If \( \mathrm{A}=\begin{bmatrix}-2 & 2 & 0\\3 & 1 & 4\end{bmatrix} \) and \( \mathrm{B}=\begin{bmatrix}2 & 0 & -2\\7 & 1 & 6\end{bmatrix} \) are two matrices such that −3A + 5B + 2C = 0 then matrix C is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \begin{bmatrix}-8 & 3 & 5\\-13 & -1 & -9\end{bmatrix} \)
Here 0 is the 2 × 3 zero matrix O. Making C the subject:
\( 2C=3A-5B \)
\( \displaystyle C=\frac{1}{2}(3A-5B) \)
\( \displaystyle 3A=\begin{bmatrix}-6 & 6 & 0\\9 & 3 & 12\end{bmatrix} \)
\( \displaystyle 5B=\begin{bmatrix}10 & 0 & -10\\35 & 5 & 30\end{bmatrix} \)
\( \displaystyle 3A-5B=\begin{bmatrix}-16 & 6 & 10\\-26 & -2 & -18\end{bmatrix} \)
\( \displaystyle C=\frac{1}{2}\begin{bmatrix}-16 & 6 & 10\\-26 & -2 & -18\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}-8 & 3 & 5\\-13 & -1 & -9\end{bmatrix} \)
Shortcut: The options differ already in the first entry. Only \( c_{11} \) is needed: \( c_{11}=\frac{3(-2)-5(2)}{2}=-8 \), and Option (4) is the only one starting with −8.