CBSE 2025 · Region 6 · Set 1 · Q26 · 3 marks
Let $\displaystyle \mathrm{A}=\left[\begin{array}{r}1 \\ 4 \\ -2\end{array}\right]$ and $\displaystyle \mathrm{C}=\left[\begin{array}{rrr}3 & 4 & 2 \\ 12 & 16 & 8 \\ -6 & -8 & -4\end{array}\right]$ be two matrices. Then, find the matrix B if $\displaystyle \mathrm{AB}=\mathrm{C}$.
Marking-scheme solution
(a)
Let $\displaystyle \mathrm{B}=\left[\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z}\end{array}\right]$
\[\mathrm{AB}=\mathrm{C} \Rightarrow\left[\begin{array}{ccc}
\mathrm{x} & \mathrm{y} & \mathrm{z} \\
4 \mathrm{x} & 4 \mathrm{y} & 4 \mathrm{z} \\
-2 \mathrm{x} & -2 \mathrm{y} & -2 \mathrm{z}
\end{array}\right]=\left[\begin{array}{rrr}
3 & 4 & 2 \\
12 & 16 & 8 \\
-6 & -8 & -4
\end{array}\right] \text { which }
\]
gives $\displaystyle \mathrm{x}=3, \mathrm{y}=4$ and $\displaystyle \mathrm{z}=2$
\[\mathrm{B}=\left[\begin{array}{lll}
3 & 4 & 2
\end{array}\right]
\]
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.