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Q2(ii):
Write Minors and Cofactors of the elements of following determinant: (ii) $\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}$
Solution :
Given Determinant Setup
Let the matrix $A$ be defined as:
$A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{pmatrix}$Definitions
- Minor ($M_{ij}$): The determinant of the sub-matrix obtained by deleting the $i^{th}$ row and $j^{th}$ column. [Definition: $M_{ij} = \det(A_{ij})$]
- Cofactor ($C_{ij}$): Defined as $(-1)^{i+j} M_{ij}$. [Definition: $C_{ij} = (-1)^{i+j} M_{ij}$]
Step 1: Calculating Minors ($M_{ij}$)
For each element $a_{ij}$, we compute the $2 \times 2$ determinant:
- $M_{11} = \begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix} = (5)(2) - (-1)(1) = 10 + 1 = 11$
- $M_{12} = \begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix} = (3)(2) - (-1)(0) = 6 - 0 = 6$
- $M_{13} = \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = (3)(1) - (5)(0) = 3 - 0 = 3$
- $M_{21} = \begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix} = (0)(2) - (4)(1) = 0 - 4 = -4$
- $M_{22} = \begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix} = (1)(2) - (4)(0) = 2 - 0 = 2$
- $M_{23} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1 - 0 = 1$
- $M_{31} = \begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix} = (0)(-1) - (4)(5) = 0 - 20 = -20$
- $M_{32} = \begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (4)(3) = -1 - 12 = -13$
- $M_{33} = \begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix} = (1)(5) - (0)(3) = 5 - 0 = 5$
Step 2: Calculating Cofactors ($C_{ij}$)
Applying the formula $C_{ij} = (-1)^{i+j} M_{ij}$:
| Cofactor | Calculation | Result |
|---|---|---|
| $C_{11}$ | $(-1)^{1+1}(11)$ | $11$ |
| $C_{12}$ | $(-1)^{1+2}(6)$ | $-6$ |
| $C_{13}$ | $(-1)^{1+3}(3)$ | $3$ |
| $C_{21}$ | $(-1)^{2+1}(-4)$ | $4$ |
| $C_{22}$ | $(-1)^{2+2}(2)$ | $2$ |
| $C_{23}$ | $(-1)^{2+3}(1)$ | $-1$ |
| $C_{31}$ | $(-1)^{3+1}(-20)$ | $-20$ |
| $C_{32}$ | $(-1)^{3+2}(-13)$ | $13$ |
| $C_{33}$ | $(-1)^{3+3}(5)$ | $5$ |
Final Solution: The set of Minors is $\{11, 6, 3, -4, 2, 1, -20, -13, 5\}$ and the set of Cofactors is $\{11, -6, 3, 4, 2, -1, -20, 13, 5\}$.
More Questions from Class 12 Mathematics Determinants EXERCISE 4.3
- Q1(i): Write Minors and Cofactors of the elements of following determinant: (i) $\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$
- Q1(ii): Write Minors and Cofactors of the elements of following determinant: (ii) $\begin{vmatrix} a & c \\ b & d \end{vmatrix}$
- Q2(i): Write Minors and Cofactors of the elements of following determinant: (i) $\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$
- Q3: Using Cofactors of elements of second row, evaluate $\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$.
- Q4: Using Cofactors of elements of third column, evaluate $\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}$.
- 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
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