If A = begin{bmatrix} 2 & 0 & -1 5 & 1 & 0 | vedclass

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(B) We know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix of order $3 	imes 3$.$\begin{bmatrix} 2 & 0 & -1 \ 5 & 1 & 0 \ 0 & 1 & 3 \en

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$-15, 5$

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(B) We know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix of order $3 	imes 3$.$\begin{bmatrix} 2 & 0 & -1 \ 5 & 1 & 0 \ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} 3 & -1 & 1 \ \alpha & 6 & -5 \ \beta & -2 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$Multiplying the matrices on the left side:Row $1$,Column $1$: $(2)(3) + (0)(\alpha) + (-1)(\beta) = 6 - \beta$Since the result must be $I$,$6 - \beta = 1$,which gives $\beta = 5$.Row $2$,Column $1$: $(5)(3) + (1)(\alpha) + (0)(\beta) = 15 + \alpha$Since the result must be $I$,$15 + \alpha = 0$,which gives $\alpha = -15$.Thus,the values are $\alpha = -15$ and $\beta = 5$.

If $A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix}$,then the values of $\alpha$ and $\beta$ are respectively:

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