Question

Using properties of determinants, prove that
$\left| \begin{aligned} & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3x \\\ & 1+3y\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\\ & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\\ \end{aligned} \right|\,=9\left( 3xyz+xy+yz+zx \right)$

Answer 1

Hint: Consider the given determinant given in the left-hand side of the expression. Then apply the property in the Row 2 i.e. ${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}$. Replace ${{R}_{2}}$ by $\left( {{R}_{2}}-{{R}_{1}} \right)$. Then Apply ${{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$ ie. Replace Row 3 by Row 3 minus Row 1. After that, expand the determinant along Row 1 to get the required expression.

Complete step-by-step answer:

First we will consider the given determinant in the left-hand side of the expression,

We have given determinant $\left| \begin{aligned} & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3x \\\ & 1+3y\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\\ & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\\ \end{aligned} \right|\,$ .
Now, replace row 2 by row 2 minus row 1 i.e. replace ${{R}_{2}}$ by ${{R}_{2}}-{{R}_{1}}$ .
Hence on applying ${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}$on the above given determinant, we have:

& 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3x \\\ & \left( 1+3y\, \right)-1\,\,\,\,\,\,\,\,\,\,\,\left( \,1-1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \,1-\left( \,1+3x \right) \right) \\\ & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\\ \end{aligned} \right|\,$$ $$\Rightarrow \left| \begin{aligned} & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3x \\\ & 1+3y-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1-1-3x \\\ & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\\ \end{aligned} \right|\,$$ $$\Rightarrow \left| \begin{aligned} & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3x \\\ & 3y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3x \\\ & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\\ \end{aligned} \right|.................(i)$$ Now, replace Row 3 by Row 3 minus Row 1 i.e. replace ${{R}_{3}}$ by${{R}_{3}}-{{R}_{1}}$ Hence, on applying ${{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$ on the determinant given in equation (i), we have $\left| \begin{aligned} & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3x \\\ & 3y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3x \\\ & \left( 1-1\, \right)\,\,\,\,\,\,\,\,\left( \,1+3z \right)-1\,\,\,\,\,\,\,\,1-\left( 1+3x \right) \\\ \end{aligned} \right|\,$ $\Rightarrow \left| \begin{aligned} & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3x \\\ & 3y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3x \\\ & 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3z-1\,\,\,\,\,\,\,\,\,\,\,\,1-1-3x \\\ \end{aligned} \right|\,$ $\Rightarrow \left| \begin{aligned} & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1+3x \\\ & 3y\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3x \\\ & 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,3z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3x \\\ \end{aligned} \right|\,.........................(ii)$ Now, expand the above determinant along Row 1. On expanding the determinant given in equation (ii), along ${{R}_{1}}$ , we have- $\begin{aligned} & \Delta =1\left[ 0\left( -3x \right)-\left( -3x \right)\left( 3z \right) \right]-1\left[ 3y\left( -3x \right)-0\left( -3x \right) \right] \\\ & \,\,\,\,\,\,+\left( 1+3x \right)\left[ \left( 3y \right)\left( 3x \right)-0\left( 0 \right) \right] \\\ & \Delta =1\left[ 0-\left( -9xz \right) \right]-1\left[ -9xy-0 \right]+\left( 1+3x \right)\left( 9yz \right) \\\ & \Delta =1\left[ +9xz \right]-1\left[ -9xy \right]+\left( 1+3x \right)\left( 9yz \right) \\\ \end{aligned}$ On expanding the above expression, we get, $\begin{aligned} & \Delta =9xz+9xy+1\left( 9yz \right)+3x\left( 9yz \right) \\\ & \Delta =9xz+9xy+9yz+27xyz \\\ \end{aligned}$ Taking 9 common in the above expression, $$\begin{aligned} & \Delta =9\left( xz+xy+yz+3xyz \right) \\\ & \Delta =9\left( 3xyz+xy+yz+zx \right)=RHS \\\ \end{aligned}$$ Note: Alternative method: Students can also solve, this question by applying ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\,\,\And \,{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$ on the given determinant and then by expanding the determinant along ${{R}_{1}}$ . $LHS=\left| \begin{aligned} & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3x \\\ & 1+3y\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\\ & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+3z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\\ \end{aligned} \right|\,$ Applying, ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\,\,\And \,{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$ on the above given determinant, we have: $\left| \begin{aligned} & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3x \\\ & 1+3y\,\,\,\,\,\,-3y\,\,\,\,\,\,\,\,\,\,-3y \\\ & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\\ \end{aligned} \right|\,$ Expanding the above determinant along ${{R}_{1}}$, we have $\Rightarrow 1\times \left( 9yz \right)+3x\left( 3z+9yz+3y \right)$ $\Rightarrow 9\left( 3xyz+xy+yz+zx \right)=RHS$