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Q2(i):
Write Minors and Cofactors of the elements of following determinant: (i) $\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$
Solution :
Initial Matrix Definition
Let the given determinant be denoted by $A$:
$A = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$
Theoretical Framework
- Minor ($M_{ij}$): The determinant of the sub-matrix obtained by deleting the $i$-th row and $j$-th column of the original determinant. [Per definition of Matrix Minors]
- Cofactor ($C_{ij}$): Defined as $C_{ij} = (-1)^{i+j} M_{ij}$. [Per Laplace expansion criteria]
Step 1: Calculating Minors ($M_{ij}$)
| Element | Calculation | Result ($M_{ij}$) |
|---|---|---|
| $a_{11}$ | $M_{11} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0)$ | $1$ |
| $a_{12}$ | $M_{12} = \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = (0)(1) - (0)(0)$ | $0$ |
| $a_{13}$ | $M_{13} = \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} = (0)(0) - (1)(0)$ | $0$ |
| $a_{21}$ | $M_{21} = \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = (0)(1) - (0)(0)$ | $0$ |
| $a_{22}$ | $M_{22} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0)$ | $1$ |
| $a_{23}$ | $M_{23} = \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} = (1)(0) - (0)(0)$ | $0$ |
| $a_{31}$ | $M_{31} = \begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix} = (0)(0) - (0)(1)$ | $0$ |
| $a_{32}$ | $M_{32} = \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} = (1)(0) - (0)(0)$ | $0$ |
| $a_{33}$ | $M_{33} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0)$ | $1$ |
Step 2: Calculating Cofactors ($C_{ij}$)
Applying $C_{ij} = (-1)^{i+j} M_{ij}$:
- $C_{11} = (-1)^{1+1} (1) = 1(1) = 1$
- $C_{12} = (-1)^{1+2} (0) = -1(0) = 0$
- $C_{13} = (-1)^{1+3} (0) = 1(0) = 0$
- $C_{21} = (-1)^{2+1} (0) = -1(0) = 0$
- $C_{22} = (-1)^{2+2} (1) = 1(1) = 1$
- $C_{23} = (-1)^{2+3} (0) = -1(0) = 0$
- $C_{31} = (-1)^{3+1} (0) = 1(0) = 0$
- $C_{32} = (-1)^{3+2} (0) = -1(0) = 0$
- $C_{33} = (-1)^{3+3} (1) = 1(1) = 1$
Visualization of Identity Determinant Structure
Final Solution: The set of Minors is $\{1, 0, 0, 0, 1, 0, 0, 0, 1\}$ and the set of Cofactors is $\{1, 0, 0, 0, 1, 0, 0, 0, 1\}$.
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(ii): Write Minors and Cofactors of the elements of following determinant: (ii) $\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}$
- 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
CBSE Solutions for Class 12 Mathematics Determinants
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