If A = begin{bmatrix} 2 & -1 -1 & 2 end{bmat | vedclass

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

Question

(C) Given the equation $A^2 - 4A + 3I = 0$.Subtract $3I$ from both sides: $A^2 - 4A = -3I$.Multiply by $A^{-1}$ on both sides: $A^{-1}(A^2 - 4A) = A^{

Vedclass · Accepted Answer

$\frac{1}{3} \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix}$

Vedclass · Answer

(C) Given the equation $A^2 - 4A + 3I = 0$.Subtract $3I$ from both sides: $A^2 - 4A = -3I$.Multiply by $A^{-1}$ on both sides: $A^{-1}(A^2 - 4A) = A^{-1}(-3I)$.This simplifies to $A - 4I = -3A^{-1}$.Therefore,$A^{-1} = -\frac{1}{3}(A - 4I) = \frac{1}{3}(4I - A)$.Substitute $A = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$:$A^{-1} = \frac{1}{3} \left( \begin{bmatrix} 4 & 0 \ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix} ight) = \frac{1}{3} \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix}$.

If $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$ such that $A^2 - 4A + 3I = 0$,where $I$ is a unit matrix of order $2$,then $A^{-1}$ is

Similar Questions

If $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$,such that $A^{2} - 4A + 3I = 0$,then $A^{-1} =$

If $A = \begin{bmatrix} 1 & -\tan(\theta/2) \\ \tan(\theta/2) & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & \tan(\theta/2) \\ -\tan(\theta/2) & 1 \end{bmatrix}$ then $A^{-1}B$ is equal to

Let $A$ be a $3 \times 3$ real matrix. If $\det(2 \operatorname{Adj}(2 \operatorname{Adj}(\operatorname{Adj}(2 A))))=2^{41}$,then the value of $\det(A^{2})$ is equal to ..... .

If $A(\operatorname{adj} A)=5 I$ where $I$ is the identity matrix of order $3$,then $|\operatorname{adj} A|$ is equal to

The inverse of a matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module