Matrix Operations
Question:
Let \[ A=\begin{bmatrix}2 & 4 \\ 3 & 2\end{bmatrix}, \quad B=\begin{bmatrix}1 & 3 \\ -2 & 5\end{bmatrix}, \quad C=\begin{bmatrix}-2 & 5 \\ 3 & 4\end{bmatrix} \] Find:
(i) \(2A – 3B\)
(ii) \(B – 4C\)
(iii) \(3A – C\)
(iv) \(3A – 2B + 3C\)
Let \[ A=\begin{bmatrix}2 & 4 \\ 3 & 2\end{bmatrix}, \quad B=\begin{bmatrix}1 & 3 \\ -2 & 5\end{bmatrix}, \quad C=\begin{bmatrix}-2 & 5 \\ 3 & 4\end{bmatrix} \] Find:
(i) \(2A – 3B\)
(ii) \(B – 4C\)
(iii) \(3A – C\)
(iv) \(3A – 2B + 3C\)
Solution:
(i) \(2A – 3B\)
\[ 2A=\begin{bmatrix}4 & 8 \\ 6 & 4\end{bmatrix}, \quad 3B=\begin{bmatrix}3 & 9 \\ -6 & 15\end{bmatrix} \] \[ 2A-3B= \begin{bmatrix} 4-3 & 8-9 \\ 6-(-6) & 4-15 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 12 & -11 \end{bmatrix} \](ii) \(B – 4C\)
\[ 4C=\begin{bmatrix}-8 & 20 \\ 12 & 16\end{bmatrix} \] \[ B-4C= \begin{bmatrix} 1-(-8) & 3-20 \\ -2-12 & 5-16 \end{bmatrix} = \begin{bmatrix} 9 & -17 \\ -14 & -11 \end{bmatrix} \](iii) \(3A – C\)
\[ 3A=\begin{bmatrix}6 & 12 \\ 9 & 6\end{bmatrix} \] \[ 3A-C= \begin{bmatrix} 6-(-2) & 12-5 \\ 9-3 & 6-4 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix} \](iv) \(3A – 2B + 3C\)
\[ 2B=\begin{bmatrix}2 & 6 \\ -4 & 10\end{bmatrix}, \quad 3C=\begin{bmatrix}-6 & 15 \\ 9 & 12\end{bmatrix} \] \[ 3A – 2B + 3C= \begin{bmatrix} 6-2-6 & 12-6+15 \\ 9-(-4)+9 & 6-10+12 \end{bmatrix} \] \[ = \begin{bmatrix} -2 & 21 \\ 22 & 8 \end{bmatrix} \]Final Answers:
\[ (i)\ \begin{bmatrix}1 & -1 \\ 12 & -11\end{bmatrix}, \quad (ii)\ \begin{bmatrix}9 & -17 \\ -14 & -11\end{bmatrix} \] \[ (iii)\ \begin{bmatrix}8 & 7 \\ 6 & 2\end{bmatrix}, \quad (iv)\ \begin{bmatrix}-2 & 21 \\ 22 & 8\end{bmatrix} \]