Question

If ${△}_1=\left|\begin{array}{ccc}x& a& b\\ b& x& a\\ a& b& x\end{array}\right|$ and ${△}_2=\left|\begin{array}{cc}x& b\\ a& x\end{array}\right|$ are the given determinants, then

• A. ${∆}_1=3{\left({∆}_2\right)}^2$
• B. $\frac{d}{dx}{∆}_1=3\left({∆}_2\right)$
• C. $\frac{d}{dx}{∆}_1=3{\left({∆}_2\right)}^2$
• D. ${∆}_1=3{\left({∆}_2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Answer 1

The correct option is B

$\frac{d}{dx}{∆}_1=3\left({∆}_2\right)$

Step 1. Find the value of determinant ${△}_1$:

$\begin{array}{rcl}{△}_1& =& \left|\begin{array}{ccc}x& a& b\\ b& x& a\\ a& b& x\end{array}\right|\\ & =& x(x^2-ab)–a(xb–a^2)+b(b^2–ax)\\ & =& x^3–axb–axb+a^3+b^3–abx\\ & =& x^3–3axb+a^3+b^3\end{array}$

Differentiate it with respect to $x$:

$\begin{array}{rcl}\frac{d}{dx}{∆}_1& =& 3x^2–3ab\\ & =& 3\left(x^2-ab\right)\end{array}$

Step 2. Find the value of determinant ${△}_2$:

$\begin{array}{rcl}{△}_2& =& \left|\begin{array}{cc}x& b\\ a& x\end{array}\right|\\ & =& x^2-ab\end{array}$

$\begin{array}{rcl}\therefore \frac{d}{dx}{∆}_1& =& 3x^2–3ab\\ & =& 3\left(x^2-ab\right)\\ & =& \mathbf{3}{\mathbf{△}}_{\mathbf{2}}\end{array}$

Hence, Option ‘B’ is Correct.