JEE Advanced · Mathematics · 18. Matrices
Paragraph:
Let \(a\), b and \(c\) be three real numbers satisfying \(\left[\begin{array}{lll}a & b & c\end{array}\right]\left[\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]\)Question:
If the point \(P(a, b, c)\), with reference to Eq. (E), lies on the plane \(2 x+y+z=1\), then the value of \(7 a+b+c\) is
- A
0
- B
12
- C
7
- D
6
Answer & Solution
Correct Answer
(D)
6
Step-by-step Solution
Detailed explanation
\[
\begin{aligned}
& \text { Given, }[a b c]_{1 \times 3}\left[\begin{array}{lll}
1 & 9 & 7 \\
8 & 2 & 7 \\
7 & 3 & 7
\end{array}\right]_{3 \times 3}=\left[\begin{array}{lll}
0 & 0 & 0
\end{array}\right] \\
& \Rightarrow \quad\left[\begin{array}{c}
a+8 b+7 c \\
9 a+2 b+3 c \\
7 a+7 b+7 c
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right] \\
& \Rightarrow \quad a+8 b+7 c=0 \\
& \Rightarrow 9 a+2 b+3 c=0 \\
& \Rightarrow \quad a+b+c=0 \\
&
\end{aligned}
\]
On multiplying Eq. (iii) by 2 , then subtract from Eq. (ii), we get
\[
7 a+c=0
\]
Again multiplying Eq. (iii) by 3 , then subtract from Eq. (ii), we get
\[
\begin{array}{ll}
& 6 a-b=0 \\
\therefore \quad & b=6 a \text { and } c=-7 a
\end{array}
\]
As \((a, b, c)\) lies on \(2 x+y+z=1\)
\[
\begin{array}{lc}
\Rightarrow & 2 a+b+c=1 \\
\Rightarrow & 2 a+6 a-7 a=1 \\
\Rightarrow & a=1, b=6 \text { and } c=-7 \\
\therefore & 7 a+b+c=7+6-7=6
\end{array}
\]
\begin{aligned}
& \text { Given, }[a b c]_{1 \times 3}\left[\begin{array}{lll}
1 & 9 & 7 \\
8 & 2 & 7 \\
7 & 3 & 7
\end{array}\right]_{3 \times 3}=\left[\begin{array}{lll}
0 & 0 & 0
\end{array}\right] \\
& \Rightarrow \quad\left[\begin{array}{c}
a+8 b+7 c \\
9 a+2 b+3 c \\
7 a+7 b+7 c
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right] \\
& \Rightarrow \quad a+8 b+7 c=0 \\
& \Rightarrow 9 a+2 b+3 c=0 \\
& \Rightarrow \quad a+b+c=0 \\
&
\end{aligned}
\]
On multiplying Eq. (iii) by 2 , then subtract from Eq. (ii), we get
\[
7 a+c=0
\]
Again multiplying Eq. (iii) by 3 , then subtract from Eq. (ii), we get
\[
\begin{array}{ll}
& 6 a-b=0 \\
\therefore \quad & b=6 a \text { and } c=-7 a
\end{array}
\]
As \((a, b, c)\) lies on \(2 x+y+z=1\)
\[
\begin{array}{lc}
\Rightarrow & 2 a+b+c=1 \\
\Rightarrow & 2 a+6 a-7 a=1 \\
\Rightarrow & a=1, b=6 \text { and } c=-7 \\
\therefore & 7 a+b+c=7+6-7=6
\end{array}
\]
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