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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
If $\text{A}=\begin{bmatrix}1&-1&0\\ 2&3&4\\ 0&1&2\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$ are two square matrices, find AB and hence solve the system of linear equations:
x - y = 3, 2x + 3y + 4z = 17, y + 2z = 7
✓

Answer

Here,
$\text{A}=\begin{bmatrix}1&-1&0\\ 2&3&4\\ 0&1&2\end{bmatrix}\text{and}​​$$\text{B}=\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$
Now,
$\text{AB}=\begin{bmatrix}1&-1&0\\ 2&3&4\\ 0&1&2\end{bmatrix}\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}6&0&0\\ 0&6&0\\ 0&0&6\end{bmatrix}$
$\Rightarrow\text{AB}=6\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}$
$\Rightarrow\text{AB}=6\text{I}_{3}$
$\Rightarrow\frac{1}{6}\text{AB}=\text{I}_{3}$
$\Rightarrow\big(\frac{1}{6}\text{B}\big)\text{A}=\text{I}_{3}\ (\because\text{AB}=\text{AB})$
$\Rightarrow\text{A}^{-1}=\frac{1}{6}\text{B}$
$\Rightarrow\text{A}^{-1}=\frac{1}{6}\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\text{X}=\frac{1}{6}\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}\begin{bmatrix}3\\ 17\\ 7\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}6+34-28\\ -12+34-28\\ 6-17+35\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}12\\ -6\\ 24\end{bmatrix}$
$\therefore$ x = 2, y = -1 and z = 4

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