Finding Matrix X (3×2)
Question:
If \[ A=\begin{bmatrix}8 & 0 \\ 4 & -2 \\ 3 & 6\end{bmatrix}, \quad B=\begin{bmatrix}2 & -2 \\ 4 & 2 \\ -5 & 1\end{bmatrix} \] find matrix \(X\) of order \(3\times2\) such that: \[ 2A + 3X = 5B \]
If \[ A=\begin{bmatrix}8 & 0 \\ 4 & -2 \\ 3 & 6\end{bmatrix}, \quad B=\begin{bmatrix}2 & -2 \\ 4 & 2 \\ -5 & 1\end{bmatrix} \] find matrix \(X\) of order \(3\times2\) such that: \[ 2A + 3X = 5B \]
Solution:
Step 1: Rearrange the equation
\[ 3X = 5B – 2A \]Step 2: Compute \(5B\)
\[ 5B = \begin{bmatrix} 10 & -10 \\ 20 & 10 \\ -25 & 5 \end{bmatrix} \]Step 3: Compute \(2A\)
\[ 2A = \begin{bmatrix} 16 & 0 \\ 8 & -4 \\ 6 & 12 \end{bmatrix} \]Step 4: Compute \(5B – 2A\)
\[ = \begin{bmatrix} 10-16 & -10-0 \\ 20-8 & 10-(-4) \\ -25-6 & 5-12 \end{bmatrix} = \begin{bmatrix} -6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix} \]Step 5: Divide by 3
\[ X = \frac{1}{3} \begin{bmatrix} -6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix} = \begin{bmatrix} -2 & -\frac{10}{3} \\ 4 & \frac{14}{3} \\ -\frac{31}{3} & -\frac{7}{3} \end{bmatrix} \]Final Answer:
\[ \boxed{ \begin{bmatrix} -2 & -\frac{10}{3} \\ 4 & \frac{14}{3} \\ -\frac{31}{3} & -\frac{7}{3} \end{bmatrix} } \]