Let A be a 3 × 3 real matrix such that A ( 1 0 1 ) = 2 ( 1 0 0 ) , A ( - 1 0 1 ) = 4 ( - 1 0 1 ) , A ( 0 1 0 ) = 2 ( 0 1 0 ) . Then, the system ( A - 3 I ) ( x y z ) = ( 1 2 3 ) has [2024]

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Solution

Q.
Let $A$ be a $3 \times 3$ real matrix such that $A (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) = 2 (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), A (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}) = 4 (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}), A (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) = 2 (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) .$

Then, the system $(A - 3 I)$ $(\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix})$ has [2024]

1 exactly two solutions

2 unique solution

3 infinitely many solutions

4 no solution

Ans.
(2)

Let $A = [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}]$

Given, $A [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} 2 \\ 0 \\ 2 \end{matrix}] \Rightarrow [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}] [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} 2 \\ 0 \\ 2 \end{matrix}] \Rightarrow \begin{matrix} x_{1} + z_{1} = 2 \\ x_{2} + z_{2} = 0 \\ x_{3} + z_{3} = 2 \end{matrix}}$ ...(i)

$A [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} - 4 \\ 0 \\ 4 \end{matrix}] \Rightarrow [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}] [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} - 4 \\ 0 \\ 4 \end{matrix}] \Rightarrow \begin{matrix} - x_{1} + z_{1} = - 4 \\ - x_{2} + z_{2} = 0 \\ - x_{3} + z_{3} = 4 \end{matrix}}$ (ii)

and $A [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}] = [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}] [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}]$

$\Rightarrow y_{1} = 0, y_{2} = 2, y_{3} = 0$

Solving (i) and (ii), we get $x_{1} = 3, x_{2} = 0, x_{3} = - 1, y_{1} = 0, y_{2} = 2, y_{3} = 0$ and $z_{1} = - 1, z_{2} = 0, z_{3} = 3$

$\Rightarrow A = [\begin{matrix} 3 & 0 & - 1 \\ 0 & 2 & 0 \\ - 1 & 0 & 3 \end{matrix}]$

Now, $[\begin{matrix} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$

$\Rightarrow - z = 3, - y = 2 and - x = 1$

$\Rightarrow x = - 1, y = - 2 and z = - 3$

Hence, the given system has unique solution.

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