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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Given $\text{A}=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix},\text{B}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}$ , find BA and use this to solve the system of equations $y + 2z = 7, x - y = 3, 2x + 3y + 4z = 17$
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Answer

Here,
$\text{A}=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}$
$\text{BA}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}2+4+0&2-2+0&-4+4+0\\4-12+8&4+6-4&-8-12+20\\0-4+4&0+2-2&0-4+10\end{bmatrix}$
$=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}$
$\Rightarrow\text{BA}=6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{BA}=6\text{I}_3$
$\Rightarrow\text{B}\Big(\frac{1}{6}\text{A}\Big)=\text{I}_3$
$\Rightarrow\text{B}^{-1}=\frac{1}{6}\text{A}$
$\Rightarrow\text{B}^{-1}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}$
Now, BX = C
where, $\text{B}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{C}=\begin{bmatrix}3\\17\\7\end{bmatrix}$
$\therefore X = B^{-1}C$
$\Rightarrow\text{X}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\begin{bmatrix}3\\17\\7\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}6+34-28\\-12+34-28\\6-17+35\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}12\\-6\\24\end{bmatrix}$
$\therefore$ x = 2, y = -1 and z = 4.

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