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Mathematics · Matrices

JEE Main 2024 — 31 January, Shift 2 — Question 80

Let A be a 3×33 \times 3 real matrix such that

$A(101)\begin{pmatrix}1\\0\\1\end{pmatrix} =2(011)\begin{pmatrix}0\\1\\1\end{pmatrix},\quad A(−101)\begin{pmatrix}-1\\0\\1\end{pmatrix} =4(−110)\begin{pmatrix}-1\\1\\0\end{pmatrix},\quad A(001)\begin{pmatrix}0\\0\\1\end{pmatrix} =2$$\begin{pmatrix}0\1\0\end{pmatrix}$$$

Then, the system

$(A-3I)(xyz)\begin{pmatrix}x\\y\\z\end{pmatrix} =$$\begin{pmatrix}1\2\3\end{pmatrix}$$$ has

  1. Option A:

    unique solution

    Correct
  2. Option B:

    exactly two solutions

  3. Option C:

    no solution

  4. Option D:

    infinitely many solutions

Answer: A

Step-by-step solution

$\text{Sol. Let } A=

x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{pmatrix}$$$ $\text{Given }\; A$$\begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix}$$ = $$\begin{pmatrix} 2\\ 0\\ 2 \end{pmatrix}$$ \qquad (1)$ $\therefore $$\begin{pmatrix} x_1+z_1\\ x_2+z_2\\ x_3+z_3 \end{pmatrix}$$ = $$\begin{pmatrix} 2\\ 0\\ 2 \end{pmatrix}$$$ $x_1+z_1=2 \qquad (2)$ $x_2+z_2=0 \qquad (3)$ $x_3+z_3=0 \qquad (4)$ $\text{Given }\; A$$\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}$$ = $$\begin{pmatrix} -4\\ 0\\ 4 \end{pmatrix}$$$ $\therefore $$\begin{pmatrix} -x_1+z_1\\ -x_2+z_2\\ -x_3+z_3 \end{pmatrix}$$ = $$\begin{pmatrix} -4\\ 0\\ 4 \end{pmatrix}$$$ $-x_1+z_1=-4 \qquad (5)$ $-x_2+z_2=0 \qquad (6)$ $-x_3+z_3=4$ $\text{Given }\; A$$\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}$$ = $$\begin{pmatrix} 0\\ 2\\ 0 \end{pmatrix}$$$ $\therefore $$\begin{pmatrix} y_1\\ y_2\\ y_3 \end{pmatrix}$$ = $$\begin{pmatrix} 0\\ 2\\ 0 \end{pmatrix}$$$ $\therefore\ y_1=0,\ y_2=2,\ y_3=0$ $\text{From (2), (3), (4), (5), (6) and (7)}$ $x_1=3,\ x_2=0,\ x_3=-1$ $y_1=0,\ y_2=2,\ y_3=0$ $z_1=-1,\ z_2=0,\ z_3=3$ $\therefore\ A= $$\begin{pmatrix} 3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3 \end{pmatrix}$$$ $\therefore\ \text{Now } (A-3I) $$\begin{pmatrix} x\\ y\\ z \end{pmatrix}$$ = $$\begin{pmatrix} -1\\ 2\\ 3 \end{pmatrix}$$$ $$$\begin{pmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{pmatrix}$$ $$\begin{pmatrix} x\\ y\\ z \end{pmatrix}$$ = $$\begin{pmatrix} -1\\ 2\\ 3 \end{pmatrix}$$$ $$$\begin{pmatrix} -z\\ -y\\ -x \end{pmatrix}$$ = $$\begin{pmatrix} -1\\ 2\\ 3 \end{pmatrix}$$$ $z=-1,\ y=-2,\ x=-3$

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
31 January, Shift 2
Subject
Mathematics
Chapter
Matrices
Topic
solving System of Linear Equations using Matrices

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