JEE Advanced · Mathematics · 19. Determinants
Paragraph:
Let \(A\) be the set of all \(3 \times 3\) symmetric matrices all of whose entries are either 0 or 1 . Five of these entries are 1 and four of them are 0 .Question:
The number of matrices \(A\) in \(A\) for which the system of linear equations \(A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\) is is inconsistent, is
- A
0
- B
more than 2
- C
2
- D
1
Answer & Solution
Correct Answer
(B)
more than 2
Step-by-step Solution
Detailed explanation
Given, \(A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\)
Case I \(\left[\begin{array}{lll}1 & a & b \\ a & 1 & c \\ b & c & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\)
\(a, b, c\) are selected from \(1,0,0\).
\(\Rightarrow x+a y+b z=1 \Rightarrow a x+y+c z=0\) \(b x+c y+z=0\)
(i) If \(a=1, b=c=0\), then \(x+y=1\) Inconsistent system of equation \(x+y=0\)
(ii) If \(a=0=c, b=1\), then \(x+z=1\), \(y=0\)
Inconsistent system of equation
\[
x+z=0
\]
(iii) If \(c=1, a=b=0\), then \(x=1, z=0\), \(y=0\)
Case II
(i) \(\left[\begin{array}{lll}1 & a & b \\ a & 0 & c \\ b & c & 0\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\)
\(a, b, c\) are selected from \(1,1,0\).
\[
\begin{aligned}
\Rightarrow x+a y+b z & =1 \Rightarrow a x+c z=0 \\
b x+c y & =0
\end{aligned}
\]
Clearly, in all three cases, solutions are possible, so system is consistent.
\[
\begin{gathered}
\text { (ii) }\left[\begin{array}{lll}
0 & a & b \\
a & 1 & c \\
b & c & 0
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
0 \\
0
\end{array}\right] \\
\Rightarrow a y+b z=1 \Rightarrow a x+y+c z=0 \\
b x+c y=0
\end{gathered}
\]
Clearly, \(b=0, a=c=1\) gives
\[
y=1 ; x+y+z=0
\]
Inconsistent system \(y=0\)
More than 2 matrices are possible.
Hence, option (b) is correct.
Case I \(\left[\begin{array}{lll}1 & a & b \\ a & 1 & c \\ b & c & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\)
\(a, b, c\) are selected from \(1,0,0\).
\(\Rightarrow x+a y+b z=1 \Rightarrow a x+y+c z=0\) \(b x+c y+z=0\)
(i) If \(a=1, b=c=0\), then \(x+y=1\) Inconsistent system of equation \(x+y=0\)
(ii) If \(a=0=c, b=1\), then \(x+z=1\), \(y=0\)
Inconsistent system of equation
\[
x+z=0
\]
(iii) If \(c=1, a=b=0\), then \(x=1, z=0\), \(y=0\)
Case II
(i) \(\left[\begin{array}{lll}1 & a & b \\ a & 0 & c \\ b & c & 0\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\)
\(a, b, c\) are selected from \(1,1,0\).
\[
\begin{aligned}
\Rightarrow x+a y+b z & =1 \Rightarrow a x+c z=0 \\
b x+c y & =0
\end{aligned}
\]
Clearly, in all three cases, solutions are possible, so system is consistent.
\[
\begin{gathered}
\text { (ii) }\left[\begin{array}{lll}
0 & a & b \\
a & 1 & c \\
b & c & 0
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
0 \\
0
\end{array}\right] \\
\Rightarrow a y+b z=1 \Rightarrow a x+y+c z=0 \\
b x+c y=0
\end{gathered}
\]
Clearly, \(b=0, a=c=1\) gives
\[
y=1 ; x+y+z=0
\]
Inconsistent system \(y=0\)
More than 2 matrices are possible.
Hence, option (b) is correct.
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