# Multiplication Of Determinants

Property - 7 :  Multiplication of determinants

Suppose we have two 2 × 2 determinants

${{\Delta }_{1}}=\left| \ \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\\end{matrix}\ \right| \qquad \qquad{{\Delta }_{2}}=\left| \ \begin{matrix} {{\alpha }_{1}} & {{\beta }_{1}} \\ {{\alpha }_{2}} & {{\beta }_{2}} \\\end{matrix}\ \right|$

and we wish to find $${{\Delta }_{1}}{{\Delta }_{2}}.$$ By expansion,

${{\Delta }_{1}}={{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}},\ \ {{\Delta }_{2}}={{\alpha }_{1}}{{\beta }_{2}}-{{\alpha }_{2}}{{\beta }_{1}}$

so

${{\Delta }_{1}}{{\Delta }_{2}}=\left( {{a}_{1}}{{\alpha }_{1}}{{b}_{2}}{{\beta }_{2}}+{{a}_{2}}{{\alpha }_{2}}{{b}_{1}}{{\beta }_{1}} \right)-\left( {{a}_{1}}{{\alpha }_{1}}{{b}_{2}}{{\beta }_{1}}+{{a}_{2}}{{\alpha }_{1}}{{b}_{1}}{{\beta }_{2}} \right) \qquad \qquad \dots (1)$

To find a more enlightening expression, we introduce the term $$\left( {{a}_{1}}{{a}_{2}}{{\alpha }_{1}}{{\alpha }_{2}}+{{b}_{1}}{{b}_{2}}{{\beta }_{1}}{{\beta }_{2}} \right):$$  add and subtract this. Adding this to the first bracket in (1) will generate the factors $$\left( {{a}_{1}}{{\alpha }_{1}}+{{b}_{1}}{{\beta }_{1}} \right)\left( {{a}_{2}}{{\alpha }_{2}}+{{b}_{2}}{{\beta }_{2}} \right).$$ Make sure you verify this. Similarly, when we subtract the same term from (1), it will combine with the second bracket to generate the factors $$\left( {{a}_{1}}{{\alpha }_{2}}+{{b}_{1}}{{\beta }_{2}} \right)\left( {{a}_{2}}{{\alpha }_{1}}+{{b}_{2}}{{\beta }_{1}} \right).$$ Therefore, (1) can be rearranged as

\begin{align} {{\Delta }_{1}}{{\Delta }_{2}}&=\left( {{a}_{1}}{{\alpha }_{1}}+{{b}_{1}}{{\beta }_{1}} \right)\left( {{a}_{2}}{{\alpha }_{2}}+{{b}_{2}}{{\beta }_{2}} \right)-\left( {{a}_{1}}{{\alpha }_{2}}+{{b}_{1}}{{\beta }_{2}} \right)\left( {{a}_{2}}{{\alpha }_{1}}+{{b}_{2}}{{\beta }_{1}} \right) \\ \\ & =\left| \ \begin{matrix} {{a}_{1}}{{\alpha }_{1}}+{{b}_{1}}{{\beta }_{1}} & {{a}_{1}}{{\alpha }_{2}}+{{b}_{1}}{{\beta }_{2}} \\ {{a}_{2}}{{\alpha }_{1}}+{{b}_{2}}{{\beta }_{1}} & {{a}_{2}}{{\alpha }_{2}}+{{b}_{2}}{{\beta }_{2}} \\\end{matrix}\ \right| \\ \end{align}

Look carefully at the term in $${{\Delta }_{1}}{{\Delta }_{2}}$$ at the (1, 1) position. It is a ‘sort of’ product of the row (a1, b1) from $${{\Delta }_{1}}\ \text{and}\ \left( {{\alpha }_{1}},{{\beta }_{1}} \right)\ \text{from}\ {{\Delta }_{2}}:$$

${{a}_{1}}{{\alpha }_{1}}+{{b}_{1}}{{\beta }_{1}}=\left( {{a}_{1}},{{b}_{1}} \right)\times \left( {{\alpha }_{1}},{{\beta }_{1}} \right)$

Similarly, at (1, 2) :

