Question

Let \[X = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}, A = \begin{bmatrix}1 & - 1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}3 \\ 1 \\ 4\end{bmatrix}\] . If AX = B, then X is equal to

 

पर्याय

• \[\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}\]

• \[\begin{bmatrix}- 1 \\ - 2 \\ - 3\end{bmatrix}\]

• \[\begin{bmatrix}- 1 \\ - 2 \\ - 3\end{bmatrix}\]

• \[\begin{bmatrix}- 1 \\ 2 \\ 3\end{bmatrix}\]

• \[\begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix}\] 

Answer 1

(a) \[\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}\]

Here,
\[ A = \begin{bmatrix}1 & - 1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{bmatrix}, X = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}\text{ and }B = \begin{bmatrix}3 \\ 1 \\ 4\end{bmatrix} \left(\text{ Given }\right)\]
\[\left| A \right|=1 \left( 0 - 2 \right) + 1\left( 2 - 3 \right) + 2\left( 4 - 0 \right)\]
\[ = - 2 - 1 + 8\]
\[ = 5\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then, }\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}0 & 1 \\ 2 & 1\end{vmatrix} = - 2, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ 3 & 1\end{vmatrix} = 1, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 0 \\ 3 & 2\end{vmatrix} = 4\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & 2 \\ 2 & 1\end{vmatrix} = 5, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 2 \\ 3 & 1\end{vmatrix} = - 5, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 1 \\ 3 & 2\end{vmatrix} = - 5\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & 2 \\ 0 & 1\end{vmatrix} = - 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 2 \\ 2 & 1\end{vmatrix} = 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 1 \\ 2 & 0\end{vmatrix} = 2\]
\[adj A = \begin{bmatrix}- 2 & 1 & 4 \\ 5 & - 5 & - 5 \\ - 1 & 3 & 2\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 2 & 5 & - 1 \\ 1 & - 5 & 3 \\ 4 & - 5 & 2\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{5}\begin{bmatrix}- 2 & 5 & - 1 \\ 1 & - 5 & 3 \\ 4 & - 5 & 2\end{bmatrix}\]
\[ \therefore X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{5}\begin{bmatrix}- 2 & 5 & - 1 \\ 1 & - 5 & 3 \\ 4 & - 5 & 2\end{bmatrix}\begin{bmatrix}3 \\ 1 \\ 4\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{5}\begin{bmatrix}- 6 + 5 - 4 \\ 3 - 5 + 12 \\ 12 - 5 + 8\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} = \frac{1}{5}\begin{bmatrix}- 5 \\ 10 \\ 15\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} = \begin{bmatrix}- 1 \\ 2 \\ 3\end{bmatrix}\]
\[\]