Let A be a 3×3 real matrix such that A 1 0 1 =2 1 0 1 , A −1 0 1 =4 −1 0 1 , A 0 1 0 =2 0 1 0 . Then, the system A−3I x y z = 1 2 3 has

Correct answer: (A) unique solution. Let, A= x_1 y_1 z_1 x_2 y_2 z_2 x_3 y_3 z_3 It is given that, A 1 0 1 = 2 0 2 ⇒ x_1+z_1 x_2+z_2 x_3+z_3 = 2 0 2 ⇒x_1+z_1=2, x_2+z_2=0, x_3+z_3=0 ...i Also, A −1 0 1 = −4 0 4 ⇒ −x_1+z_1 −x_2+z_2 −x_3+z_3 = -4 0 4 ⇒−x_1+z_1=−4, −x_2+z_2=0, −x_3+z_3=4 ...ii Now, A 0 1 0 = 0 2 0 ⇒ y_1 y_2 y_3 = 0 2 0 ⇒y_1=0, y_2=2, y_3=0 ...iii Using i and ii, ⇒x_1=3, x_2=0, x_3=−1 ⇒z_1=−1, z_2=0, z_3=3 ∴A= 3 0 −1 0 2 0 −1 0 3 Now, A−3I x y z = −1 2 3 ⇒ 0 0 −1 0 −1 0 −1 0 0 x y z = −1 2 3 ⇒ −z −y −x = 1 2 3 ⇒z=−1, y=−2, x=−3

Q48 / 90 · JEE Main 2024 (31 Jan Shift 2) · Mathematics

Let A be a $3 \times 3$ real matrix such that $A (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) = 2 (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}), A (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}) = 4 (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}), A (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) = 2 (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix})$ . Then, the system $(A - 3 I) (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix})$ has

Correct answer
A
unique solution
B
exactly two solutions
C
no solution
D
infinitely many solutions

Solution

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

It is given that, $A [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} 2 \\ 0 \\ 2 \end{matrix}]$

$\Rightarrow [\begin{matrix} x_{1} + z_{1} \\ x_{2} + z_{2} \\ x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 2 \\ 0 \\ 2 \end{matrix}]$

$\Rightarrow x_{1} + z_{1} = 2, x_{2} + z_{2} = 0, x_{3} + z_{3} = 0 . . . (i)$

Also, $A [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} - 4 \\ 0 \\ 4 \end{matrix}]$

$\Rightarrow [\begin{matrix} - x_{1} + z_{1} \\ - x_{2} + z_{2} \\ - x_{3} + z_{3} \end{matrix}] = [\begin{matrix} - 4 \\ 0 \\ 4 \end{matrix}]$

$\Rightarrow - x_{1} + z_{1} = - 4, - x_{2} + z_{2} = 0, - x_{3} + z_{3} = 4 . . . (i i)$

Now, $A [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}]$

$\Rightarrow [\begin{matrix} y_{1} \\ y_{2} \\ y_{3} \end{matrix}] = [\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}]$

$\Rightarrow y_{1} = 0, y_{2} = 2, y_{3} = 0 . . . (i i i)$

Using $(i) and (i i)$ ,

$\Rightarrow x_{1} = 3, x_{2} = 0, x_{3} = - 1$

$\Rightarrow z_{1} = - 1, z_{2} = 0, z_{3} = 3$

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

Now, $(A - 3 I) [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} - 1 \\ 2 \\ 3 \end{matrix}]$

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

$\Rightarrow [\begin{matrix} - z \\ - y \\ - x \end{matrix}] = [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$

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

Hence, the correct answer is option: 'A'.

Let A be a 3×3 real matrix such that A101=2101, A−101=4−101, A010=2010. Then, the system A−3Ixyz=123 has

Solution

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