Let A = begin{bmatrix} a & 1 1 & b end{bmatr | vedclass

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(C) Given $x^2 - 4x + 2 = 0$. Since $a$ and $b$ are roots, $a+b = 4$ and $ab = 2$.Matrix $A = \begin{bmatrix} a & 1 \ 1 & b \end{bmatrix}$. The deter

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$4$

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(C) Given $x^2 - 4x + 2 = 0$. Since $a$ and $b$ are roots, $a+b = 4$ and $ab = 2$.Matrix $A = \begin{bmatrix} a & 1 \ 1 & b \end{bmatrix}$. The determinant $|A| = ab - 1 = 2 - 1 = 1$.The inverse $A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{1} \begin{bmatrix} b & -1 \ -1 & a \end{bmatrix} = \begin{bmatrix} b & -1 \ -1 & a \end{bmatrix}$.Now, $A + A^{-1} = \begin{bmatrix} a & 1 \ 1 & b \end{bmatrix} + \begin{bmatrix} b & -1 \ -1 & a \end{bmatrix} = \begin{bmatrix} a+b & 0 \ 0 & a+b \end{bmatrix}$.Since $a+b = 4$, we have $A + A^{-1} = \begin{bmatrix} 4 & 0 \ 0 & 4 \end{bmatrix} = 4I_2$.Comparing with $A + A^{-1} = kI_2$, we get $k = 4$.

Let $A = \begin{bmatrix} a & 1 \\ 1 & b \end{bmatrix}$, where $a$ and $b$ are the roots of the equation $x^2 - 4x + 2 = 0$. If $A + A^{-1} = kI_2$, then the value of $k$ is . . . . . .

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