Mark the tick against the correct answer in the following: $\left|\begin{array}{lll} 1^{2} & 2^{2} & 3^{2} \\ 2^{2} & 3^{2} & 4^{2} \\ 3^{2} & 4^{2} & 5^{2} \end{array}\right|=?$ A. 8 B. $-8$ C. 16 D. 142

Mark the tick against the correct answer in the following: $\left|\begin{array}{lll} 1^{2} & 2^{2} & 3^{2} \\ 2^{2} & 3^{2} & 4^{2} \\ 3^{2} & 4^{2} & 5^{2} \end{array}\right|=?$
A. 8
B. $-8$
C. 16
D. 142

CBSE | Class 12 | Determinants | Maths | RS Aggarwal

Solution:

Option(A)
To find: Value of $\left|\begin{array}{lll}1^{2} & 2^{2} & 3^{2} \\ 2^{2} & 3^{2} & 4^{2} \\ 3^{2} & 4^{2} & 5^{2}\end{array}\right|$
We have, $\left|\begin{array}{lll}1^{2} & 2^{2} & 3^{2} \\ 2^{2} & 3^{2} & 4^{2} \\ 3^{2} & 4^{2} & 5^{2}\end{array}\right|$
$\Rightarrow\left|\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right|$
Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}$
$\Rightarrow\left|\begin{array}{ccc} 8 & 12 & 16 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right|$
Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$
$\Rightarrow\left|\begin{array}{ccc} 8 & 12 & 16 \\ 4 & 3 & 0 \\ 9 & 16 & 25 \end{array}\right|$
Taking 4 common from $R_{1}$
$\Rightarrow 4\left|\begin{array}{ccc} 2 & 3 & 4 \\ 4 & 3 & 0 \\ 9 & 16 & 25 \end{array}\right|$
Applying $\mathbf{R}_{1} \rightarrow \mathbf{R}_{1}-\mathbf{R}_{2}$
$\Rightarrow 4\left|\begin{array}{ccc} -2 & 0 & 4 \\ 4 & 3 & 0 \\ 9 & 16 & 25 \end{array}\right|$
Taking $-2$ common from $\mathrm{R}_{1}$
$\Rightarrow(4)(-2)\left|\begin{array}{ccc} 1 & 0 & -2 \\ 4 & 3 & 0 \\ 9 & 16 & 25 \end{array}\right|$
Applying $\mathbf{R}_{1} \rightarrow 9 \mathbf{R}_{1}$
$\Rightarrow \frac{-8}{9}\left|\begin{array}{ccc} 9 & 0 & -18 \\ 4 & 3 & 0 \\ 9 & 16 & 25 \end{array}\right|$
Applying $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}$
$\Rightarrow \frac{-8}{9}\left|\begin{array}{ccc} 9 & 0 & -18 \\ 4 & 3 & 0 \\ 0 & 16 & 43 \end{array}\right|$
Taking 9 common from $R_{1}$
$\Rightarrow-8\left|\begin{array}{ccc} 1 & 0 & -2 \\ 4 & 3 & 0 \\ 0 & 16 & 43 \end{array}\right|$
Expanding along $R_{1}$
$\Rightarrow-8[1[(3)(43)-(16)(0)]-0[(4)(43)-(0)(0)]-2[(4)(16)-(3)(0)]]$
$\begin{array}{l} \Rightarrow-8[[(129)-(0)]-2[(64)-(0)]] \\ \Rightarrow-8[129-128] \\ \Rightarrow-8 \end{array}$