If matrix A = begin{bmatrix} 1 & 2 3 & 5 end | vedclass

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(D) Given $A = \begin{bmatrix} 1 & 2 \ 3 & 5 \end{bmatrix}$. The determinant $|A| = (1)(5) - (2)(3) = 5 - 6 = -1$. Since $|A| 
eq 0$,$A^{-1}$ exists

Vedclass · Accepted Answer

$-11$

Vedclass · Answer

(D) Given $A = \begin{bmatrix} 1 & 2 \ 3 & 5 \end{bmatrix}$. The determinant $|A| = (1)(5) - (2)(3) = 5 - 6 = -1$. Since $|A| 
eq 0$,$A^{-1}$ exists. $A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{-1} \begin{bmatrix} 5 & -2 \ -3 & 1 \end{bmatrix} = \begin{bmatrix} -5 & 2 \ 3 & -1 \end{bmatrix}$. Given $A^{-1} = \alpha I + \beta A$,we have: $\begin{bmatrix} -5 & 2 \ 3 & -1 \end{bmatrix} = \alpha \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 2 \ 3 & 5 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 2\beta \ 3\beta & \alpha + 5\beta \end{bmatrix}$. Comparing the elements: $2\beta = 2 \Rightarrow \beta = 1$. $\alpha + \beta = -5 \Rightarrow \alpha + 1 = -5 \Rightarrow \alpha = -6$. Check: $3\beta = 3(1) = 3$ (Correct) and $\alpha + 5\beta = -6 + 5(1) = -1$ (Correct). Thus,$\alpha + \beta + \alpha\beta = -6 + 1 + (-6)(1) = -5 - 6 = -11$.

If matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}$ and $A^{-1} = \alpha I + \beta A$ where $I$ is a unit matrix of order $2$ and $\alpha, \beta$ are constants,then the value of $\alpha + \beta + \alpha \beta$ is

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