\[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\], find AB. Hence, solve the system of equations: x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

Here,\[ A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\]\[AB = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix} \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\]\[ \Rightarrow AB = \begin{bmatrix}7 + 4 + 0 & 2 - 2 + 0 & - 6 + 6 + 0 \\ 14 - 2 - 12 & 4 + 1 + 6 & - 12 - 3 + 15 \\ 0 + 4 - 4 & 0 - 2 + 2 & 0 + 6 + 5\end{bmatrix}\]\[ = \begin{bmatrix}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{bmatrix}\]\[ \Rightarrow AB = 11\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\]\[AB = 11 I_3 \]\[ \Rightarrow \frac{1}{11}AB = I_3 \]\[ \Rightarrow \left( \frac{1}{11}B \right)A = I_3 \]\[ \Rightarrow A^{- 1} = \frac{1}{11}B\]\[ \Rightarrow A^{- 1} = \frac{1}{11}\begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\]\[X = A^{- 1} B\]\[X = \frac{1}{11}\begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\begin{bmatrix}10 \\ 8 \\ 7\end{bmatrix}\]\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{11}\begin{bmatrix}70 + 16 - 42 \\ - 20 + 9 - 21 \\ - 40 + 16 + 35\end{bmatrix}\]\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{11}\begin{bmatrix}44 \\ - 33 \\ 11\end{bmatrix}\]\[ \therefore x = 4, y = - 3\text{ and }z = 1\]

A = ⎡ ⎢ ⎣ 1 − 2 0 2 1 3 0 − 2 1 ⎤ ⎥ ⎦ and B = ⎡ ⎢ ⎣ 7 2 − 6 − 2 1 − 3 − 4 2 5 ⎤ ⎥ ⎦ , Find Ab. Hence, Solve the System of Equations: X − 2y = 10, 2x + Y + 3z = 8 and −2y + Z = 7 - Mathematics

Question

A = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}] and B = [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}]

, find AB. Hence, solve the system of equations: x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

Solution

Here,
$A = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}] and B = [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}]$
$A B = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}] [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}]$
$\Rightarrow A B = [\begin{matrix} 7 + 4 + 0 & 2 - 2 + 0 & - 6 + 6 + 0 \\ 14 - 2 - 12 & 4 + 1 + 6 & - 12 - 3 + 15 \\ 0 + 4 - 4 & 0 - 2 + 2 & 0 + 6 + 5 \end{matrix}]$
$= [\begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix}]$
$\Rightarrow A B = 11 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$
$A B = 11 I_{3}$
$\Rightarrow \frac{1}{11} A B = I_{3}$
$\Rightarrow (\frac{1}{11} B) A = I_{3}$
$\Rightarrow A^{- 1} = \frac{1}{11} B$
$\Rightarrow A^{- 1} = \frac{1}{11} [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}]$
$X = A^{- 1} B$
$X = \frac{1}{11} [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}] [\begin{matrix} 10 \\ 8 \\ 7 \end{matrix}]$
$[\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{11} [\begin{matrix} 70 + 16 - 42 \\ - 20 + 9 - 21 \\ - 40 + 16 + 35 \end{matrix}]$
$[\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{11} [\begin{matrix} 44 \\ - 33 \\ 11 \end{matrix}]$
$∴ x = 4, y = - 3 and z = 1$

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Applications of Determinants and Matrices

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Q 8.3 Q 8.2 Q 8.4

Chapter 8: Solution of Simultaneous Linear Equations - Exercise 8.1 [Page 16]

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Chapter 8 Solution of Simultaneous Linear Equations
Exercise 8.1 | Q 8.3 | Page 16

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A = ⎡ ⎢ ⎣ 1 − 2 0 2 1 3 0 − 2 1 ⎤ ⎥ ⎦ and B = ⎡ ⎢ ⎣ 7 2 − 6 − 2 1 − 3 − 4 2 5 ⎤ ⎥ ⎦ , Find Ab. Hence, Solve the System of Equations: X − 2y = 10, 2x + Y + 3z = 8 and −2y + Z = 7 - Mathematics

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A = ⎡ ⎢ ⎣ 1 − 2 0 2 1 3 0 − 2 1 ⎤ ⎥ ⎦ and B = ⎡ ⎢ ⎣ 7 2 − 6 − 2 1 − 3 − 4 2 5 ⎤ ⎥ ⎦ , Find Ab. Hence, Solve the System of Equations: X − 2y = 10, 2x + Y + 3z = 8 and −2y + Z = 7 - Mathematics

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