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

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

Download practice questions along with solutions for FREE:
Determinants and Matrices
grade 11 | Answers Set 2
Determinants and Matrices
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Determinants and Matrices
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Download practice questions along with solutions for FREE:
Determinants and Matrices
grade 11 | Answers Set 2
Determinants and Matrices
grade 11 | Questions Set 1
Determinants and Matrices
grade 11 | Answers Set 1
Determinants and Matrices
grade 11 | Questions Set 2
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