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

The Correct Option is B

\(A = \begin{bmatrix}       a & b & c  \\[0.3em]  d & e  & f \\[0.3em] g & h & i      \end{bmatrix}\)
\(A\begin{bmatrix}      1  \\[0.3em]1 \\[0.3em] 0 \end{bmatrix}\)=\(\begin{bmatrix}      1  \\[0.3em]1 \\[0.3em] 0 \end{bmatrix}\)⇒\(\begin{bmatrix}       a & b & c  \\[0.3em]  d & e  & f \\[0.3em] g & h & i      \end{bmatrix}\)=\(\begin{bmatrix}      1  \\[0.3em]1 \\[0.3em] 0 \end{bmatrix}\)
\(⇒ a+b=1\)
\(⇒ d+e=1\)
\(⇒ g+h=0\)
\(A\begin{bmatrix}      1  \\[0.3em]0 \\[0.3em] 1 \end{bmatrix}\)=\(\begin{bmatrix}      -1  \\[0.3em]0 \\[0.3em] 1 \end{bmatrix}\)⇒ \(\begin{bmatrix}       a & b & c  \\[0.3em]  d & e  & f \\[0.3em] g & h & i      \end{bmatrix}\) \(\begin{bmatrix}      1  \\[0.3em]0 \\[0.3em] 1 \end{bmatrix}\)=\(\begin{bmatrix}      -1  \\[0.3em]0 \\[0.3em] 1 \end{bmatrix}\)
\(⇒ a+c=−1\)
\(⇒ d+f=0\)
\(⇒ g+i=1\)
\(A\begin{bmatrix}      0  \\[0.3em]0 \\[0.3em] 1 \end{bmatrix}\)=\(\begin{bmatrix}      1  \\[0.3em]1 \\[0.3em] 2 \end{bmatrix}\) ⇒ \(\begin{bmatrix}       a & b & c  \\[0.3em]  d & e  & f \\[0.3em] g & h & i      \end{bmatrix}\)\(\begin{bmatrix}      0  \\[0.3em]0 \\[0.3em] 1 \end{bmatrix}\)=
\(⇒c=1\)
\(⇒f=1\)
\(⇒  i=2\)
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.