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Solution for If a = ⎡ ⎢ ⎣ 1 − 1 0 2 3 4 0 1 2 ⎤ ⎥ ⎦ and B = ⎡ ⎢ ⎣ 2 2 − 4 − 4 2 − 4 2 − 1 5 ⎤ ⎥ ⎦ Are Two Square Matrices, Find Ab and Hence Solve the System of Linear Equations: X − - CBSE (Commerce) Class 12 - Mathematics

Question

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

Solution

Here,
$A = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}\text{ and }B = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}$
Now,
$AB = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix} \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}$
$\Rightarrow AB = \begin{bmatrix}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{bmatrix}$
$\Rightarrow AB = 6\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$
$\Rightarrow AB = 6 I_3$
$\Rightarrow \frac{1}{6}AB = I_3$
$\Rightarrow \left( \frac{1}{6}B \right)A = I_3 \left( \because AB = BA \right)$
$\Rightarrow A^{- 1} = \frac{1}{6}B$
$\Rightarrow A^{- 1} = \frac{1}{6}\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}$
$X = A^{- 1} B$
$X = \frac{1}{6}\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}\begin{bmatrix}3 \\ 17 \\ 7\end{bmatrix}$
$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{6}\begin{bmatrix}6 + 34 - 28 \\ - 12 + 34 - 28 \\ 6 - 17 + 35\end{bmatrix}$
$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{6}\begin{bmatrix}12 \\ - 6 \\ 24\end{bmatrix}$
$\therefore x = 2, y = - 1\text{ and }z = 4$

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Solution If a = ⎡ ⎢ ⎣ 1 − 1 0 2 3 4 0 1 2 ⎤ ⎥ ⎦ and B = ⎡ ⎢ ⎣ 2 2 − 4 − 4 2 − 4 2 − 1 5 ⎤ ⎥ ⎦ Are Two Square Matrices, Find Ab and Hence Solve the System of Linear Equations: X − Concept: Determinant of a Square Matrix.
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