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

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

Let X = ⎡ ⎢ ⎣ X 1 X 2 X 3 ⎤ ⎥ ⎦ , a = ⎡ ⎢ ⎣ 1 − 1 2 2 0 1 3 2 1 ⎤ ⎥ ⎦ and B = ⎡ ⎢ ⎣ 3 1 4 ⎤ ⎥ ⎦ . If Ax = B, Then X is Equal to (A) ⎡ ⎢ ⎣ - Mathematics

Question

Let $X = [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}], A = [\begin{matrix} 1 & - 1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{matrix}] and B = [\begin{matrix} 3 \\ 1 \\ 4 \end{matrix}]$ . If AX = B, then X is equal to

Options

$[\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$
$[\begin{matrix} - 1 \\ - 2 \\ - 3 \end{matrix}]$
$[\begin{matrix} - 1 \\ - 2 \\ - 3 \end{matrix}]$
$[\begin{matrix} - 1 \\ 2 \\ 3 \end{matrix}]$
$[\begin{matrix} 0 \\ 2 \\ 1 \end{matrix}]$

MCQ

Solution

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

Here,
$A = [\begin{matrix} 1 & - 1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{matrix}], X = [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] and B = [\begin{matrix} 3 \\ 1 \\ 4 \end{matrix}] (Given)$
$| A | = 1 (0 - 2) + 1 (2 - 3) + 2 (4 - 0)$
$= - 2 - 1 + 8$
$= 5$
${Let C}_{i j} {be the cofactors of the elements a}_{i j} in A = [a_{i j}] . Then,$
$C_{11} = {(- 1)}^{1 + 1} | \begin{matrix} 0 & 1 \\ 2 & 1 \end{matrix} | = - 2, C_{12} = {(- 1)}^{1 + 2} | \begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix} | = 1, C_{13} = {(- 1)}^{1 + 3} | \begin{matrix} 2 & 0 \\ 3 & 2 \end{matrix} | = 4$
$C_{21} = {(- 1)}^{2 + 1} | \begin{matrix} - 1 & 2 \\ 2 & 1 \end{matrix} | = 5, C_{22} = {(- 1)}^{2 + 2} | \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} | = - 5, C_{23} = {(- 1)}^{2 + 3} | \begin{matrix} 1 & - 1 \\ 3 & 2 \end{matrix} | = - 5$
$C_{31} = {(- 1)}^{3 + 1} | \begin{matrix} - 1 & 2 \\ 0 & 1 \end{matrix} | = - 1, C_{32} = {(- 1)}^{3 + 2} | \begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix} | = 3, C_{33} = {(- 1)}^{3 + 3} | \begin{matrix} 1 & - 1 \\ 2 & 0 \end{matrix} | = 2$
$a d j A = {[\begin{matrix} - 2 & 1 & 4 \\ 5 & - 5 & - 5 \\ - 1 & 3 & 2 \end{matrix}]}^{T}$
$= [\begin{matrix} - 2 & 5 & - 1 \\ 1 & - 5 & 3 \\ 4 & - 5 & 2 \end{matrix}]$
$\Rightarrow A^{- 1} = \frac{1}{| A |} a d j A$
$= \frac{1}{5} [\begin{matrix} - 2 & 5 & - 1 \\ 1 & - 5 & 3 \\ 4 & - 5 & 2 \end{matrix}]$
$∴ X = A^{- 1} B$
$\Rightarrow X = \frac{1}{5} [\begin{matrix} - 2 & 5 & - 1 \\ 1 & - 5 & 3 \\ 4 & - 5 & 2 \end{matrix}] [\begin{matrix} 3 \\ 1 \\ 4 \end{matrix}]$
$\Rightarrow X = \frac{1}{5} [\begin{matrix} - 6 + 5 - 4 \\ 3 - 5 + 12 \\ 12 - 5 + 8 \end{matrix}]$
$\Rightarrow [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = \frac{1}{5} [\begin{matrix} - 5 \\ 10 \\ 15 \end{matrix}]$
$\Rightarrow [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} - 1 \\ 2 \\ 3 \end{matrix}]$

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Applications of Determinants and Matrices

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Q 3 Q 2 Q 4

Chapter 8: Solution of Simultaneous Linear Equations - Exercise 8.4 [Page 22]

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RD Sharma Mathematics [English] Class 12

Chapter 8 Solution of Simultaneous Linear Equations
Exercise 8.4 | Q 3 | Page 22

Let X = ⎡ ⎢ ⎣ X 1 X 2 X 3 ⎤ ⎥ ⎦ , a = ⎡ ⎢ ⎣ 1 − 1 2 2 0 1 3 2 1 ⎤ ⎥ ⎦ and B = ⎡ ⎢ ⎣ 3 1 4 ⎤ ⎥ ⎦ . If Ax = B, Then X is Equal to (A) ⎡ ⎢ ⎣ - Mathematics

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Let X = ⎡ ⎢ ⎣ X 1 X 2 X 3 ⎤ ⎥ ⎦ , a = ⎡ ⎢ ⎣ 1 − 1 2 2 0 1 3 2 1 ⎤ ⎥ ⎦ and B = ⎡ ⎢ ⎣ 3 1 4 ⎤ ⎥ ⎦ . If Ax = B, Then X is Equal to (A) ⎡ ⎢ ⎣ - Mathematics

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