Examples On Row And Column Transformations Set-1

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Example - 6 

 Evaluate

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

Solution:

\[\begin{align}   \Delta &=\left| \ \begin{matrix}   a+b+c & b+c+a & c+a+b  \\   b & c & a  \\   c & a & b  \\\end{matrix}\  \right| \qquad \qquad  \begin{matrix}   {{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}  \\   \left( \text{Justify this transformation} \right)  \\\end{matrix} \\  &  \\  & =\left( a+b+c \right)\ \left| \ \begin{matrix}   1 & 1 & 1  \\   b & c & a  \\   c & a & b  \\\end{matrix}\  \right| \\  &  \\  & =\left( a+b+c \right)\ \left| \ \begin{matrix}   1 & 0 & 0  \\   b & c-b & a-b  \\   c & a-c & b-c  \\\end{matrix}\  \right|  \qquad \qquad \begin{matrix}   {{C}_{2}}\to {{C}_{2}}-{{C}_{1}}  \\   {{C}_{3}}\to {{C}_{3}}-{{C}_{1}}  \\\end{matrix} \\ \end{align}\]

Expanding along R1,

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

Note that if a, b, c are positive and unequal, then this determinant is always positive, an important fact that should be committed to memory.

Example - 7

Evaluate

 \(\Delta =\left| \ \begin{matrix}   a & bc & abc  \\   b & ca & abc  \\   c & ab & abc  \\\end{matrix}\  \right|\ \)

Solution:

\[\begin{align}   \Delta &=\frac{1}{abc}\left| \ \begin{matrix}   {{a}^{2}} & abc & {{a}^{2}}bc  \\   {{b}^{2}} & abc & a{{b}^{2}}c  \\   {{c}^{2}} & abc & ab{{c}^{2}}  \\\end{matrix}\  \right| \qquad \begin{matrix}   {{R}_{1}}\to a{{R}_{1}}  \\   {{R}_{2}}\to b{{R}_{2}}  \\   {{R}_{3}}\to c{{R}_{3}}  \\\end{matrix}\\\ \\  & =\frac{{{\left( abc \right)}^{2}}}{abc}\left| \ \begin{matrix}   {{a}^{2}} & 1 & a  \\   {{b}^{2}} & 1 & b  \\   {{c}^{2}} & 1 & c  \\\end{matrix}\  \right| \\ \\ & =abc{{\left( -1 \right)}^{2}}\left| \ \begin{matrix}   1 & a & {{a}^{2}}  \\   1 & b & {{b}^{2}}  \\   1 & c & {{c}^{2}}  \\\end{matrix}\  \right| \qquad\begin{matrix}   {{C}_{1}}\leftrightarrow {{C}_{2}}  \\   \text{and then}  \\   {{C}_{2}}\leftrightarrow {{C}_{3}}  \\\end{matrix} \\ \\& =abc\left( a-b \right)\left( b-c \right)\left( c-a \right) \\ \end{align}\]

In the last step, we have used the result from Example - 4.

Example - 8

Let  \(a,b,c\ \text{be}\ \text{the}\ {{p}^{\text{th}}},{{q}^{\text{th}}},{{r}^{\text{th}}}\) terms respectively of a G.P. Evaluate

\[\Delta =\left| \ \begin{matrix}   \log a & p & 1  \\   \log b & q & 1  \\   \log c & r & 1  \\\end{matrix}\  \right|\]

Solution: Let the first term of the G.P. be f and the common ratio be t then \(a=f\ {{t}^{p-1}}\ \ \ \Rightarrow \ \ \log a=\log f+\left( p-1 \right)\log t\), . Similar raltions can be written for \(\log b\ \text{and}\ \log c\) :

\[\begin{align}   \Delta &=\left| \ \begin{matrix}   \log f+\left( p-1 \right)\log t & p & 1  \\   \log f+\left( q-1 \right)\log t & q & 1  \\   \log f+\left( r-1 \right)\log t & r & 1  \\\end{matrix}\  \right| \\\\  & =\left| \ \begin{matrix}   \left( p-1 \right)\log t & p & 1  \\   \left( q-1 \right)\log t & q & 1  \\   \left( r-1 \right)\log t & r & 1  \\\end{matrix}\  \right| \qquad \qquad  {{C}_{1}}\to {{C}_{1}}-\left( \log t \right){{C}_{3}} \\\\  & =\log t\left| \ \begin{matrix}   p-1 & p & 1  \\   q-1 & q & 1  \\   r-1 & r & 1  \\\end{matrix}\  \right| \\\\  & =\log t\left| \ \begin{matrix}   p & p & 1  \\   q & q & 1  \\   r & r & 1  \\\end{matrix}\  \right| \qquad \qquad \qquad  {{C}_{1}}\to {{C}_{1}}+{{C}_{3}} \\\\  & =0 \\ \end{align}\]

Example – 9

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

Solution: The general approach in evaluating a determinant is to obtain as many zeroes as possible in a given row (or column) and then expanding the determinant along that row (or column). Note carefully the appropriate steps in this particular case:

\[\begin{align}   \Delta & =\left| \ \begin{matrix}   0 & c & b  \\   -2a & c+a & a  \\   -2a & a & a+b  \\\end{matrix}\  \right| \qquad \qquad \begin{matrix}   {{C}_{1}}\to {{C}_{1}}-\left( {{C}_{ 2}}+{{C}_{3}} \right)  \\   \left( \text{Justify this transformation} \right)  \\\end{matrix} \\\\  & =-2a\left| \ \begin{matrix}   0 & c & b  \\   1 & c+a & a  \\   1 & a & a+b  \\\end{matrix}\  \right| \\\\  & =-2a\left| \ \begin{matrix}   0 & c & b  \\   0 & c & -b  \\   1 & a & a+b  \\\end{matrix}\ \right| \qquad \qquad \qquad{{R}_{2}}\to {{R}_{2}}-{{R}_{3}} \\ \end{align}\]

Now, we expand along C1 :

\[\Delta =-2a\left( -bc-bc \right)=4abc\]

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Determinants and Matrices
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Determinants and Matrices
grade 11 | Answers Set 2
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Determinants and Matrices
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