CBSE 2023 · Region 2 · Set 1 · Q32 · 5 marks
If $\displaystyle \mathrm{A}=\left[\begin{array}{rrr}-3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{array}\right], \mathrm{B}=\left[\begin{array}{rrr}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right]$, then find AB and use it to solve the following system of equations: \[\begin{aligned} & \mathrm{x}-2 \mathrm{y}=3 \\ & 2 \mathrm{x}-\mathrm{y}-\mathrm{z}=2 \\ & -2 \mathrm{y}+\mathrm{z}=3 \end{aligned} \]If $\displaystyle f(\alpha)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]$, prove that $\displaystyle f(\alpha) \cdot f(-\beta)=f(\alpha-\beta)$
If $\displaystyle \mathrm{A}=\left[\begin{array}{rrr}-3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{array}\right], \mathrm{B}=\left[\begin{array}{rrr}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right]$, then find AB and use it to solve the following system of equations: \[\begin{aligned} & \mathrm{x}-2 \mathrm{y}=3 \\ & 2 \mathrm{x}-\mathrm{y}-\mathrm{z}=2 \\ & -2 \mathrm{y}+\mathrm{z}=3 \end{aligned} \]
If $\displaystyle f(\alpha)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]$, prove that $\displaystyle f(\alpha) \cdot f(-\beta)=f(\alpha-\beta)$
Marking-scheme solution
(a)
$\displaystyle \mathrm{AB}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
$\displaystyle \Rightarrow \mathrm{B}^{-1}=\mathrm{A}$
The given system of equations can be written as:
$$\mathbf{B}^{\mathrm{T}} \cdot \mathbf{X}=\mathbf{C}, \mathbf{X}=\left[\begin{array}{l}
\mathrm{x}
\mathrm{y}
\mathrm{z}
\end{array}\right], \mathbf{C}=\left[\begin{array}{l}
3
2
\end{array}\right]
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.