If \(A = \begin{bmatrix}1&5\\7&12\end{bmatrix},B=\begin{bmatrix}9&1\\7&8\end{bmatrix}\) and \(C\) are three matrices such that \(3A +5B+2C = 0\), then the matrix \(C\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\begin{bmatrix}-24&-10\\-28&-38\end{bmatrix}\)
Here 0 is the zero matrix O. Solving for C:
\(2C = -(3A + 5B)\)
By scalar multiplication and addition, entry by entry:
\( \displaystyle 3A = \begin{bmatrix}3 & 15\\ 21 & 36\end{bmatrix} \)
\( \displaystyle 5B = \begin{bmatrix}45 & 5\\ 35 & 40\end{bmatrix} \)
\( \displaystyle 3A + 5B = \begin{bmatrix}48 & 20\\ 56 & 76\end{bmatrix} \)
\( \displaystyle C = -\frac{1}{2}\begin{bmatrix}48 & 20\\ 56 & 76\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}-24 & -10\\ -28 & -38\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}-48&-20\\-56&-76\end{bmatrix}\) — Incorrect. This is \(-(3A + 5B) = 2C\); the division by 2 is missing.
Option (3) \(\begin{bmatrix}-48&20\\56&-76\end{bmatrix}\) — Incorrect. It is not halved, and its entries 20 and 56 have the wrong sign.
Option (4) \(\begin{bmatrix}24&10\\28&38\end{bmatrix}\) — Incorrect. This is \(\frac{1}{2}(3A + 5B) = -C\); the sign is wrong.
NCERT Exemplar:
Matrices
