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Model Paper 6 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 65 Marks
Question
Given $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{ccc}-4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1\end{array}\right]$ find $AB$ and use this result in solving the following system of equations. $x - y + z =4 , x - 2y - 2z = 9 , 2x + y + 3z = 1$
✓

Answer

$x-y+z=4$
$x-2 y-2 z=9$
$2 x+y+3 z=1$
Let $A=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{array}\right] \quad X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] C=\left[\begin{array}{l} 4 \\ 9 \\ 1 \end{array}\right]$
Then, given system of equations can be rewritten as,
$AX=C$
Now, $A B=\left[\begin{array}{ccc}1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}-4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1\end{array}\right]$
$ =\left[\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}\right]$
$AB=8 $
$A^{-1}=\frac{1}{8} B\left[\begin{array}{c} \because A^{-1} A B=8 A^{-1} I \\ B=8 A^{-1} \end{array}\right] $
$\Rightarrow A^{-1}=\frac{1}{8}\left[\begin{array}{ccc} \frac{-1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{-7}{8} & \frac{1}{8} & \frac{3}{8} \\ \frac{5}{8} & \frac{-3}{8} & \frac{-1}{8} \end{array}\right]$
Now, $AX = C$,
$ \Rightarrow X=A^{-1} C \\ \Rightarrow\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{ccc} \frac{-1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{-7}{8} & \frac{1}{8} & \frac{3}{8} \\ \frac{5}{8} & \frac{-3}{8} & \frac{-1}{8} \end{array}\right]\left[\begin{array}{l} 4 \\ 9 \\ 1 \end{array}\right] $
$=\left[\begin{array}{c} \frac{-4}{2}+\frac{9}{2}+\frac{1}{2} \\ \frac{-28}{8}+\frac{9}{8}+\frac{3}{8} \\ \frac{20}{8}+\frac{-27}{8}+\frac{-1}{8} \end{array}\right]=\left[\begin{array}{c} 3 \\ -2 \\ -1 \end{array}\right] $
$\Rightarrow x=3, y=-2, z=-1$

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