${{a}_{1}}{{\alpha }_{2}}+{{b}_{1}}{{\beta }_{2}}=\left( {{a}_{1}},{{b}_{1}} \right)\times \left( {{\alpha }_{2}},{{\beta }_{2}} \right)$

At (2, 1) :

${{a}_{2}}{{\alpha }_{1}}+{{b}_{2}}{{\beta }_{1}}=\left( {{a}_{2}},{{b}_{2}} \right)\times \left( {{\alpha }_{1}},{{\beta }_{1}} \right)$

And at (2, 2) :

${{a}_{2}}{{\alpha }_{2}}+{{b}_{2}}{{\beta }_{2}}=\left( {{a}_{2}},{{b}_{2}} \right)\times \left( {{\alpha }_{2}},{{\beta }_{2}} \right)$

If the two rows of $${{\Delta }_{1}}$$ are denoted as $${{R}_{1}},{{R}_{2}}$$ and the two rows of $${{\Delta }_{2}}\ \text{as}\ R_{1}^{'},R_{2}^{'}$$, then,

${{\Delta }_{1}}{{\Delta }_{2}}=\left| \begin{gathered} & \ {{R}_{1}} \\ & \ {{R}_{2}}\ \\ \end{gathered} \right|\times \left| \begin{gathered} & \ R_{1}^{'} \\ & \ R_{2}^{'}\ \\ \end{gathered} \right|\ \ =\ \ \left| \begin{matrix} {{R}_{1}}R_{1}^{'} & {{R}_{1}}R_{2}^{'} \\ \ {{R}_{2}}R_{1}^{'} & {{R}_{2}}R_{2}^{'} \\\end{matrix}\ \right|\$

Since a determinant stays the same by interchaning the rows and columns, it should be obvious that similar to ‘row-by-row’ multiplication that we’ve encountered above, we can also have ‘row-by-column’ multiplication and ‘column-by-column’ multiplication. For example,

This is a row-by-column multiplication. Similarly, we’ll have column-by-column multiplication.

What if we have to multiply two 3 × 3 determinants?

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

In terms of rows,

${{\Delta }_{1}}=\left| \ \begin{matrix} {{R}_{1}} \\ {{R}_{2}} \\ {{R}_{3}} \\\end{matrix}\ \right|\ \ {{\Delta }_{2}}=\ \left| \ \begin{matrix} R_{1}^{'} \\ R_{2}^{'} \\ R_{3}^{'} \\\end{matrix}\ \right|$

The multiplication process is analogous to the 2 × 2 case:

${{\Delta }_{1}}{{\Delta }_{2}}=\left| \ \begin{matrix} {{R}_{1}}R_{1}^{'} & {{R}_{1}}R_{2}^{'} & {{R}_{1}}R_{3}^{'} \\ {{R}_{2}}R_{1}^{'} & {{R}_{2}}R_{2}^{'} & {{R}_{2}}R_{3}^{'} \\ {{R}_{3}}R_{1}^{'} & {{R}_{3}}R_{2}^{'} & {{R}_{3}}R_{3}^{'} \\\end{matrix}\ \right|\$

Rows are multiplied similarly as before. For example,

${{R}_{1}}R_{2}^{'}=\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)\ \times \ \left( {{\alpha }_{2}},{{\beta }_{2}},{{\gamma }_{2}} \right)=\left( {{a}_{1}}{{\alpha }_{2}}+{{b}_{1}}{{\beta }_{2}}+{{c}_{1}}{{\gamma }_{2}} \right)$

As in the 2 × 2 case, we can have row-by-column and column-by-column multiplication.

To gain a little practice, let us evaluate the numerical product of two 3 × 3 determinants:

${{\Delta }_{1}}=\left| \begin{matrix} 1 & 3 & \ 2 \\ 4 & 2 & -1 \\ 5 & -1 & 3 \\ \end{matrix} \right| \qquad {{\Delta }_{2}}=\left| \begin{matrix} 2 & -1 & \ 3 \\ 3 & 1 & 4 \\ 0 & 5 & 2 \\\end{matrix} \right|$

Note that  $${{\Delta }_{1}}=-74\ \text{and}\ {{\Delta }_{2}}=15,\ \ \text{so}\ {{\Delta }_{1}}{{\Delta }_{2}}$$ should be –1110.

