CBSE 2024 · Region 5 · Set 1 · Q32 · 5 marks
If $\displaystyle \mathrm{A}=\left[\begin{array}{ccc}1 & 2 & -3 \\ 2 & 0 & -3 \\ 1 & 2 & 0\end{array}\right]$, then find $\displaystyle \mathrm{A}^{-1}$ and hence solve the following system of equations : \[\begin{aligned} & x+2 y-3 z=1 \\ & 2 x-3 z=2 \\ & x+2 y=3 \end{aligned} \]Find the product of the matrices $\displaystyle \left[\begin{array}{ccc}1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4\end{array}\right]\left[\begin{array}{ccc}-6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1\end{array}\right]$ and hence solve the system of linear equations: \[\begin{aligned} & x+2 y-3 z=-4 \\ & 2 x+3 y+2 z=2 \\ & 3 x-3 y-4 z=11 \end{aligned} \]
If $\displaystyle \mathrm{A}=\left[\begin{array}{ccc}1 & 2 & -3 \\ 2 & 0 & -3 \\ 1 & 2 & 0\end{array}\right]$, then find $\displaystyle \mathrm{A}^{-1}$ and hence solve the following system of equations : \[\begin{aligned} & x+2 y-3 z=1 \\ & 2 x-3 z=2 \\ & x+2 y=3 \end{aligned} \]
Find the product of the matrices $\displaystyle \left[\begin{array}{ccc}1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4\end{array}\right]\left[\begin{array}{ccc}-6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1\end{array}\right]$ and hence solve the system of linear equations: \[\begin{aligned} & x+2 y-3 z=-4 \\ & 2 x+3 y+2 z=2 \\ & 3 x-3 y-4 z=11 \end{aligned} \]
Marking-scheme solution
(a)
$\displaystyle |\mathrm{A}|=1(6)-2(3)-3(4)=-12 \neq 0, \therefore \mathrm{A}^{-1}$ exist
$$\begin{aligned}
\operatorname{adj} \mathrm{A} & =\left[\begin{array}{ccc}
6 & -6 & -6
-3 & 3 & -3
4 & 0 & -4
\end{array}\right]
\therefore \mathrm{A}^{-$\displaystyle 1$} & =-\frac{1}{12}\left[\begin{array}{ccc}
6 & -6 & -6
-3 & 3 & -3
4 & 0 & -4
\end{array}\right]
\end{aligned}
$$The given system of equations can be written as $\displaystyle \mathbf{A X}=\mathbf{B}, \mathbf{X}=\left[\begin{array}{l}\mathbf{x} \\
\mathbf{y} \\
\mathbf{z}\end{array}\right], \mathbf{B}=\left[\begin{array}{l}1 \\
3\end{array}\right]$
$$X=\mathrm{A}^{-$\displaystyle 1$} B \Rightarrow\left[\begin{array}{l}
x
y
z
\end{array}\right]=-\frac{1}{12}\left[\begin{array}{ccc}
6 & -6 & -6
-3 & 3 & -3
4 & 0 & -4
\end{array}\right]\left[\begin{array}{l}
1
2
\end{array}\right]=\left[\begin{array}{c}
2
1 / 2
2 / 3
\end{array}\right]
$$∴ The solution of the given system of equations is: $\displaystyle x=2, y=\frac{1}{2}, z=\frac{2}{3}$
Or
$$
\text { (b) } \begin{aligned}
& {\left[\begin{array}{ccc}
1 & 2 & -3
2 & 3 & 2
3 & -3 & -4
\end{array}\right]\left[\begin{array}{ccc}
-6 & 17 & 13
14 & 5 & -8
-15 & 9 & -1
\end{array}\right]=\left[\begin{array}{ccc}
67 & 0 & 0
0 & 67 & 0
0 & 0 & 67
\end{array}\right] }
\Rightarrow & {\left[\begin{array}{ccc}
1 & 2 & -3
2 & 3 & 2
3 & -3 & -4
\end{array}\right]^{-$\displaystyle 1$}=\frac{1}{67}\left[\begin{array}{ccc}
-6 & 17 & 13
14 & 5 & -8
-15 & 9 & -1
\end{array}\right] }
\end{aligned}
$$Solution of the system of equations is given by:
$$\begin{gathered}
{\left[\begin{array}{l}
x
y
z
\end{array}\right]=\left[\begin{array}{ccc}
1 & 2 & -3
2 & 3 & 2
3 & -3 & -4
\end{array}\right]^{-$\displaystyle 1$}\left[\begin{array}{c}
-4
2
11
\end{array}\right]=\frac{1}{67}\left[\begin{array}{ccc}
-6 & 17 & 13
14 & 5 & -8
-15 & 9 & -1
\end{array}\right]\left[\begin{array}{c}
-4
2
11
\end{array}\right]=\left[\begin{array}{c}
3
-2
\end{array}\right],}
\therefore x=$\displaystyle 3$, y=-$\displaystyle 2$, z=$\displaystyle 1$
\end{gathered}
DeterminantsAdjoint and Inverse of a MatrixApplylong_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.