KCET · Maths · Determinants
If \(a x^{4}+b x^{3}+c x^{2}+d x+e\) \(=\left|\begin{array}{ccc}x^{3}+3 x & x-1 & x+3 \\ x+1 & -2 x & x-4 \\ x-3 & x+4 & 3 x\end{array}\right|\), then e is equal to
- A 1
- B 0
- C 2
- D \(-1\)
Answer & Solution
Correct Answer
(B) 0
Step-by-step Solution
Detailed explanation
Given, \(a x^{4}+b x^{3}+c x^{2}+d x+e\)
\[
=\left|\begin{array}{ccc}
x^{3}+3 x & x-1 & x+3 \\
x+1 & -2 x & x-4 \\
x-3 & x+4 & 3 x
\end{array}\right|
\]
On putting \(x=0\) both sides, we get
\[
e=\left|\begin{array}{rrr}
0 & -1 & 3 \\
1 & 0 & -4 \\
-3 & 4 & 0
\end{array}\right|
\]
Expanding along first row, we get
\[
\begin{aligned}
&=0(0+16)-(-1)(0-12)+3(4-0) \\
&=0-12+12=0
\end{aligned}
\]
\[
=\left|\begin{array}{ccc}
x^{3}+3 x & x-1 & x+3 \\
x+1 & -2 x & x-4 \\
x-3 & x+4 & 3 x
\end{array}\right|
\]
On putting \(x=0\) both sides, we get
\[
e=\left|\begin{array}{rrr}
0 & -1 & 3 \\
1 & 0 & -4 \\
-3 & 4 & 0
\end{array}\right|
\]
Expanding along first row, we get
\[
\begin{aligned}
&=0(0+16)-(-1)(0-12)+3(4-0) \\
&=0-12+12=0
\end{aligned}
\]
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