Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

If $\left|\begin{array}{rrr}-a^2 & a b & a c \\ a b & -b^2 & b c \\ a c & b c & -c^2\end{array}\right|=\lambda a^2 b^2 c^2$, then the value of $\lambda$ is :

Options:

4

3

2

1

Correct Answer:

4

Explanation:

$\left|\begin{array}{rrr}-a^2 & a b & a c \\ a b & -b^2 & b c \\ a c & b c & -c^2\end{array}\right|=\lambda a^2 b^2 c^2$

taking factor a, b,c from R1, R2, R3 respectively (rows of matrix)

so  $a b c\left|\begin{array}{ccc}-a & b & c \\ a & -b & c \\ a & b & -c\end{array}\right|=\lambda a^2 b^2 c^2$

taking out factor a, b, c from C1, C2, C3 respectively (Columns of matrix)

$\Rightarrow a^2 b^2 c^2\left|\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right|=\lambda a^2 b^2 c^2$

$\Rightarrow \lambda = \left|\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right| \Rightarrow \lambda=\left|\begin{array}{rrr}
-1 & 0 & 0 \\
1 & 0 & 2 \\
1 & 2 & 0
\end{array}\right| $

using operations 

C2 → C1 + C2

C3 → C1 + C3

calculating determinant across R1 row

$\lambda=-1\left|\begin{array}{ll}0 & 2 \\ 2 & 0\end{array}\right|+0+0=-1(-4)$

$\lambda = 4$