Question
Using properties of determinants, prove that $\begin{vmatrix}\text{a}^2+2\text{a} & 2\text{a}+1 & 1 \\2\text{a}+1 & \text{a}+2 & 1\\3 & 3 & 1 \end{vmatrix}=(\text{a}-1)^3.$

Answer

$\text{LHS}=\begin{vmatrix}\text{a}^2+2\text{a} & 2\text{a}+1 & 1 \\2\text{a}+1 & \text{a}+2 & 1\\3 & 3 & 1 \end{vmatrix}$
$\text{R}_2\rightarrow\text{R}_2-\text{R}_1,\text{R}_3\rightarrow\text{R}_3-\text{R}_1$
$=\begin{vmatrix}\text{a}^2+2\text{a} & 2\text{a}+1 & 1 \\1-\text{a}^2 & -\text{a}+1 & 0\\3-\text{a}^2-2\text{a} & 3-2\text{a}-1 & 0 \end{vmatrix}$
$=\begin{vmatrix}\text{a}^2+2\text{a} & 2\text{a}+1 & 1 \\1-\text{a}^2 & 1-\text{a} & 0\\3-\text{a}^2-2\text{a} & 2-2\text{a} & 0 \end{vmatrix}$
Expanding along $C_3$
$=1\big[(1-\text{a}^2)(2-2\text{a})-(1-\text{a})(3-\text{a}^2-2\text{a})\big]$
$=2(1-\text{a})(1-\text{a})(1+\text{a})-(1-\text{a})(3-\text{a}^2-2\text{a})$
$=(1-\text{a})\big[2(1-\text{a}^2)-3+\text{a}^2+2\text{a}\big]$
$=(1-\text{a})(2\text{a}-\text{a}^2-1)$
$=(\text{a}-1)^3$
$=\text{RHS}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free