NCERT Solutions Class-12-Chapter-4-Determinants
Excercise-4.3
Note:
Minor (\(M_{ij}\)): The determinant of the submatrix obtained by deleting the \(i\)-th row and \(j\)-th column.
Cofactor (\(A_{ij}\)): \( A_{ij} = (-1)^{i+j} M_{ij} \).
The value of a determinant can be found by summing the product of elements of any row (or column) with their corresponding cofactors.
Q1
Write Minors and Cofactors of the elements of following determinants:
(i) \( \begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix} \)
(ii) \( \begin{vmatrix} a & c \\ b & d \end{vmatrix} \)
(i) \( \begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix} \)
(ii) \( \begin{vmatrix} a & c \\ b & d \end{vmatrix} \)
Solution:
(i) \( \begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix} \)
Minors:
- \( M_{11} = 3 \)
- \( M_{12} = 0 \)
- \( M_{21} = -4 \)
- \( M_{22} = 2 \)
Cofactors (\( A_{ij} = (-1)^{i+j} M_{ij} \)):
- \( A_{11} = (-1)^{1+1}(3) = 3 \)
- \( A_{12} = (-1)^{1+2}(0) = 0 \)
- \( A_{21} = (-1)^{2+1}(-4) = 4 \)
- \( A_{22} = (-1)^{2+2}(2) = 2 \)
(ii) \( \begin{vmatrix} a & c \\ b & d \end{vmatrix} \)
Minors:
- \( M_{11} = d \)
- \( M_{12} = b \)
- \( M_{21} = c \)
- \( M_{22} = a \)
Cofactors:
- \( A_{11} = d \)
- \( A_{12} = -b \)
- \( A_{21} = -c \)
- \( A_{22} = a \)
Q2
Write Minors and Cofactors of the elements of following determinants:
(i) \( \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \)
(ii) \( \begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix} \)
(i) \( \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \)
(ii) \( \begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix} \)
Solution:
(i) Identity Matrix
Minors:
- \( M_{11} = 1, M_{12} = 0, M_{13} = 0 \)
- \( M_{21} = 0, M_{22} = 1, M_{23} = 0 \)
- \( M_{31} = 0, M_{32} = 0, M_{33} = 1 \)
Cofactors:
- \( A_{11} = 1, A_{12} = 0, A_{13} = 0 \)
- \( A_{21} = 0, A_{22} = 1, A_{23} = 0 \)
- \( A_{31} = 0, A_{32} = 0, A_{33} = 1 \)
(ii) \( \begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix} \)
Minors:
- \( M_{11} = 10 - (-1) = 11 \)
- \( M_{12} = 6 - 0 = 6 \)
- \( M_{13} = 3 - 0 = 3 \)
- \( M_{21} = 0 - 4 = -4 \)
- \( M_{22} = 2 - 0 = 2 \)
- \( M_{23} = 1 - 0 = 1 \)
- \( M_{31} = 0 - 20 = -20 \)
- \( M_{32} = -1 - 12 = -13 \)
- \( M_{33} = 5 - 0 = 5 \)
Cofactors:
- \( A_{11} = 11, A_{12} = -6, A_{13} = 3 \)
- \( A_{21} = 4, A_{22} = 2, A_{23} = -1 \)
- \( A_{31} = -20, A_{32} = 13, A_{33} = 5 \)
Q3
Using Cofactors of elements of second row, evaluate \( \Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix} \).
Solution:
Second row elements: \( a_{21}=2, a_{22}=0, a_{23}=1 \).
Expansion formula: \( \Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} \).
\[ \begin{aligned} M_{21} &= \begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = 9 - 16 = -7 \Rightarrow A_{21} = -(-7) = 7 \\ M_{22} &= \begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = 15 - 8 = 7 \Rightarrow A_{22} = 7 \\ M_{23} &= \begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = 10 - 3 = 7 \Rightarrow A_{23} = -7 \end{aligned} \]
\[ \Delta = 2(7) + 0(7) + 1(-7) = 14 + 0 - 7 = 7 \]
Q4
Using Cofactors of elements of third column, evaluate \( \Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix} \).
Solution:
Third column elements: \( a_{13}=yz, a_{23}=zx, a_{33}=xy \).
Formula: \( \Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33} \).
- \( M_{13} = z - y \Rightarrow A_{13} = (z-y) \)
- \( M_{23} = z - x \Rightarrow A_{23} = -(z-x) = (x-z) \)
- \( M_{33} = y - x \Rightarrow A_{33} = (y-x) \)
\[ \begin{aligned} \Delta &= yz(z-y) + zx(x-z) + xy(y-x) \\ &= yz^2 - y^2z + zx^2 - z^2x + xy^2 - x^2y \\ &= (yz^2 - z^2x) + (zx^2 - x^2y) + (xy^2 - y^2z) \\ &= -z^2(x-y) + x^2(z-y) - y^2(z-x) \\ \dots &\text{ Factoring yields: } \\ &= (x-y)(y-z)(z-x) \end{aligned} \]
Q5
If \( \Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \) and \( A_{ij} \) is Cofactors of \( a_{ij} \), then value of \( \Delta \) is given by- \( a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33} \)
- \( a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31} \)
- \( a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13} \)
- \( a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31} \)
- \( a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33} \)
- \( a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31} \)
- \( a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13} \)
- \( a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31} \)
Answer: (D)
Solution:
The value of a determinant is the sum of the product of elements of any row (or column) with their corresponding cofactors.
- Option (A) mixes Row 1 elements with Row 3 cofactors. (Value is 0).
- Option (B) mixes Row 1 elements with Column 1 cofactors. Incorrect indices.
- Option (C) mixes Row 2 elements with Row 1 cofactors. (Value is 0).
- Option (D) uses Column 1 elements \( a_{11}, a_{21}, a_{31} \) with their corresponding cofactors \( A_{11}, A_{21}, A_{31} \).
Therefore, (D) is the correct expansion along the first column.