If matrix A = frac{1}{11} begin{bmatrix} -1 & | vedclass

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(C) We know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix of order $3 	imes 3$. Given $A = \frac{1}{11} \begin{bmatrix} -1 & 7 & -24 \

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$-3, 0, 1$

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(C) We know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix of order $3 	imes 3$. Given $A = \frac{1}{11} \begin{bmatrix} -1 & 7 & -24 \ 2 & a & 4 \ 2 & -3 & 15 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 3 & 3 & 4 \ 2 & -3 & 4 \ b & -1 & c \end{bmatrix}$. So,$A \cdot A^{-1} = \frac{1}{11} \begin{bmatrix} -1 & 7 & -24 \ 2 & a & 4 \ 2 & -3 & 15 \end{bmatrix} \begin{bmatrix} 3 & 3 & 4 \ 2 & -3 & 4 \ b & -1 & c \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$. Multiplying the matrices: Row $1 	imes$ Column $1$: $\frac{1}{11} [(-1)(3) + (7)(2) + (-24)(b)] = 1 \implies -3 + 14 - 24b = 11 \implies 11 - 24b = 11 \implies b = 0$. Row $2 	imes$ Column $2$: $\frac{1}{11} [(2)(3) + (a)(-3) + (4)(-1)] = 1 \implies 6 - 3a - 4 = 11 \implies 2 - 3a = 11 \implies -3a = 9 \implies a = -3$. Row $3 	imes$ Column $3$: $\frac{1}{11} [(2)(4) + (-3)(4) + (15)(c)] = 1 \implies 8 - 12 + 15c = 11 \implies -4 + 15c = 11 \implies 15c = 15 \implies c = 1$. Thus,the values are $a = -3, b = 0, c = 1$.

If matrix $A = \frac{1}{11} \begin{bmatrix} -1 & 7 & -24 \\ 2 & a & 4 \\ 2 & -3 & 15 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 3 & 3 & 4 \\ 2 & -3 & 4 \\ b & -1 & c \end{bmatrix}$,then the values of $a, b, c$ respectively are ......

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