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If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are in $\mathbf{A} . P$, then the determinant $\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$ is
(a) 0
(b) 1
(c) $\mathrm{x}$
(d) $2 \mathrm{x}$
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are in $\mathbf{A} . P$, then the determinant $\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$ is
(a) 0
(b) 1
(c) $\mathrm{x}$
(d) $2 \mathrm{x}$
Solution:
1912 Upvotes
Verified Answer
(a) Let $\Delta=\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$
$=\frac{1}{2}\left|\begin{array}{ccc}
x+2 & x+3 & x+2 a \\
2 x+6 & 2 x+8 & 2 x+4 b \\
x+4 & x+5 & x+2 c
\end{array}\right|$
Applying $\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1-\mathrm{R}_3$
$\Delta=\frac{1}{2}\left|\begin{array}{ccc}
\mathrm{x}+2 & \mathrm{x}+3 & \mathrm{x}+2 \mathrm{a} \\
0 & 0 & 0+2(2 \mathrm{~b}-\mathrm{a}-\mathrm{c}) \\
\mathrm{x}+4 & \mathrm{x}+5 & \mathrm{x}+2 \mathrm{c}
\end{array}\right|$
But $a, b, c$ are in A.P $2 b=a+c$ or $2 b-a-c=0$
$\Delta=\frac{1}{2}\left|\begin{array}{ccc}
x+2 & x+3 & x+2 a \\
0 & 0 & 0 \\
x+4 & x+1 & x+2 c
\end{array}\right|=0$
$=\frac{1}{2}\left|\begin{array}{ccc}
x+2 & x+3 & x+2 a \\
2 x+6 & 2 x+8 & 2 x+4 b \\
x+4 & x+5 & x+2 c
\end{array}\right|$
Applying $\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1-\mathrm{R}_3$
$\Delta=\frac{1}{2}\left|\begin{array}{ccc}
\mathrm{x}+2 & \mathrm{x}+3 & \mathrm{x}+2 \mathrm{a} \\
0 & 0 & 0+2(2 \mathrm{~b}-\mathrm{a}-\mathrm{c}) \\
\mathrm{x}+4 & \mathrm{x}+5 & \mathrm{x}+2 \mathrm{c}
\end{array}\right|$
But $a, b, c$ are in A.P $2 b=a+c$ or $2 b-a-c=0$
$\Delta=\frac{1}{2}\left|\begin{array}{ccc}
x+2 & x+3 & x+2 a \\
0 & 0 & 0 \\
x+4 & x+1 & x+2 c
\end{array}\right|=0$
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