If A = begin{bmatrix} 1 & -2 & 2 2 & -6 & 5 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To find the adjoint of matrix $A$,we first find the cofactor of each element of $A$.Let $C_{ij}$ be the cofactor of the element $a_{ij}$.$C_{11} =

Vedclass · Accepted Answer

$\begin{bmatrix} -24 & 8 & 2 \ 17 & -6 & -1 \ 30 & -10 & -2 \end{bmatrix}$

Vedclass · Answer

(C) To find the adjoint of matrix $A$,we first find the cofactor of each element of $A$.Let $C_{ij}$ be the cofactor of the element $a_{ij}$.$C_{11} = +((-6)(4) - (5)(0)) = -24$$C_{12} = -((2)(4) - (5)(5)) = -(8 - 25) = 17$$C_{13} = +((2)(0) - (-6)(5)) = 30$$C_{21} = -((-2)(4) - (2)(0)) = -(-8) = 8$$C_{22} = +((1)(4) - (2)(5)) = 4 - 10 = -6$$C_{23} = -((1)(0) - (-2)(5)) = -(0 + 10) = -10$$C_{31} = +((-2)(5) - (2)(-6)) = -10 + 12 = 2$$C_{32} = -((1)(5) - (2)(2)) = -(5 - 4) = -1$$C_{33} = +((1)(-6) - (-2)(2)) = -6 + 4 = -2$The cofactor matrix $C$ is $\begin{bmatrix} -24 & 17 & 30 \ 8 & -6 & -10 \ 2 & -1 & -2 \end{bmatrix}$.The adjoint of $A$ is the transpose of the cofactor matrix,$\operatorname{Adj} A = C^T = \begin{bmatrix} -24 & 8 & 2 \ 17 & -6 & -1 \ 30 & -10 & -2 \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & -2 & 2 \\ 2 & -6 & 5 \\ 5 & 0 & 4 \end{bmatrix}$,then $\operatorname{Adj} A = $

Similar Questions

For matrices $A$ and $B$,if $AB = 4I$,then $A^{-1}$ is equal to:

The adjoint of the matrix $N = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}$ is

If $A$ is a square matrix of order $3 \times 3$ and $\operatorname{det}(A) = 3$,then the value of $\operatorname{det}(3A^{-1})$ is:

If $A = \begin{bmatrix} 1 & \cot \frac{\theta}{2} \\ -\cot \frac{\theta}{2} & 1 \end{bmatrix}$,then $A^{-1} =$

Let $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$ and $A^{-1} = \alpha I + \beta A$,where $\alpha, \beta \in \mathbb{R}$ and $I$ is the identity matrix of order $2$. Then $4(\alpha - \beta)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module