If A = begin{bmatrix} 2 & -3 5 & -7 end{bmat | vedclass

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

Question

(C) Given $A = \begin{bmatrix} 2 & -3 \ 5 & -7 \end{bmatrix}$.First,we find the determinant $|A| = (2)(-7) - (-3)(5) = -14 + 15 = 1$.Since $|A| 
eq

Vedclass · Accepted Answer

$\begin{bmatrix} 25 & -15 \ 25 & -20 \end{bmatrix}$

Vedclass · Answer

(C) Given $A = \begin{bmatrix} 2 & -3 \ 5 & -7 \end{bmatrix}$.First,we find the determinant $|A| = (2)(-7) - (-3)(5) = -14 + 15 = 1$.Since $|A| 
eq 0$,$A^{-1}$ exists.The inverse is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{1} \begin{bmatrix} -7 & 3 \ -5 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 3 \ -5 & 2 \end{bmatrix}$.Now,calculate $2A = 2 \begin{bmatrix} 2 & -3 \ 5 & -7 \end{bmatrix} = \begin{bmatrix} 4 & -6 \ 10 & -14 \end{bmatrix}$.Next,calculate $3A^{-1} = 3 \begin{bmatrix} -7 & 3 \ -5 & 2 \end{bmatrix} = \begin{bmatrix} -21 & 9 \ -15 & 6 \end{bmatrix}$.Finally,$2A - 3A^{-1} = \begin{bmatrix} 4 & -6 \ 10 & -14 \end{bmatrix} - \begin{bmatrix} -21 & 9 \ -15 & 6 \end{bmatrix} = \begin{bmatrix} 4 - (-21) & -6 - 9 \ 10 - (-15) & -14 - 6 \end{bmatrix} = \begin{bmatrix} 25 & -15 \ 25 & -20 \end{bmatrix}$.

If $A = \begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix}$,then $2A - 3A^{-1} = $

Similar Questions

If $A = \begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{bmatrix}$,then $|A| |adj A|$ is equal to

Let $A = \begin{bmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{bmatrix}$. If $AA^T = I_3$,then $|p|$ is

If $A=\begin{bmatrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{bmatrix}$,$B=\begin{bmatrix} 1 & 0 \\ i & 1 \end{bmatrix}$,$i=\sqrt{-1}$,and $Q=A^{T}BA$,then the inverse of the matrix $AQ^{2021}A^{T}$ is equal to:

If $A = \begin{bmatrix} k & 2 \\ -2 & -k \end{bmatrix}$,then $A^{-1}$ does not exist if $k =$

If $A$ is a $n \times n$ matrix,then $adj(adj \,A) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module