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KCET · Maths · Determinants

If \(A=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right|\) and \(B=\left|\begin{array}{cc}x & 1 \\ 1 & x\end{array}\right|\), then \(\frac{d A}{d x}\) is equal to

  1. A \(3 B+1\)
  2. B \(3 B\)
  3. C \(-3 B\)
  4. D \(1-3 B\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3 B\)

Step-by-step Solution

Detailed explanation

\(\begin{array}{ll}
\text { Given, } & A=\left|\begin{array}{lll}
x & 1 & 1 \\
1 & x & 1 \\
1 & 1 & x
\end{array}\right| \\
\Rightarrow & A=x\left(x^{2}-1\right)-(x-1)+(1-x) \\
\Rightarrow & A=x^{3}-x-x+1+1-x \\
\Rightarrow & A=x^{3}-3 x+2
\end{array}\)
On differentiating w.r.t. \(x\), we get
\(\frac{d A}{d x}=3 x^{2}-3..(i)\)
Also, given \(\quad B=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right|\)
\(\begin{array}{cc}
\Rightarrow & B=x^{2}-1 \\
\Rightarrow & 3 B=3 x^{2}-3...(ii)
\end{array}\)
From Eqs. (i) and (ii), we get
\(\frac{d A}{d x}=3 B\)
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