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Q2:
Find adjoint of the matrix $\begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$
Solution :
Given Matrix and Objective
Let $A$ be the given $3 \times 3$ matrix:
$A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$
The objective is to find the adjoint of matrix $A$, denoted as $\text{adj}(A)$, which is defined as the transpose of the cofactor matrix $C$, where $C_{ij} = (-1)^{i+j} M_{ij}$. [Per the definition of the Adjoint of a Matrix].
Step 1: Calculating the Cofactors ($C_{ij}$)
We calculate the cofactors for each element $a_{ij}$ of matrix $A$.
- $C_{11} = + \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = (3)(1) - (0)(5) = 3$
- $C_{12} = - \begin{vmatrix} 2 & 5 \\ -2 & 1 \end{vmatrix} = -[(2)(1) - (-2)(5)] = -[2 + 10] = -12$
- $C_{13} = + \begin{vmatrix} 2 & 3 \\ -2 & 0 \end{vmatrix} = (2)(0) - (-2)(3) = 6$
- $C_{21} = - \begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} = -[(-1)(1) - (0)(2)] = -[-1] = 1$
- $C_{22} = + \begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} = (1)(1) - (-2)(2) = 1 + 4 = 5$
- $C_{23} = - \begin{vmatrix} 1 & -1 \\ -2 & 0 \end{vmatrix} = -[(1)(0) - (-2)(-1)] = -[0 - 2] = 2$
- $C_{31} = + \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = (-1)(5) - (3)(2) = -5 - 6 = -11$
- $C_{32} = - \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = -[(1)(5) - (2)(2)] = -[5 - 4] = -1$
- $C_{33} = + \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (2)(-1) = 3 + 2 = 5$
Step 2: Constructing the Cofactor Matrix and Transposing
The cofactor matrix $C$ is formed by placing the calculated cofactors in their respective positions:
$C = \begin{bmatrix} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{bmatrix}$
By definition, $\text{adj}(A) = C^T$ (the transpose of the cofactor matrix). We swap the rows and columns [Per matrix transposition rules]:
$\text{adj}(A) = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}$
Visual Representation of Matrix Transformation
Final Solution: $\text{adj}(A) = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}$
More Questions from Class 12 Mathematics Determinants EXERCISE 4.4
- Q1: Find adjoint of the matrix $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
- Q3: Verify $A (adj A) = (adj A) A = |A| I$ for the matrix $\begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$
- Q4: Verify $A (adj A) = (adj A) A = |A| I$ for the matrix $\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$
- Q5: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$
- Q6: Find the inverse of the matrix (if it exists): $\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}$
- Q7: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}$
- Q8: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$
- Q9: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}$
- Q10: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$
- Q11: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$
- Q12: Let $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$. Verify that $(AB)^{-1} = B^{-1} A^{-1}$.
- Q13: If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 – 5A + 7I = O$. Hence find $A^{-1}$.
- Q14: For the matrix $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$, find the numbers $a$ and $b$ such that $A^2 + aA + bI = O$.
- Q15: For the matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$ Show that $A^3– 6A^2 + 5A + 11 I = O$. Hence, find $A^{-1}$.
- Q16: If $A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$ Verify that $A^3 – 6A^2 + 9A – 4I = O$ and hence find $A^{-1}$
- Q17: Let $A$ be a nonsingular square matrix of order $3 \times 3$. Then $|adj A|$ is equal to
- Q18: If $A$ is an invertible matrix of order 2, then det $(A^{-1})$ is equal to
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