# Examples On Row And Column Transformations Set-2

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Example - 10

Evaluate

$$\Delta =\left| \ \begin{matrix} b+c & a+b & a \\ c+a & b+c & b \\ a+b & c+a & c \\\end{matrix}\ \right|$$

Solution: By $${{C}_{1}}\to {{C}_{1}}+{{C}_{3}}$$ and then factoring out  $$\left( a+b+c \right)$$, we have

\begin{align} \Delta & =\left( a+b+c \right)\left| \ \begin{matrix} 1 & a+b & a \\ 1 & b+c & b \\ 1 & c+a & c \\\end{matrix}\ \right| \\\\ & =\left( a+b+c \right)\left| \ \begin{matrix} 1 & a+b & a \\ 0 & c-a & b-a \\ 0 & c-b & c-a \\\end{matrix}\ \right| \qquad \begin{matrix} {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\ {{R}_{3}}\to {{R}_{3}}-R{{ }_{1}} \\\end{matrix} \\\\ & =\left( a+b+c \right)\left( {{\left( c-a \right)}^{2}}-\left( b-a \right)\left( a-b \right) \right) \\\\ & =\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right) \\\\ & ={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc \\ \end{align}

Example - 11

Evaluate

$\Delta =\left| \ \begin{matrix} ax-by-cz & ay+bx & cx+az \\ ay+bx & by-cz-ax & bz+cy \\ cx+az & bz+cy & cz-ax-by \\\end{matrix}\ \right|$

Solution: Let us try to reduce C1 to a simpler form. By the transformation $${{C}_{1}}\to a{{C}_{1}}+b{{C}_{2}}+c{{C}_{3}},$$ we generate the factor $$\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)$$ in all the terms of $${{C}_{1}}:$$

\begin{align}\Delta &=\frac{1}{a}\left| \ \begin{matrix} x\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right) & ay+bx & cx+az \\ y\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right) & by-cz-ax & bz+cy \\ z\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right) & bz+cy & cz-ax-by \\\end{matrix}\ \right| \qquad \qquad\left(\begin{gathered}\text{Note that there}\\ \text{is a factor of }\frac{1}{a}\end{gathered} \right)\\\\&=\frac{1}{a}\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left| \ \begin{matrix} x & ay+bx & cx+az \\ y & by-cz-ax & bz+cy \\ z & bz+cy & cz-ax-by \\\end{matrix}\ \right| \end{align}

Using a similar artifice, we can simplify R1 now. By $${{R}_{1}}\to x{{R}_{1}}+y{{R}_{2}}+z{{R}_{3}},$$ we have

\begin{align} \Delta &=\frac{1}{ax}\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left| \ \begin{matrix} {{x}^{2}}+{{y}^{2}}+{{z}^{2}} & b\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right) & c\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right) \\ y & by-cz-ax & bz+cy \\ z & bz+cy & cz-ax-by \\\end{matrix}\ \right| \\ \\ & =\frac{1}{ax}\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\left| \ \begin{matrix} 1 & b & c \\ y & by-cz-ax & bz+cy \\ z & bz+cy & cz-ax-by \\\end{matrix}\ \right| \\ \end{align}

Now, we simplify C1 further by the row transformations $${{R}_{2}}\to {{R}_{2}}-y{{R}_{1}}\ \text{and}\ {{R}_{3}}\to {{R}_{3}}-z{{R}_{1}}:$$

$\Delta =\frac{1}{axyz}\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\left| \ \begin{matrix} 1 & b & c \\ 0 & -cz-ax & bz \\ 0 & cy & -ax-by \\\end{matrix}\ \right|$

Finally, we expand along C1 and simplify

$\Delta =0\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\left( ax+by+cz \right)$

Example - 12

Find $$\theta \in \left( 0,\frac{\pi }{2} \right)$$ satisfying

$\left| \ \begin{matrix} 1+{{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 4\sin 4\theta \\ {{\cos }^{2}}\theta & 1+{{\sin }^{2}}\theta & 4\sin 4\theta \\ {{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 1+4\sin 4\theta \\\end{matrix}\ \right|=0$

Solution: Using the transformation $${{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}$$ and factoring out the common factor $$\left( 2+4\sin 4\theta \right)\ \text{from}\ {{C}_{1}}$$, we obtain.

