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प्रश्न
Given \[A = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}, B = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}\] , find BA and use this to solve the system of equations y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17
उत्तर
Here,
\[ A = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}\]
\[BA = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}\]
\[ \Rightarrow BA = \begin{bmatrix}2 + 4 + 0 & 2 - 2 + 0 & - 4 + 4 + 0 \\ 4 - 12 + 8 & 4 + 6 - 4 & - 8 - 12 + 20 \\ 0 - 4 + 4 & 0 + 2 - 2 & 0 - 4 + 10\end{bmatrix}\]
\[ = \begin{bmatrix}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{bmatrix}\]
\[ \Rightarrow BA = 6\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow BA = 6 I_3 \]
\[ \Rightarrow B\left( \frac{1}{6}A \right) = I_3 \]
\[ \Rightarrow B^{- 1} = \frac{1}{6}A\]
\[ \Rightarrow B^{- 1} = \frac{1}{6}\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}\]
Now, BX = C
\[\text{ where, }B = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }C = \begin{bmatrix}3 \\ 17 \\ 7\end{bmatrix}\]
\[ \therefore X = B^{- 1} C\]
\[ \Rightarrow X = \frac{1}{6}\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}\begin{bmatrix}3 \\ 17 \\ 7\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{6}\begin{bmatrix}6 + 34 - 28 \\ - 12 + 34 - 28 \\ 6 - 17 + 35\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{6}\begin{bmatrix}12 \\ - 6 \\ 24\end{bmatrix}\]
\[ \therefore x = 2, y = - 1\text{ and }z = 4 .\]
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