Matrix A = begin{bmatrix} 1 & 0 & -k 2 & 1 & | vedclass

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(D) matrix $A$ is invertible if and only if its determinant $|A| 
eq 0$. Calculating the determinant of matrix $A$ by expanding along the second colu

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All real $k$

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(D) matrix $A$ is invertible if and only if its determinant $|A| 
eq 0$. Calculating the determinant of matrix $A$ by expanding along the second column: $|A| = 0 \cdot \begin{vmatrix} 2 & 3 \ k & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -k \ k & 1 \end{vmatrix} - 0 \cdot \begin{vmatrix} 1 & -k \ 2 & 3 \end{vmatrix}$ $|A| = 1 \cdot (1 - (-k^2)) = 1 + k^2$. Since $k^2 \geq 0$ for all real $k$,$1 + k^2 \geq 1$. Therefore,$|A|$ is always greater than or equal to $1$,which means $|A|$ can never be $0$. Hence,the matrix $A$ is invertible for all real values of $k$.

Matrix $A = \begin{bmatrix} 1 & 0 & -k \\ 2 & 1 & 3 \\ k & 0 & 1 \end{bmatrix}$ is invertible for

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