Here you will learn how to find the determinant of matrix 2×2 with examples.
Let’s begin –
Determinant of Matrix 2×2
If A = \(\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\) is a square matrix of 2×2,
then \(a_{11}a_{22} – a_{12}a_{21}\) is called the determinant of A.
i.e. | A | = \(\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}\)
= \(a_{11}a_{22} – a_{12}a_{21}\)
Thus, the determinant of a square matrix of order 2 is equal to the product of the diagonal elements minus the product of off-diagonal elements.
Example 1 : find the determinant of \(\begin{vmatrix} 5 & 4 \\ -2 & 3 \end{vmatrix}\).
Solution : Let | A | = \(\begin{vmatrix} 5 & 4 \\ -2 & 3 \end{vmatrix}\)
By definition, we obtain
| A | = ( \(5\times 3\)) – (\(4\times -2\)) = 15 + 8 = 23
Example 2 : find the determinant of \(\begin{vmatrix} sinx & cosx \\ -cosx & sinx \end{vmatrix}\).
Solution : Let | A | = \(\begin{vmatrix} sinx & cosx \\ -cosx & sinx \end{vmatrix}\)
By definition, we obtain
| A | = ( \(sin^2x\)) – (\(-cos^2x\)) = \(sin^2x\) + \(cos^2x\) = 1
Example 3 : find the determinant of \(\begin{vmatrix} x – 1 & 1 \\ x^3 & x^2 + x + 1 \end{vmatrix}\).
Solution : Let | A | = \(\begin{vmatrix} x – 1 & 1 \\ x^3 & x^2 + x + 1 \end{vmatrix}\)
By definition, we obtain
| A | = (x – 1)( \(x^2 + x + 1\)) – (\(x^3\))
= \(x^3 – 1\) – \(x^3\) = -1
Example 4 : find the determinant of \(\begin{vmatrix} x^2 + xy + y^2 & x + y \\ x^2 – xy + y^2 & x – y \end{vmatrix}\).
Solution : Let | A | = \(\begin{vmatrix} x^2 + xy + y^2 & x + y \\ x^2 – xy + y^2 & x – y \end{vmatrix}\)
By definition, we obtain
| A | = ( \(x^2 + xy + y^2\))(x – y) – (\( x^2 – xy + y^2\))(x + y)
= (\(x^3 – y^3\)) – (\(x^3 + y^3\)) = \(-2y^3\)
Example 5 : find the determinant of \(\begin{vmatrix} 1 & log_ba \\ log_ab & 1 \end{vmatrix}\).
Solution : Let | A | = \(\begin{vmatrix} 1 & log_ab \\ log_ab & 1 \end{vmatrix}\)
By definition, we obtain
| A | = 1 – ( \(log_ab \times log_ba\)) = 1 – 1 = 0