Question Given the matrices A=[} -1&1 1&1 ] solve for the matrix X in the equation XA+5B=C

To solve for the matrix X in the equation $XA + 5B = C$, we need to isolate X on one side of the equation.<br /><br />First, let's rewrite the equation as $XA = C - 5B$.<br /><br />Next, we need to find the inverse of matrix A, denoted as $A^{-1}$. The inverse of a matrix is a matrix that, when multiplied with the original matrix, results in the identity matrix.<br /><br />Once we have the inverse of matrix A, we can multiply both sides of the equation $XA = C - 5B$ by $A^{-1}$ to isolate X.<br /><br />So, $X = A^{-1}(C - 5B)$.<br /><br />Now, let's calculate the inverse of matrix A:<br /><br />$A = \begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix}$<br /><br />To find the inverse of a 2x2 matrix, we swap the diagonal elements and change the sign of the off-diagonal elements:<br /><br />$A^{-1} = \begin{bmatrix} 1 & -1 \\ -1 & -1 \end{bmatrix}$<br /><br />Now, let's calculate $C - 5B$:<br /><br />$C = \begin{bmatrix} 0 & -1 \\ 6 & 2 \end{bmatrix}$<br /><br />$5B = \begin{bmatrix} -30 & -5 \\ 25 & 10 \end{bmatrix}$<br /><br />$C - 5B = \begin{bmatrix} 0 - (-30) & -1 - (-5) \\ 6 - 25 & 2 - 10 \end{bmatrix} = \begin{bmatrix} 30 & 4 \\ -19 & -8 \end{bmatrix}$<br /><br />Finally, let's calculate $X = A^{-1}(C - 5B)$:<br /><br />$X = \begin{bmatrix} 1 & -1 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 30 & 4 \\ -19 & -8 \end{bmatrix} = \begin{bmatrix} 1(30) + (-1)(-19) & 1(4) + (-1)(-8) \\ (-1)(30) + (-1)(-19) & (-1)(4) + (-1)(-8) \end{bmatrix} = \begin{bmatrix} 49 & 12 \\ -11 & -4 \end{bmatrix}$<br /><br />Therefore, the matrix X in the equation $XA + 5B = C$ is $\begin{bmatrix} 49 & 12 \\ -11 & -4 \end{bmatrix}$.