Question

Let A be a $3×3$ real matrix such that $A\begin{pmatrix} 1 \\\1 \\\ 0 \end{pmatrix}$=$\begin{pmatrix} 1 \\\1 \\\ 0 \end{pmatrix}$; $A\begin{pmatrix} 1 \\\0 \\\ 1 \end{pmatrix}$=$A\begin{pmatrix} -1 \\\0 \\\ 1 \end{pmatrix}$ and $A\begin{pmatrix} 0 \\\0 \\\ 1 \end{pmatrix}$=$\begin{pmatrix} 1 \\\1 \\\ 2 \end{pmatrix}$
If $X = [x_1, x_2, x_3]^T$and $I$ is an identity matrix of order $3$, then the system $[A−2I]X$= $\begin{pmatrix} 4 \\\1 \\\ 1 \end{pmatrix}$ has:

• A. No solution
• B. Infinitely many solutions
• C. Unique solution
• D. Exactly two solutions

Answer 1

Answer: (B)

Infinitely many solutions

Answer 2

$A = \begin{bmatrix} a & b & c \\\[0.3em] d & e & f \\\[0.3em] g & h & i \end{bmatrix}$
$A\begin{bmatrix} 1 \\\[0.3em]1 \\\[0.3em] 0 \end{bmatrix}$=$\begin{bmatrix} 1 \\\[0.3em]1 \\\[0.3em] 0 \end{bmatrix}$⇒$\begin{bmatrix} a & b & c \\\[0.3em] d & e & f \\\[0.3em] g & h & i \end{bmatrix}$=$\begin{bmatrix} 1 \\\[0.3em]1 \\\[0.3em] 0 \end{bmatrix}$
$⇒ a+b=1$
$⇒ d+e=1$
$⇒ g+h=0$
$A\begin{bmatrix} 1 \\\[0.3em]0 \\\[0.3em] 1 \end{bmatrix}$=$\begin{bmatrix} -1 \\\[0.3em]0 \\\[0.3em] 1 \end{bmatrix}$⇒ $\begin{bmatrix} a & b & c \\\[0.3em] d & e & f \\\[0.3em] g & h & i \end{bmatrix}$ $\begin{bmatrix} 1 \\\[0.3em]0 \\\[0.3em] 1 \end{bmatrix}$=$\begin{bmatrix} -1 \\\[0.3em]0 \\\[0.3em] 1 \end{bmatrix}$
$⇒ a+c=−1$
$⇒ d+f=0$
$⇒ g+i=1$
$A\begin{bmatrix} 0 \\\[0.3em]0 \\\[0.3em] 1 \end{bmatrix}$=$\begin{bmatrix} 1 \\\[0.3em]1 \\\[0.3em] 2 \end{bmatrix}$ ⇒ \begin{bmatrix} a & b & c \\\[0.3em] d & e & f \\\[0.3em] g & h & i \end{bmatrix}$$\begin{bmatrix} 0 \\\[0.3em]0 \\\[0.3em] 1 \end{bmatrix}=
$⇒c=1$
$⇒f=1$
$⇒ i=2$
On solving,
$a = –2, b = 3, c = 1, d = –1, e = 2, f = 1, g = –1,h = 1, i = 2$
$A = \begin{bmatrix} -2 & 3 & 1 \\\[0.3em] -1 & 2 & 1 \\\[0.3em] -1& 1 & 2 \end{bmatrix}$ $⇒ A=2I$ $\begin{bmatrix} -4 & 3 & 1 \\\[0.3em] -1 & 0 & 1 \\\[0.3em] -1& 1 & 0 \end{bmatrix}$
(A−2I)x=$$\begin{bmatrix} 4 \\\[0.3em]1 \\\[0.3em] 1 \end{bmatrix}
$⇒ –4x_1 \+ 3x_2 + x_3 = 4$ …(i)
$⇒ –x_1 + x_3 = 1$ …(ii)
$⇒ –x_1 + x_2 = 1$ …(iii)
So 3(iii) + (ii) = (i)
∴ Infinite solution

So, the correct option (B): Infinitely many solutions.