants: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 6\u2013Determinants<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 12 Maths Chapter 6\u2013Determinants PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 12 Maths :<\/strong><\/h2>\n\n\n\n