Question

If A is a square matrix of order 2 and if $\left| A \right|=-5$ then find the value of $\left| 3A \right|$ .

Answer 1

Now we are given with a square matrix of order 2. Now we know that the determinant of a order 2 matrix is given by $\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \\\ {{a}_{21}} & {{a}_{22}} \\\ \end{matrix} \right|=\left( {{a}_{11}}{{a}_{22}}-{{a}_{21}}{{a}_{12}} \right)$ hence using this we will get the first equation. Now again calculate the determinant of 3A and simplify. Now substituting the value from the first equation we will get the value of determinant of 3A.

Complete step by step answer:
Now let us first understand matrices and determinants.
Now matrices are nothing but rectangular arrays consisting of rows and columns.
Hence we have a matrix A with m rows and n columns as $A=\left[ \begin{matrix} {{a}_{1}} & ... & {{a}_{m}} \\\ \vdots & \ddots & \vdots \\\ {{a}_{n}} & ... & {{a}_{mn}} \\\ \end{matrix} \right]$ .
Now if the matrix has an equal number of rows and columns then we say the matrix is a square matrix.
Hence if a matrix has n rows and n columns the order of the matrix is called n or n × n.
For a square matrix we can define determinant.
Determinant is calculated in a very particular manner.
For matrix of order 2 the determinant is given by,
$\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \\\ {{a}_{21}} & {{a}_{22}} \\\ \end{matrix} \right|=\left( {{a}_{11}}{{a}_{22}}-{{a}_{21}}{{a}_{12}} \right)$ .
Now let us consider a 2 × 2 matrix A such that the determinant of A is – 5.
Hence we have, $\left| A \right|=\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \\\ {{a}_{21}} & {{a}_{22}} \\\ \end{matrix} \right|=\left( {{a}_{11}}{{a}_{22}}-{{a}_{21}}{{a}_{12}} \right)=-5........\left( 1 \right)$
Now consider the matrix 3A.
$3A=\left[ \begin{matrix} 3{{a}_{11}} & 3{{a}_{12}} \\\ 3{{a}_{21}} & 3{{a}_{22}} \\\ \end{matrix} \right]$
Now let us calculate the determinant of this matrix.
$\begin{aligned} & \left| \begin{matrix} 3{{a}_{11}} & 3{{a}_{12}} \\\ 3{{a}_{21}} & 3{{a}_{22}} \\\ \end{matrix} \right|=\left( 3{{a}_{11}}3{{a}_{22}}-3{{a}_{21}}3{{a}_{12}} \right) \\\ & \Rightarrow \left| \begin{matrix} 3{{a}_{11}} & 3{{a}_{12}} \\\ 3{{a}_{21}} & 3{{a}_{22}} \\\ \end{matrix} \right|=\left( 9{{a}_{11}}{{a}_{22}}-9{{a}_{21}}{{a}_{12}} \right) \\\ & \Rightarrow \left| \begin{matrix} 3{{a}_{11}} & 3{{a}_{12}} \\\ 3{{a}_{21}} & 3{{a}_{22}} \\\ \end{matrix} \right|=9\left( {{a}_{11}}{{a}_{22}}-{{a}_{21}}{{a}_{12}} \right) \\\ \end{aligned}$
Now from equation (1) we get,
$\Rightarrow \left| \begin{matrix} 3{{a}_{11}} & 3{{a}_{12}} \\\ 3{{a}_{21}} & 3{{a}_{22}} \\\ \end{matrix} \right|=9\left( -5 \right)=-45$
Hence we get $\left| 3A \right|=-45$.

Note: Now note that we can directly find the value of the determinant of matrix 3A. the determinant is given by ${{3}^{n}}\left| A \right|$ where n is the order of the matrix. In general we have that if r is a scalar and A is a matrix of order n then $\left| rA \right|={{r}^{n}}\left| A \right|$ .