KCET · Maths · Determinants
The constant term of the polynomial \(\left|\begin{array}{ccc}x+3 & x & x+2 \\ x & x+1 & x-1 \\ x+2 & 2 x & 3 x+1\end{array}\right|\) is
- A 0
- B 2
- C \(-1\)
- D 1
Answer & Solution
Correct Answer
(C) \(-1\)
Step-by-step Solution
Detailed explanation
\[
\left|\begin{array}{ccc}
x+3 & x & x+2 \\
x & x+1 & x-1 \\
x+2 & 2 x & 3 x+1
\end{array}\right|=f(x)
\]
Applying \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\) and \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\)
\[
f(x)=\left|\begin{array}{ccc}
x+3 & x & x+2 \\
-3 & 1 & -3 \\
-1 & x & 2 x-1
\end{array}\right|
\]
Applying \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3}\) and \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}\)
\[
f(x)=\left|\begin{array}{ccc}
1 & -2 & x+2 \\
0 & 4 & -3 \\
-2 x & 1-x & 2 x-1
\end{array}\right|
\]
Expand w.r.t. ' \(\mathrm{C}_{1}\),
\[
f(x)=[4(2 x-1)+3(1-x)]
\]
\[
+(-2 \mathrm{x})[6-4(\mathrm{x}+2)]
\]
\(f(x)=[8 x-4+3-3 x]+[-2 x][-4 x-2]\)
\(f(x)=(5 x-1)+\left(8 x^{2}+4 x\right)\)
\[
f(x)=8 x^{2}+9 x-1
\]
Hence, the constant term of quadratic equation is \(-1\).
\left|\begin{array}{ccc}
x+3 & x & x+2 \\
x & x+1 & x-1 \\
x+2 & 2 x & 3 x+1
\end{array}\right|=f(x)
\]
Applying \(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\) and \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\)
\[
f(x)=\left|\begin{array}{ccc}
x+3 & x & x+2 \\
-3 & 1 & -3 \\
-1 & x & 2 x-1
\end{array}\right|
\]
Applying \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3}\) and \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}\)
\[
f(x)=\left|\begin{array}{ccc}
1 & -2 & x+2 \\
0 & 4 & -3 \\
-2 x & 1-x & 2 x-1
\end{array}\right|
\]
Expand w.r.t. ' \(\mathrm{C}_{1}\),
\[
f(x)=[4(2 x-1)+3(1-x)]
\]
\[
+(-2 \mathrm{x})[6-4(\mathrm{x}+2)]
\]
\(f(x)=[8 x-4+3-3 x]+[-2 x][-4 x-2]\)
\(f(x)=(5 x-1)+\left(8 x^{2}+4 x\right)\)
\[
f(x)=8 x^{2}+9 x-1
\]
Hence, the constant term of quadratic equation is \(-1\).
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