10) Solve each equation.. Check your answers. a) 2[0 1] [3 -4] -3X=[} 9&-6 [1&-2 ] I b 1/2X+[5 -1]=2[3 0] [0 2/3] [1 2]

Let's solve each equation step by step.<br /><br />### Part (a)<br /><br />Given:<br />\[ 2 \begin{bmatrix} 0 & 1 \\ 3 & -4 \end{bmatrix} - 3X = \begin{bmatrix} 9 & -6 \\ 1 & -2 \end{bmatrix} \]<br /><br />First, multiply the matrix by 2:<br />\[ 2 \begin{bmatrix} 0 & 1 \\ 3 & -4 \end{bmatrix} = \begin{bmatrix} 0 \cdot 2 & 1 \cdot 2 \\ 3 \cdot 2 & -4 \cdot 2 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ 6 & -8 \end{bmatrix} \]<br /><br />So the equation becomes:<br />\[ \begin{bmatrix} 0 & 2 \\ 6 & -8 \end{bmatrix} - 3X = \begin{bmatrix} 9 & -6 \\ 1 & -2 \end{bmatrix} \]<br /><br />Next, isolate \( -3X \):<br />\[ -3X = \begin{bmatrix} 9 & -6 \\ 1 & -2 \end{bmatrix} - \begin{bmatrix} 0 & 2 \\ 6 & -8 \end{bmatrix} \]<br /><br />Subtract the matrices:<br />\[ -3X = \begin{bmatrix} 9 - 0 & -6 - 2 \\ 1 - 6 & -2 - (-8) \end{bmatrix} = \begin{bmatrix} 9 & -8 \\ -5 & 6 \end{bmatrix} \]<br /><br />Now, divide by -3 to solve for \( X \):<br />\[ X = \frac{1}{-3} \begin{bmatrix} 9 & -8 \\ -5 & 6 \end{bmatrix} = \begin{bmatrix} \frac{9}{-3} & \frac{-8}{-3} \\ \frac{-5}{-3} & \frac{6}{-3} \end{bmatrix} = \begin{bmatrix} -3 & \frac{8}{3} \\ \frac{5}{3} & -2 \end{bmatrix} \]<br /><br />So, the solution for part (a) is:<br />\[ X = \begin{bmatrix} -3 & \frac{8}{3} \\ \frac{5}{3} & -2 \end{bmatrix} \]<br /><br />### Part (b)<br /><br />Given:<br />\[ \frac{1}{2}X + \begin{bmatrix} 5 & -1 \\ 0 & \frac{2}{3} \end{bmatrix} = 2 \begin{bmatrix} 3 & 0 \\ 1 & 2 \end{bmatrix} \]<br /><br />First, multiply the matrix by 2:<br />\[ 2 \begin{bmatrix} 3 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 \cdot 3 & 2 \cdot 0 \\ 2 \cdot 1 & 2 \cdot 2 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 2 & 4 \end{bmatrix} \]<br /><br />So the equation becomes:<br />\[ \frac{1}{2}X + \begin{bmatrix} 5 & -1 \\ 0 & \frac{2}{3} \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 2 & 4 \end{bmatrix} \]<br /><br />Next, isolate \( \frac{1}{2}X \):<br />\[ \frac{1}{2}X = \begin{bmatrix} 6 & 0 \\ 2 & 4 \end{bmatrix} - \begin{bmatrix} 5 & -1 \\ 0 & \frac{2}{3} \end{bmatrix} \]<br /><br />Subtract the matrices:<br />\[ \frac{1}{2}X = \begin{bmatrix} 6 - 5 & 0 - (-1) \\ 2 - 0 & 4 - \frac{2}{3} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 2 & \frac{10}{3} \end{bmatrix} \]<br /><br />Now, multiply by 2 to solve for \( X \):<br />\[ X = 2 \begin{bmatrix} 1 & 1 \\ 2 & \frac{10}{3end} = \begin{bmatrix} 2 \cdot 1 & 2 \cdot 1 \\ 2