$\left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|$ is equal…
$\left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|$ is equal to
- $3 x^2+4 x+5$
- $x^3+8 x+2$
- $0$
- $-2$
Solution
Let $\Delta=\left|\begin{array}{lll}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|$
Apply operations $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1$, we get
$
\Delta=\left|\begin{array}{ccc}
x+2 & x+3 & x+5 \\
2 & 3 & 4 \\
6 & 8 & 10
\end{array}\right|
$
Again, apply operation $C_2 \rightarrow C_2-C_1$, $C_3 \rightarrow C_3-C_1$, we get
$
\Delta=\left|\begin{array}{ccc}
x+2 & 1 & 3 \\
2 & 1 & 2 \\
6 & 2 & 4
\end{array}\right|
$
Expand along $R_1$, we get
$
\begin{aligned}
\Delta & =(x+2)(4-4)-1(8-12)+3(4-6) \\
& =0+4+3(-2)=4-6=-2
\end{aligned}
$
Asked in: AP EAMCET 2013
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