Now, let us evaluated $${{\Delta }_{1}}{{\Delta }_{2}}$$ through row-by-row multiplication:

We have,

 \begin{align} {{R}_{1}}R_{1}^{'}&=\left( 1\ \ 3\ \ 2 \right)\left( 2\ \ -1\ \ 3 \right) \\ & =5 \\ \end{align} \begin{align} {{R}_{1}}R_{2}^{'}&=\left( 1\ \ 3\ \ 2 \right)\left( 3\ \ 1\ \ 4 \right) \\ & =14 \\ \end{align} \begin{align} {{R}_{1}}R_{3}^{'}&=\left( 1\ \ 3\ \ 2 \right)\left( 0\ \ 5\ \ 2 \right) \\ & =19 \\ \end{align} \begin{align} {{R}_{2}}R_{1}^{'}&=\left( 4\ \ 2\ \ -1 \right)\left( 2\ \ -1\ \ 3 \right) \\ & =3 \\ \end{align} \begin{align} {{R}_{2}}R_{2}^{'}&=\left( 4\ \ 2\ \ -1 \right)\left( 3\ \ 1\ \ 4 \right) \\ & =10 \\ \end{align} \begin{align} {{R}_{2}}R_{3}^{'}&=\left( 4\ \ 2\ \ -1 \right)\left( 0\ \ 5\ \ 2 \right) \\ & =8 \\ \end{align} \begin{align} {{R}_{3}}R_{1}^{'} &=\left( 5\ \ -1\ \ 3 \right)\left( 2\ \ -1\ \ 3 \right) \\ & =20 \\ \end{align} \begin{align} {{R}_{3}}R_{2}^{'} &=\left( 5\ \ -1\ \ 3 \right)\left( 3\ \ 1\ \ 4 \right) \\ & =26 \\ \end{align} \begin{align} \ {{R}_{3}}R_{3}^{'}&=\left( 5\ \ -1\ \ 3 \right)\left( 0\ \ 5\ \ 2 \right) \\ & =1 \\ \end{align}

So,

\begin{align} {{\Delta }_{1}}{{\Delta }_{2}}&=\left| \ \begin{matrix} 5 & 14 & 19 \\ 3 & 10 & 8 \\ 20 & 26 & 1 \\\end{matrix}\ \right|=-990+2198-2318 \\\\ & =-1110 \\ \end{align}

As an exercise, evaluate the same product through row-by-column and column-by-column multiplication.

Example - 14

Evaluate

$\Delta =\left| \ \begin{matrix} {{\left( {{a}_{1}}-{{b}_{1}} \right)}^{2}} & {{\left( {{a}_{1}}-{{b}_{2}} \right)}^{2}} & {{\left( {{a}_{1}}-{{b}_{3}} \right)}^{2}} \\ {{\left( {{a}_{2}}-{{b}_{1}} \right)}^{2}} & {{\left( {{a}_{2}}-{{b}_{2}} \right)}^{2}} & {{\left( {{a}_{2}}-{{b}_{3}} \right)}^{2}} \\ {{\left( {{a}_{3}}-{{b}_{1}} \right)}^{2}} & {{\left( {{a}_{3}}-{{b}_{2}} \right)}^{2}} & {{\left( {{a}_{3}}-{{b}_{3}} \right)}^{2}} \\\end{matrix}\ \right|$

Solution: The trick is in recognizing the fact that $$\Delta$$ can be expressed as a product of two 3 × 3 determinants $${{\Delta }_{1}}\ \text{and}\ {{\Delta }_{2}}$$ . To arrive at this product, we note how elements of row 1 can be expressed as products of rows.

\begin{align} & {{\left( {{a}_{1}}-{{b}_{1}} \right)}^{2}}=a_{1}^{2}-2{{a}_{1}}{{b}_{1}}+b_{1}^{2}=\left( a_{1}^{2}\ \ \ 2{{a}_{1}}\ \ \ 1 \right)\ \times \ \left( 1\ \ -{{b}_{1}}\ \ \ b_{1}^{2} \right) \\ & \\ & {{\left( {{a}_{1}}-{{b}_{2}} \right)}^{2}}=a_{1}^{2}-2{{a}_{1}}{{b}_{2}}+b_{2}^{2}=\left( a_{1}^{2}\ \ \ 2{{a}_{1}}\ \ \ 1 \right)\ \times \ \left( 1\ \ -{{b}_{2}}\ \ \ b_{2}^{2} \right) \\ & \\ & {{\left( {{a}_{1}}-{{b}_{3}} \right)}^{2}}=a_{1}^{2}-2{{a}_{1}}{{b}_{3}}+b_{3}^{2}=\left( a_{1}^{2}\ \ \ 2{{a}_{1}}\ \ \ 1 \right)\ \times \ \left( 1\ \ -{{b}_{3}}\ \ \ b_{3}^{2} \right) \\ \end{align}

