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

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(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

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$\frac{2}{11}$

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(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 $a+b=$ . . . . . . .

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