# Row And Column Transformations

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Property - 6 : Row and column transformations

The transformations property is the most widely used property to simplify determinants. This says that:

The value of a determinant does not change when any row (or column) is multiplied by a scalar (a real number) and is then added to or subtracted from any other row (or column).

For example, lets consider

$\Delta =\left| \ \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\end{matrix}\ \right|\$

Lets multiply the 3rd row by $$\lambda$$ and add to R1. This operation can be succintly denoted as $${{R}_{1}}\to {{R}_{1}}+\lambda {{R}_{3}}.$$ Our property tells us that the determinant’s value stays the same. Indeed:

\begin{align} \Delta& =\left| \ \begin{matrix} {{a}_{1}}+\lambda {{c}_{1}} & {{a}_{2}}+\lambda {{c}_{2}} & {{a}_{3}}+\lambda {{c}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\end{matrix}\ \right| \qquad \qquad \qquad \qquad{{R}_{1}}\to {{R}_{1}}+\lambda {{C}_{3}} \\ & =\left| \ \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\end{matrix}\ \right|\ \ +\ \ \left| \begin{matrix} \lambda {{c}_{1}} & \lambda {{c}_{2}} & \lambda {{c}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\end{matrix} \right|\ \ \ \ \ \ \qquad \qquad \quad \left( \text{Splitting along }{{R}_{1}} \right) \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \downarrow \\ & \qquad \qquad \qquad \qquad \qquad \qquad \lambda \left| \ \begin{matrix} {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\end{matrix}\ \right|\ \ \ \qquad \qquad \quad \left(\text{This has two rows identical, so its value is 0 } \right) \\ \end{align}

Let us see how this property helps us in simplifying the evaluation of determinants.

Example - 4

Evaluate

$$\Delta =\left| \ \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\\end{matrix}\ \right|$$

Solution:

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

Theses two transformations could have been done in single step, and later on, we’ll combine such steps.

Now, we expand along C1:

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

Example - 5

Evaluate

$$\Delta =\left| \ \begin{matrix} 1 & a & {{a}^{3}} \\ 1 & b & {{b}^{3}} \\ 1 & c & {{c}^{3}} \\\end{matrix}\ \right|$$

Solution:

$\Delta =\left| \ \begin{matrix} 1 & a & {{a}^{3}} \\ 0 & b-a & {{b}^{3}}-{{a}^{3}} \\ 0 & c-a & {{c}^{3}}-{{a}^{3}} \\\end{matrix}\ \right| \qquad \begin{matrix} {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\ {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\\end{matrix}$

Expanding along C1,

\begin{align} \Delta & =\left( b-a \right)\left( {{c}^{3}}-{{a}^{3}} \right)-\left( {{b}^{3}}-{{a}^{3}} \right)\left( c-a \right) \\ & =\left( b-a \right)\left( c-a \right)\left\{ \left( {{c}^{2}}+{{a}^{2}}+ac \right)-\left( {{b}^{2}}+{{a}^{2}}+ab \right) \right\} \\ & =\left( a-b \right)\left( b-c \right)\left( c-a \right)\left( a+b+c \right) \\ \end{align}

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