Question

Find the value of determinant
$\left| \begin{matrix} 2 & 4 \\ -5 & -1 \\ \end{matrix} \right|$.

Answer 1

Hint: The given determinant is a determinant of order two. To find the value of this determinant we will first multiply diagonal elements ( \[2\] and \[ - 1\] ) then off-diagonal elements ( \[ - 5\] and \[4\] ) . The determinant of this matrix is equal to the product of the diagonal element minus the product of off-diagonal elements.

Complete answer:
For a square matrix, the number of rows is equal to the number of columns, say \[n\] , and also it is called a square matrix of order \[n\] . A square matrix of order \[n\] is also called a n-rowed square matrix. If $A = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}} \\ {{a_{21}}}&{{a_{22}}} \end{array}} \right)$
 is a square matrix of order \[2\] as there are two rows and two columns in this matrix where the elements \[{a_{11}}\] and \[{a_{22}}\] are called the diagonal elements and the line along which they lie is called the principal diagonal or leading diagonal of the matrix, then the expression \[{a_{11}}{a_{22}} - {a_{12}}{a_{21}}\] is defined as the determinant of \[A\] .
That is, $\left| A \right| = $
$\left| \begin{matrix} {a_{11}} & {a_{12}}\\ {a_{21}} & {a_{22}} \\ \end{matrix} \right|$.
\[ \Rightarrow \left| A \right| = {a_{11}}{a_{22}} - {a_{12}}{a_{21}}\]
The determinant of a square matrix of order \[2\] is equal to the product of the diagonal element minus the product of off-diagonal elements.
Now, let $\left| \begin{matrix} 2 & 4 \\ -5 & -1 \\ \end{matrix} \right|$.
Here, we have \[{a_{11}} = 2\] , \[{a_{12}} = 4\] , \[{a_{21}} = - 5\] , \[{a_{22}} = - 1\]
We know, \[\left| A \right| = {a_{11}}{a_{22}} - {a_{12}}{a_{21}}\]
Putting the given values
\[ \Rightarrow \left| A \right| = \left( 2 \right)\left( { - 1} \right) - \left( 4 \right)\left( { - 5} \right)\]
On solving,
\[ \Rightarrow \left| A \right| = \left( { - 2} \right) - \left( { - 20} \right)\]
\[ \Rightarrow \left| A \right| = - 2 + 20\]
On further simplifying,
\[ \Rightarrow \left| A \right| = 18\]
Therefore, the value of determinant $\left| \begin{matrix} 2 & 4 \\ -5 & -1 \\ \end{matrix} \right|$. is \[18\] .

Note:
The given matrix has \[2\] rows \[2\] columns i.e., the number of rows is equal to the number columns, so it is a square matrix. One important point to note is that we can find the determinant of square matrices only. So, in this case since it is a square matrix, we can find its determinant. One of the major mistakes one makes is they simply go on finding the determinant without observing if it is a square matrix or not.