Matrix Operation: \(2A – 3B + 4C\)
Question:
Let \[ A=\begin{bmatrix}-1 & 0 & 2 \\ 3 & 1 & 4\end{bmatrix}, \quad B=\begin{bmatrix}0 & -2 & 5 \\ 1 & -3 & 1\end{bmatrix}, \quad C=\begin{bmatrix}1 & -5 & 2 \\ 6 & 0 & -4\end{bmatrix} \] Compute \(2A – 3B + 4C\).
Let \[ A=\begin{bmatrix}-1 & 0 & 2 \\ 3 & 1 & 4\end{bmatrix}, \quad B=\begin{bmatrix}0 & -2 & 5 \\ 1 & -3 & 1\end{bmatrix}, \quad C=\begin{bmatrix}1 & -5 & 2 \\ 6 & 0 & -4\end{bmatrix} \] Compute \(2A – 3B + 4C\).
Solution:
All matrices are of order \(2 \times 3\), so the operation is defined.
Step 1: Compute \(2A\)
\[ 2A= \begin{bmatrix} -2 & 0 & 4 \\ 6 & 2 & 8 \end{bmatrix} \]Step 2: Compute \(3B\)
\[ 3B= \begin{bmatrix} 0 & -6 & 15 \\ 3 & -9 & 3 \end{bmatrix} \]Step 3: Compute \(4C\)
\[ 4C= \begin{bmatrix} 4 & -20 & 8 \\ 24 & 0 & -16 \end{bmatrix} \]Step 4: Compute \(2A – 3B\)
\[ = \begin{bmatrix} -2-0 & 0-(-6) & 4-15 \\ 6-3 & 2-(-9) & 8-3 \end{bmatrix} = \begin{bmatrix} -2 & 6 & -11 \\ 3 & 11 & 5 \end{bmatrix} \]Step 5: Add \(4C\)
\[ = \begin{bmatrix} -2+4 & 6-20 & -11+8 \\ 3+24 & 11+0 & 5-16 \end{bmatrix} = \begin{bmatrix} 2 & -14 & -3 \\ 27 & 11 & -11 \end{bmatrix} \]Final Answer:
\[ \boxed{ \begin{bmatrix} 2 & -14 & -3 \\ 27 & 11 & -11 \end{bmatrix} } \]