$\left( 2+4\sin 4\theta \right)\left| \ \begin{matrix} 1 & {{\sin }^{2}}\theta & 4\sin 4\theta \\ 1 & 1+{{\sin }^{2}}\theta & 4\sin 4\theta \\ 1 & {{\sin }^{2}}\theta & 1+4\sin 4\theta \\\end{matrix}\ \right|=0$

Using  $${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\ \text{and}\ {{R}_{3}}\to {{R}_{3}}-{{R}_{1}},$$ we have

$2\left( 1+2\sin 4\theta \right)\left| \ \begin{matrix} 1 & {{\sin }^{2}}\theta & 4\sin 4\theta \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\\end{matrix}\ \right|=0$

Explanding along C1, we have

\begin{align} 2\left( 1+2\sin 4\theta \right)&=0 \\ \sin 4\theta &=-\frac{1}{2} \\ \Rightarrow \;\;\quad 4\theta &=\pi +\frac{\pi }{6}=\frac{7\pi }{6} \\ \Rightarrow \qquad \theta &=\frac{7\pi }{24} \\ \end{align}

Example - 13

Evaluate the determinant

$\Delta =\left| \ \begin{matrix} a & b & c & d \\ b & c & d & a \\ c & d & a & b \\ d & a & b & c \\\end{matrix}\ \right|$

Solution: By $${{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}+{{C}_{4}}$$ and then factoring out $$\left( a+b+c+d \right),$$ we have

\begin{align} \Delta &=\left( a+b+c+d \right)\left| \ \begin{matrix} 1 & b & c & d \\ 1 & c & d & a \\ 1 & d & a & b \\ 1 & a & b & c \\\end{matrix}\ \right| \\\\ & =\left( a+b+c+d \right)\left| \ \begin{matrix} 1 & b & c & d \\ 0 & c-b & d-c & a-d \\ 0 & d-b & a-c & b-d \\ 0 & a-b & b-c & c-d \\\end{matrix}\ \right| \qquad \qquad \begin{matrix} {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\ {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\ {{R}_{4}}\to {{R}_{4}}-{{R}_{1}} \\\end{matrix} \\ \end{align}

Expanding along C1 , we have

$\Delta =\left( a+b+c+d \right)\ \left| \ \begin{matrix} c-b & d-c & a-d \\ d-b & a-c & b-d \\ a-b & b-c & c-d \\\end{matrix}\ \right|$

Applying $${{C}_{1}}\to {{C}_{1}}+{{C}_{3}}$$ and factoring out $$\left( a-b+c-d \right)$$, we have

\begin{align} \Delta &=\left( a+b+c+d \right)\left( a-b+c-d \right)\left| \ \begin{matrix} 1 & d-c & a-d \\ 0 & a-c & b-d \\ 1 & b-c & c-d \\\end{matrix}\ \right| \\ \\ & =\left( a+b+c+d \right)\left( a-b+c-d \right)\left| \ \begin{matrix} 1 & d-c & a-d \\ 0 & a-c & b-d \\ 0 & b-d & c-a \\\end{matrix}\ \right| \qquad \qquad {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\ \end{align}

Finally, expanding the 3 × 3 determinant along C1, we have

\begin{align} \Delta &=\left( a+b+c+d \right)\left( a-b+c-d \right)\left\{ -{{\left( a-c \right)}^{2}}-{{\left( b-d \right)}^{2}} \right\} \\ & =-\left( a+b+c+d \right)\left( a-b+c-d \right)\left( {{\left( a-c \right)}^{2}}+{{\left( b-d \right)}^{2}} \right) \\ \end{align}