# Examples On Row And Column Transformations Set-1

Go back to  'Determinants and Matrices'

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 :

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