1. **Problem statement:** Determine for which values of $k$ the matrix $$ \begin{bmatrix} 1 & 2 & 3 & 4 \\ 5 & 6 & k & 7 \\ 8 & 9 & 8 & 7 \\ 0 & 0 & 6 & 5 \end{bmatrix} $$ is invertible. 2. **Key fact:** A matrix is invertible if and only if its determinant is nonzero. 3. **Approach:** Calculate the determinant of the matrix as a function of $k$ and find values of $k$ for which the determinant equals zero. 4. **Calculate determinant:** Let the matrix be $A$. Using cofactor expansion along the first row: $$ \det(A) = 1 \cdot C_{11} - 2 \cdot C_{12} + 3 \cdot C_{13} - 4 \cdot C_{14} $$ where $C_{ij}$ is the $(i,j)$ cofactor. 5. Compute each minor determinant: - $C_{11} = \det \begin{bmatrix}6 & k & 7 \\ 9 & 8 & 7 \\ 0 & 6 & 5\end{bmatrix}$ Calculate: $$ = 6 \cdot \det \begin{bmatrix}8 & 7 \\ 6 & 5\end{bmatrix} - k \cdot \det \begin{bmatrix}9 & 7 \\ 0 & 5\end{bmatrix} + 7 \cdot \det \begin{bmatrix}9 & 8 \\ 0 & 6\end{bmatrix} $$ $$ = 6(8 \times 5 - 7 \times 6) - k(9 \times 5 - 0) + 7(9 \times 6 - 0) $$ $$ = 6(40 - 42) - 45k + 7(54) = 6(-2) - 45k + 378 = -12 - 45k + 378 = 366 - 45k $$ - $C_{12} = \det \begin{bmatrix}5 & k & 7 \\ 8 & 8 & 7 \\ 0 & 6 & 5\end{bmatrix}$ Calculate: $$ = 5 \cdot \det \begin{bmatrix}8 & 7 \\ 6 & 5\end{bmatrix} - k \cdot \det \begin{bmatrix}8 & 7 \\ 0 & 5\end{bmatrix} + 7 \cdot \det \begin{bmatrix}8 & 8 \\ 0 & 6\end{bmatrix} $$ $$ = 5(8 \times 5 - 7 \times 6) - k(8 \times 5 - 0) + 7(8 \times 6 - 0) $$ $$ = 5(40 - 42) - 8k(5) + 7(48) = 5(-2) - 40k + 336 = -10 - 40k + 336 = 326 - 40k $$ - $C_{13} = \det \begin{bmatrix}5 & 6 & 7 \\ 8 & 9 & 7 \\ 0 & 0 & 5\end{bmatrix}$ Calculate: $$ = 5 \cdot \det \begin{bmatrix}9 & 7 \\ 0 & 5\end{bmatrix} - 6 \cdot \det \begin{bmatrix}8 & 7 \\ 0 & 5\end{bmatrix} + 7 \cdot \det \begin{bmatrix}8 & 9 \\ 0 & 0\end{bmatrix} $$ $$ = 5(9 \times 5 - 0) - 6(8 \times 5 - 0) + 7(8 \times 0 - 0) $$ $$ = 5(45) - 6(40) + 0 = 225 - 240 = -15 $$ - $C_{14} = \det \begin{bmatrix}5 & 6 & k \\ 8 & 9 & 8 \\ 0 & 0 & 6\end{bmatrix}$ Calculate: $$ = 5 \cdot \det \begin{bmatrix}9 & 8 \\ 0 & 6\end{bmatrix} - 6 \cdot \det \begin{bmatrix}8 & 8 \\ 0 & 6\end{bmatrix} + k \cdot \det \begin{bmatrix}8 & 9 \\ 0 & 0\end{bmatrix} $$ $$ = 5(9 \times 6 - 0) - 6(8 \times 6 - 0) + k(8 \times 0 - 0) $$ $$ = 5(54) - 6(48) + 0 = 270 - 288 = -18 $$ 6. Substitute cofactors into determinant formula: $$ \det(A) = 1(366 - 45k) - 2(326 - 40k) + 3(-15) - 4(-18) $$ $$ = 366 - 45k - 652 + 80k - 45 + 72 $$ $$ = (366 - 652 - 45 + 72) + (-45k + 80k) = (-259) + 35k $$ 7. **Invertibility condition:** $$ \det(A) \neq 0 \implies -259 + 35k \neq 0 $$ $$ 35k \neq 259 $$ $$ k \neq \frac{259}{35} $$ 8. **Final answer:** The matrix is invertible for all values of $k$ except $$ k = \frac{259}{35} $$