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Solution of Simultaneous Linear Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsSolution of Simultaneous Linear Equations5 Marks
Question
$\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}$, find AB. Hence, solve the system of equations:
x - 2y = 10, 2x + y + 3z = 8 and -2y + z = 7
✓

Answer

$\text{A}=\begin{bmatrix}1&-2&0\\2&1&3\\0&-2&1\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}$
$\text{A}\times\text{B}=\begin{bmatrix}11&0&0\\ 0&11&0\\\ 0&0&11\end{bmatrix}$
AB = 11I, where I is a 3 × 3 unit matrix
$\text{A}^{-1}=\frac{1}{11}\text{B}$ [By def. of inverse]Or
Or $\frac{1}{11}=\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}$
Now, the given system of equations can be written as:
$\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}10\\ 8\\ 7\end{bmatrix}$
Or $\text{AX = B}$
$\text{X = A}^{-1}\text{B}$
Or $=\frac{1}{11}\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{11}\begin{bmatrix}44\\ -33\\ 11\end{bmatrix}=\begin{bmatrix}4\\ -3\\ 1\end{bmatrix}$
Hence, x = 4, y = -3, z = 1

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