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To prove $(kA)^{-1}=\frac{1}{k}A^{-1}$, we need to show that $(kA)(\frac{1}{k}A^{-1}) = I$ and $(\frac{1}{k}A^{-1})(kA) = I$, where $I$ is the identity matrix.
$(kA)(\frac{1}{k}A^{-1}) = k(\frac{1}{k})AA^{-1} = 1 \cdot I = I$
$(\frac{1}{k}A^{-1})(kA) = k(\frac{1}{k})A^{-1}A = 1 \cdot I = I$
Given $A=\begin{bmatrix}2&-1&1\\ -1&2&-1\\ 1&-1&2\end{bmatrix}$, we calculate the determinant of $A$ as follows: $$|A| = 2(4-1) - (-1)(-2+1) + 1(1-2) = 2(3) + 1(-1) + 1(-1) = 6 - 1 - 1 = 4$$
The matrix of cofactors is: $C = \begin{bmatrix}3&1&-1\\ 1&3&1\\ -1&1&3\end{bmatrix}$ The adjoint of $A$ is the transpose of the cofactor matrix: $adj(A) = C^T = \begin{bmatrix}3&1&-1\\ 1&3&1\\ -1&1&3\end{bmatrix}$
The inverse of $A$ is given by $A^{-1} = \frac{1}{|A|}adj(A)$. $$A^{-1} = \frac{1}{4}\begin{bmatrix}3&1&-1\\ 1&3&1\\ -1&1&3\end{bmatrix} = \begin{bmatrix}\frac{3}{4}&\frac{1}{4}&-\frac{1}{4}\\ \frac{1}{4}&\frac{3}{4}&\frac{1}{4}\\ -\frac{1}{4}&\frac{1}{4}&\frac{3}{4}\end{bmatrix}$$
Using the property $(kA)^{-1} = \frac{1}{k}A^{-1}$, we have $(3A)^{-1} = \frac{1}{3}A^{-1}$. $$(3A)^{-1} = \frac{1}{3} \cdot \frac{1}{4}\begin{bmatrix}3&1&-1\\ 1&3&1\\ -1&1&3\end{bmatrix} = \frac{1}{12}\begin{bmatrix}3&1&-1\\ 1&3&1\\ -1&1&3\end{bmatrix} = \begin{bmatrix}\frac{1}{4}&\frac{1}{12}&-\frac{1}{12}\\ \frac{1}{12}&\frac{1}{4}&\frac{1}{12}\\ -\frac{1}{12}&\frac{1}{12}&\frac{1}{4}\end{bmatrix}$$
Final Answer: $\frac{1}{12}\begin{bmatrix}3&1&-1\\ 1&3&1\\ -1&1&3\end{bmatrix}$<\/span>
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