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

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

Question

(A) Given $A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix}$.First,calculate $A^{2} = A \cdot A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} \begi

Vedclass · Accepted Answer

$\frac{1}{7} \begin{bmatrix} 2 & -1 \ 1 & 3 \end{bmatrix}$

Vedclass · Answer

(A) Given $A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix}$.First,calculate $A^{2} = A \cdot A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} (3)(3) + (1)(-1) & (3)(1) + (1)(2) \ (-1)(3) + (2)(-1) & (-1)(1) + (2)(2) \end{bmatrix} = \begin{bmatrix} 8 & 5 \ -5 & 3 \end{bmatrix}$.Now,evaluate $A^{2} - 5A + 7I = \begin{bmatrix} 8 & 5 \ -5 & 3 \end{bmatrix} - 5 \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.$= \begin{bmatrix} 8 & 5 \ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \ 0 & 7 \end{bmatrix} = \begin{bmatrix} 8-15+7 & 5-5+0 \ -5+5+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} = 0$.To find $A^{-1}$,multiply the equation $A^{2} - 5A + 7I = 0$ by $A^{-1}$:$A^{2}A^{-1} - 5AA^{-1} + 7IA^{-1} = 0$.$A - 5I + 7A^{-1} = 0$.$7A^{-1} = 5I - A$.$7A^{-1} = 5 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} 5-3 & 0-1 \ 0-(-1) & 5-2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \ 1 & 3 \end{bmatrix}$.Therefore,$A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & -1 \ 1 & 3 \end{bmatrix}$.

If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$,show that $A^{2} - 5A + 7I = 0$. Hence,find $A^{-1}$.

Similar Questions

If $A$ is a matrix of order $3$ whose determinant is equal to $6$,then $\operatorname{det}(\operatorname{adj} A) = $

If $A = \begin{bmatrix} 1 & -3 & 2 \\ -2 & 1 & 3 \\ 3 & 2 & -1 \end{bmatrix}$,then $A^2 \operatorname{Adj} A = $ (in $I$)

Matrix $A$ is a non-singular matrix and $(A-3I)(A-5I)=0$. Then,$\frac{15}{8} A^{-1} =$ . . . . . .

If $F(\alpha ) = \begin{bmatrix} \cos \alpha & - \sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $G(\beta ) = \begin{bmatrix} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ - \sin \beta & 0 & \cos \beta \end{bmatrix}$,then $[F(\alpha ) G(\beta )]^{-1} = $

If $A = \begin{bmatrix} 1 & \tan(\theta/2) \\ -\tan(\theta/2) & 1 \end{bmatrix}$ and $AB = I$,then $B = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module