If the inverse matrix of A = begin{bmatrix} 2 | vedclass

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

Question

(B) The inverse of a matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ is given by $A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \ -c & a \e

Vedclass · Accepted Answer

$\frac{2}{11}$

Vedclass · Answer

(B) The inverse of a matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ is given by $A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Given $A = \begin{bmatrix} 2 & 3 \ 1 & -4 \end{bmatrix}$, the determinant $|A| = (2)(-4) - (3)(1) = -8 - 3 = -11$.Thus, $A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4/11 & 3/11 \ 1/11 & -2/11 \end{bmatrix}$.Comparing this with the given $A^{-1} = \begin{bmatrix} a & 3/11 \ 1/11 & b \end{bmatrix}$, we get $a = 4/11$ and $b = -2/11$.Therefore, $a+b = 4/11 + (-2/11) = 2/11$.

If the inverse matrix of $A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}$ is $A^{-1} = \begin{bmatrix} a & 3/11 \\ 1/11 & b \end{bmatrix}$, then $a+b=$ . . . . . . .

Similar Questions

If $A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$,then $A^{-1}=$

$\begin{aligned} & A(\alpha, \beta)=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^\beta\end{array}\right] \\ & \Rightarrow[A(\alpha, \beta)]^{-1}=\end{aligned}$

Inverse of the matrix $\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}$ is

Suppose $n > 1$ and $A$ is a non-singular matrix of order $n$ such that $|\operatorname{adj} A| = |\operatorname{adj}(\operatorname{adj} A)|$. Then the matrix whose rank is $n$ is:

If $X = \begin{bmatrix} -x & -y \\ z & t \end{bmatrix}$,then the transpose of $\text{adj } X$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module