CBSE 2026 · Region 2 · Set 1 · Q32 · 5 marks
If $\displaystyle \mathrm{P}=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]$ and $\displaystyle \mathrm{Q}=\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$, find (QP) and hence solve the following system of equations using matrices : \[x-\mathrm{y}=3,2 x+3 \mathrm{y}+4 \mathrm{z}=17, \mathrm{y}+2 \mathrm{z}=7 \]Obtain the value of $\displaystyle \Delta=\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1 & 1+\mathrm{y} & 1 \\ 1 & 1 & 1+\mathrm{z}\end{array}\right|$ in terms of $\displaystyle x, \mathrm{y}$ and z . Further, if $\displaystyle \Delta=0$ and $\displaystyle x, \mathrm{y}, \mathrm{z}$ are non-zero real numbers, prove that $\displaystyle x^{-1}+\mathrm{y}^{-1}+\mathrm{z}^{-1}=-1$.
If $\displaystyle \mathrm{P}=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]$ and $\displaystyle \mathrm{Q}=\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$, find (QP) and hence solve the following system of equations using matrices : \[x-\mathrm{y}=3,2 x+3 \mathrm{y}+4 \mathrm{z}=17, \mathrm{y}+2 \mathrm{z}=7 \]
Obtain the value of $\displaystyle \Delta=\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1 & 1+\mathrm{y} & 1 \\ 1 & 1 & 1+\mathrm{z}\end{array}\right|$ in terms of $\displaystyle x, \mathrm{y}$ and z . Further, if $\displaystyle \Delta=0$ and $\displaystyle x, \mathrm{y}, \mathrm{z}$ are non-zero real numbers, prove that $\displaystyle x^{-1}+\mathrm{y}^{-1}+\mathrm{z}^{-1}=-1$.
Marking-scheme solution
$\displaystyle QP=\begin{bmatrix}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{bmatrix}$ i.e. $\displaystyle QP=6I$
$\displaystyle \Rightarrow P^{-1}=\dfrac{1}{6} Q$
Given system of equations is given by $\displaystyle PX=R$
where, $\displaystyle X=\begin{bmatrix}x \\ \mathrm{y} \\ \mathrm{z}\end{bmatrix}, R=\begin{bmatrix}3 \\ 17 \\ 7\end{bmatrix}$
$\displaystyle PX=R \Rightarrow X=P^{-1} R$
$\displaystyle \Rightarrow X=\dfrac{1}{6} Q \cdot R$
$\displaystyle =\dfrac{1}{6}\begin{bmatrix}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{bmatrix}\begin{bmatrix}3 \\ 17 \\ 7\end{bmatrix}$
$\displaystyle \Rightarrow \begin{bmatrix}x \\ \mathrm{y} \\ \mathrm{z}\end{bmatrix}=\dfrac{1}{6}\begin{bmatrix}12 \\ -6 \\ 24\end{bmatrix}=\begin{bmatrix}2 \\ -1 \\ 4\end{bmatrix}$
$\displaystyle \Rightarrow x=2, \mathrm{y}=-1, \mathrm{z}=4$
$\displaystyle \Delta=\begin{vmatrix}1+x & 1 & 1 \\ 1 & 1+\mathrm{y} & 1 \\ 1 & 1 & 1+\mathrm{z}\end{vmatrix}$
$\displaystyle =(1+x)[(1+\mathrm{y})(1+\mathrm{z})-1]-1[(1+\mathrm{z})-1]+1[1-(1+\mathrm{y})]$
$\displaystyle =(1+x)(1+\mathrm{z}+\mathrm{y}+\mathrm{y} \mathrm{z}-1)-\mathrm{z}-\mathrm{y}$
$\displaystyle =x \mathrm{y}+\mathrm{y} \mathrm{z}+\mathrm{z} x+x \mathrm{y} \mathrm{z}$
$\displaystyle \Delta=0 \Rightarrow x \mathrm{y}+\mathrm{y} \mathrm{z}+\mathrm{z} x+x \mathrm{y} \mathrm{z}=0$
$\displaystyle \Rightarrow \mathrm{y} \mathrm{z}+\mathrm{z} x+x \mathrm{y}=-x \mathrm{y} \mathrm{z}$
Dividing both sides by $\displaystyle x \mathrm{y} \mathrm{z}$,
$\displaystyle x^{-1}+\mathrm{y}^{-1}+\mathrm{z}^{-1}=-1$
MatricesInvertible MatricesApplylong_answerhard
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.