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

Let $A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$. Verify that $A(\text{adj } A) = (\text{adj } A) A = |A| I$.

If $P(\theta) = \begin{bmatrix} 1 & \cot \theta \\ -\cot \theta & 1 \end{bmatrix}$ and $PQ = I$,then find $(\csc^2 \theta)Q$,where $I$ is an identity matrix of order $2 \times 2$.

Find the inverse of the matrix $A = \left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right]$,if it exists.

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

If $(BA)^{-1} = C$ where $B = \begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1 \end{bmatrix}$ and $C = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix}$,then $A^{-1}$ is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module