This means that we now know how to fill $${{R}_{1}}$$ of  $${{\Delta }_{1}}$$ and the three rows of $${{\Delta }_{2}}$$

$\Delta =\left| \ \begin{matrix} a_{1}^{2} & 2{{a}_{1}} & 1 \\ \leftarrow & ? & \to \\ \leftarrow & ? & \to \\\end{matrix}\ \right|\ \ \times \ \ \left| \ \begin{matrix} 1 & -{{b}_{1}} & b_{1}^{2} \\ 1 & -{{b}_{2}} & b_{2}^{2} \\ 1 & -{{b}_{3}} & b_{3}^{2} \\\end{matrix}\ \right|$

But now, the other rows of $${{\Delta }_{1}}$$ should be evident immediately by symmetry:

$\Delta =\left| \ \begin{matrix} a_{1}^{2} & 2{{a}_{1}} & 1 \\ a_{2}^{2} & 2{{a}_{2}} & 1 \\ a_{3}^{2} & 2{{a}_{3}} & 1 \\\end{matrix}\ \right|\ \ \times \ \,\,\left| \ \begin{matrix} 1 & -{{b}_{1}} & b_{1}^{2} \\ 1 & -{{b}_{2}} & b_{2}^{2} \\ 1 & -{{b}_{3}} & b_{3}^{2} \\\end{matrix}\ \right|\ \$

These two determinants are separately very easy to evaluate; we have already done that in Example - 4:

$\Delta =2\left( {{a}_{1}}-{{a}_{2}} \right)\left( {{a}_{2}}-{{a}_{3}} \right)\left( {{a}_{3}}-{{a}_{1}} \right)\left( {{b}_{1}}-{{b}_{2}} \right)\left( {{b}_{2}}-{{b}_{3}} \right)\left( {{b}_{3}}-{{b}_{1}} \right)$

Example - 15

Evaluate  $$\Delta =\left| \ \begin{matrix} \cos \left( a-p \right) & \cos \left( a-q \right) & \cos \left( a-r \right) \\ \cos \left( b-p \right) & \cos \left( b-q \right) & \cos \left( b-r \right) \\ \cos \left( c-p \right) & \cos \left( c-q \right) & \cos \left( c-r \right) \\\end{matrix}\ \right|$$

Solution: Once again, the determinant can be split into a product. For example,

\begin{align} &\cos \left( a-p \right)=\cos a\cos p+\sin a\sin p \\ & =\left( \cos a\ \ \ \sin a\ \ \ \ 0 \right)\left( \cos p\ \ \ \sin p\ \ \ \ 0 \right) \\ \end{align}

Thus,

\begin{align} \Delta &=\left| \ \begin{matrix} \cos a & \sin a & 0 \\ \cos b & \sin b & 0 \\ \cos c & \sin c & 0 \\\end{matrix}\ \right|\ \ \times \ \ \left| \ \begin{matrix} \cos p & \sin p & 0 \\ \cos q & \sin q & 0 \\ \cos r & \sin r & 0 \\\end{matrix}\ \right| \\\\ & =0 \\ \end{align}

## TRY YOURSELF - II

Q.1      Evaluate

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

Q.2      Evaluate

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

Q.3      If a, b, c are unequal, and if

$\left| \ \begin{matrix} a & {{a}^{2}} & 1+{{a}^{3}} \\ b & {{b}^{2}} & 1+{{b}^{3}} \\ c & {{c}^{2}} & 1+{{c}^{3}} \\\end{matrix}\ \right|=0$

Find the value of abc.

Q.4      Without expanding at any stage, show that

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

Q.5      Evaluate

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