If A = begin{bmatrix} 1 & 2 & 0 0 & 1 & 2 2 | vedclass

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

Question

(B) Given the matrix $A = \begin{bmatrix} 1 & 2 & 0 \ 0 & 1 & 2 \ 2 & 0 & 1 \end{bmatrix}$.To find the adjoint of $A$,we calculate the cofactor matr

Vedclass · Accepted Answer

$\begin{bmatrix} 1 & -2 & 4 \ 4 & 1 & -2 \ -2 & 4 & 1 \end{bmatrix}$

Vedclass · Answer

(B) Given the matrix $A = \begin{bmatrix} 1 & 2 & 0 \ 0 & 1 & 2 \ 2 & 0 & 1 \end{bmatrix}$.To find the adjoint of $A$,we calculate the cofactor matrix $C = [C_{ij}]$ and then take its transpose.The cofactors are calculated as follows:$C_{11} = +\begin{vmatrix} 1 & 2 \ 0 & 1 \end{vmatrix} = 1 - 0 = 1$$C_{12} = -\begin{vmatrix} 0 & 2 \ 2 & 1 \end{vmatrix} = -(0 - 4) = 4$$C_{13} = +\begin{vmatrix} 0 & 1 \ 2 & 0 \end{vmatrix} = 0 - 2 = -2$$C_{21} = -\begin{vmatrix} 2 & 0 \ 0 & 1 \end{vmatrix} = -(2 - 0) = -2$$C_{22} = +\begin{vmatrix} 1 & 0 \ 2 & 1 \end{vmatrix} = 1 - 0 = 1$$C_{23} = -\begin{vmatrix} 1 & 2 \ 2 & 0 \end{vmatrix} = -(0 - 4) = 4$$C_{31} = +\begin{vmatrix} 2 & 0 \ 1 & 2 \end{vmatrix} = 4 - 0 = 4$$C_{32} = -\begin{vmatrix} 1 & 0 \ 0 & 2 \end{vmatrix} = -(2 - 0) = -2$$C_{33} = +\begin{vmatrix} 1 & 2 \ 0 & 1 \end{vmatrix} = 1 - 0 = 1$The cofactor matrix is $C = \begin{bmatrix} 1 & 4 & -2 \ -2 & 1 & 4 \ 4 & -2 & 1 \end{bmatrix}$.The adjoint of $A$ is the transpose of the cofactor matrix:$adj(A) = C^T = \begin{bmatrix} 1 & -2 & 4 \ 4 & 1 & -2 \ -2 & 4 & 1 \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 2 & 0 & 1 \end{bmatrix}$,then find $adj(A)$.

Similar Questions

If $\operatorname{adj} B = A$ and $|P| = |Q| = 1$, then $\operatorname{adj}(Q^{-1} B P^{-1}) = $

If the inverse matrix of $A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}$ is $A^{-1} = \begin{bmatrix} a & 3/11 \\ 1/11 & b \end{bmatrix}$, then $a+b=$ . . . . . . .

If $A = \begin{bmatrix} e^t & e^{-t} \cos t & e^{-t} \sin t \\ e^t & -e^{-t} \cos t - e^{-t} \sin t & -e^{-t} \sin t + e^{-t} \cos t \\ e^t & 2e^{-t} \sin t & -2e^{-t} \cos t \end{bmatrix}$,then $A$ is:

Let $P = \begin{bmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{bmatrix}$ where $\alpha \in R$. Suppose $Q = [q_{ij}]$ is a matrix satisfying $PQ = kI_3$ for some non-zero $k \in R$. If $q_{23} = -\frac{k}{8}$ and $|Q| = \frac{k^2}{2}$,then $\alpha^2 + k^2$ is equal to?

If $A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$,then the inverse of the matrix $A^3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module