Matrix Operations (Defined / Not Defined)
Question:
Let \[ A=\begin{bmatrix}2 & 3 \\ 5 & 7\end{bmatrix}, \quad B=\begin{bmatrix}-1 & 0 & 2 \\ 3 & 4 & 1\end{bmatrix}, \quad C=\begin{bmatrix}-1 & 2 & 3 \\ 2 & 1 & 0\end{bmatrix} \] Find:
(i) \(A+B\), \(B+C\)
(ii) \(2B+3A\), \(3C-4B\)
Let \[ A=\begin{bmatrix}2 & 3 \\ 5 & 7\end{bmatrix}, \quad B=\begin{bmatrix}-1 & 0 & 2 \\ 3 & 4 & 1\end{bmatrix}, \quad C=\begin{bmatrix}-1 & 2 & 3 \\ 2 & 1 & 0\end{bmatrix} \] Find:
(i) \(A+B\), \(B+C\)
(ii) \(2B+3A\), \(3C-4B\)
Solution:
Matrix Orders:
\(A\) is \(2 \times 2\), \(B\) is \(2 \times 3\), \(C\) is \(2 \times 3\)
(i) \(A + B\)
Not defined because matrices must have the same order.
\(A\) is \(2\times2\), \(B\) is \(2\times3\).
\(B + C\)
Both are \(2 \times 3\), so addition is possible:
\[ B+C= \begin{bmatrix} -1+(-1) & 0+2 & 2+3 \\ 3+2 & 4+1 & 1+0 \end{bmatrix} = \begin{bmatrix} -2 & 2 & 5 \\ 5 & 5 & 1 \end{bmatrix} \](ii) \(2B + 3A\)
Not defined because: \(2B\) is \(2\times3\) and \(3A\) is \(2\times2\).
\(3C – 4B\)
Both are \(2 \times 3\), so operation is defined:
\[ 3C= \begin{bmatrix} -3 & 6 & 9 \\ 6 & 3 & 0 \end{bmatrix}, \quad 4B= \begin{bmatrix} -4 & 0 & 8 \\ 12 & 16 & 4 \end{bmatrix} \] \[ 3C-4B= \begin{bmatrix} -3-(-4) & 6-0 & 9-8 \\ 6-12 & 3-16 & 0-4 \end{bmatrix} = \begin{bmatrix} 1 & 6 & 1 \\ -6 & -13 & -4 \end{bmatrix} \]Final Answers:
\(A+B\): Not Defined
\(B+C = \begin{bmatrix}-2 & 2 & 5 \\ 5 & 5 & 1\end{bmatrix}\)
\(2B+3A\): Not Defined
\(3C-4B = \begin{bmatrix}1 & 6 & 1 \\ -6 & -13 & -4\end{bmatrix}\)