If A = begin{bmatrix} 1 & -2 & 2 0 & 2 & -3 | vedclass

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

Question

(C) Given $A = \begin{bmatrix} 1 & -2 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4 \end{bmatrix}$.First,calculate the determinant $|A| = 1(8 - 6) - (-2)(0 - (-9))

Vedclass · Accepted Answer

$\begin{bmatrix} 9 & -2 & 2 \ 0 & 10 & -3 \ 3 & -2 & 12 \end{bmatrix}$

Vedclass · Answer

(C) Given $A = \begin{bmatrix} 1 & -2 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4 \end{bmatrix}$.First,calculate the determinant $|A| = 1(8 - 6) - (-2)(0 - (-9)) + 2(0 - 6) = 1(2) + 2(9) + 2(-6) = 2 + 18 - 12 = 8$.Next,find $\operatorname{adj} A$. The matrix of cofactors is:$C_{11} = +(8-6) = 2, C_{12} = -(0 - (-9)) = -9, C_{13} = +(0-6) = -6$$C_{21} = -(-8 - (-4)) = 4, C_{22} = +(4-6) = -2, C_{23} = -(-2 - (-6)) = -4$$C_{31} = +(6-4) = 2, C_{32} = -(-3-0) = 3, C_{33} = +(2-0) = 2$So,$\operatorname{adj} A = \begin{bmatrix} 2 & 4 & 2 \ -9 & -2 & 3 \ -6 & -4 & 2 \end{bmatrix}$.Then $A(I + \operatorname{adj} A) = A + A(\operatorname{adj} A) = A + |A|I$.$A + 8I = \begin{bmatrix} 1 & -2 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4 \end{bmatrix} + \begin{bmatrix} 8 & 0 & 0 \ 0 & 8 & 0 \ 0 & 0 & 8 \end{bmatrix} = \begin{bmatrix} 9 & -2 & 2 \ 0 & 10 & -3 \ 3 & -2 & 12 \end{bmatrix}$.Thus,the correct option is $C$.

If $A = \begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$,then $A(I + \operatorname{adj} A) = $

Similar Questions

If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $X$ is a $2 \times 2$ matrix such that $AX = I$,then $X =$

If $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$ and $A \cdot \text{adj}(A) = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$,then $k$ is equal to

If $A$ is a non-singular matrix,then $A(\text{adj } A) =$

If $a, b, c$ and $d$ are real numbers such that $a^2+b^2+c^2+d^2=1$ and $A=\left[\begin{array}{cc}a+ib & c+id \\ -c+id & a-ib\end{array}\right]$,then $A^{-1}$ is equal to

If $A = \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix}$,then $\text{adj}(3A^2 + 12A)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module