\( A = \left( \begin{array} { c c c } { - 2 } & { 1 } & { 0 } \\ { 1 } & { 19 } & { 1 } \\ { 4 } & { 1 } & { 0 } \end{array} \right) \quad B = \left( \begin{array} { c c } { x } & { 0 } \\ { 4 } & { 1 } \\ { 0 } & { 1 } \end{array} \right) \quad A ^ { * } B = ? \)

To find \( A^* B \), we first need to determine the conjugate transpose \( A^* \) of the matrix \( A \). The conjugate transpose (or Hermitian transpose) of a matrix involves taking the transpose and then taking the complex conjugate of each entry. However, since all entries of \( A \) are real numbers, the complex conjugate does not change anything. Given: \[ A = \begin{pmatrix} -2 & 1 & 0 \\ 1 & 19 & 1 \\ 4 & 1 & 0 \end{pmatrix} \] The transpose of \( A \) is: \[ A^T = \begin{pmatrix} -2 & 1 & 4 \\ 1 & 19 & 1 \\ 0 & 1 & 0 \end{pmatrix} \] Thus, \( A^* = A^T \): \[ A^* = \begin{pmatrix} -2 & 1 & 4 \\ 1 & 19 & 1 \\ 0 & 1 & 0 \end{pmatrix} \] Next, we compute the matrix product \( A^* B \): \[ B = \begin{pmatrix} x & 0 \\ 4 & 1 \\ 0 & 1 \end{pmatrix} \] Now we perform the multiplication: 1. First column of \( A^* B \): \[ \begin{pmatrix} -2 & 1 & 4 \\ 1 & 19 & 1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ 4 \\ 0 \end{pmatrix} = \begin{pmatrix} -2x + 4 \\ x + 76 \\ 4 \end{pmatrix} \] 2. Second column of \( A^* B \): \[ \begin{pmatrix} -2 & 1 & 4 \\ 1 & 19 & 1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 + 4 \\ 19 + 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 20 \\ 1 \end{pmatrix} \] Finally, we combine both results to form the product matrix \( A^* B \): \[ A^* B = \begin{pmatrix} -2x + 4 & 5 \\ x + 76 & 20 \\ 4 & 1 \end{pmatrix} \] Thus, the final result is: \[ A^* B = \begin{pmatrix} -2x + 4 & 5 \\ x + 76 & 20 \\ 4 & 1 \end{pmatrix} \]