{"id":5070,"date":"2021-09-05T16:34:54","date_gmt":"2021-09-05T11:04:54","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5070"},"modified":"2021-11-21T16:06:09","modified_gmt":"2021-11-21T10:36:09","slug":"determinant-of-matrix-2x2","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/determinant-of-matrix-2x2\/","title":{"rendered":"Determinant of Matrix 2×2 with Examples"},"content":{"rendered":"
Here you will learn how to find the determinant of matrix 2×2 with examples.<\/p>\n
Let’s begin –<\/p>\n
If A = \\(\\begin{bmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{bmatrix}\\) is a square matrix of 2×2,<\/p>\n
\nthen \\(a_{11}a_{22} – a_{12}a_{21}\\) is called the determinant of A.<\/p>\n
i.e. | A | = \\(\\begin{vmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{vmatrix}\\)<\/p>\n
= \\(a_{11}a_{22} – a_{12}a_{21}\\)<\/p>\n<\/blockquote>\n
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.<\/p>\n
Example 1<\/span><\/strong> : find the determinant of \\(\\begin{vmatrix} 5 & 4 \\\\ -2 & 3 \\end{vmatrix}\\).<\/p>\n
Solution <\/span><\/strong>:\u00a0Let | A | = \\(\\begin{vmatrix} 5 & 4 \\\\ -2 & 3 \\end{vmatrix}\\)<\/p>\n
By definition, we obtain<\/p>\n
| A | = ( \\(5\\times 3\\)) – (\\(4\\times -2\\)) = 15 + 8 = 23<\/p>\n
Example 2<\/span><\/strong> : find the determinant of \\(\\begin{vmatrix} sinx & cosx \\\\ -cosx & sinx \\end{vmatrix}\\).<\/p>\n
Solution <\/span><\/strong>:\u00a0\u00a0Let | A | = \\(\\begin{vmatrix} sinx & cosx \\\\ -cosx & sinx \\end{vmatrix}\\)<\/p>\n
By definition, we obtain<\/p>\n
| A | = ( \\(sin^2x\\)) – (\\(-cos^2x\\)) = \\(sin^2x\\) + \\(cos^2x\\) = 1<\/p>\n
Example 3<\/span><\/strong> : find the determinant of \\(\\begin{vmatrix} x – 1 & 1 \\\\ x^3 & x^2 + x + 1 \\end{vmatrix}\\).<\/p>\n
Solution <\/span><\/strong>: Let | A | = \\(\\begin{vmatrix} x – 1 & 1 \\\\ x^3 & x^2 + x + 1 \\end{vmatrix}\\)<\/p>\n
By definition, we obtain<\/p>\n
| A |\u00a0 = (x – 1)( \\(x^2 + x + 1\\)) – (\\(x^3\\))<\/p>\n
= \\(x^3 – 1\\) – \\(x^3\\) = -1<\/p>\n
Example 4<\/span><\/strong> : find the determinant of \\(\\begin{vmatrix} x^2 + xy + y^2 & x + y \\\\ x^2 – xy + y^2 & x – y \\end{vmatrix}\\).<\/p>\n
Solution <\/span><\/strong>: Let | A | = \\(\\begin{vmatrix} x^2 + xy + y^2 & x + y \\\\ x^2 – xy + y^2 & x – y \\end{vmatrix}\\)<\/p>\n
By definition, we obtain<\/p>\n
| A |\u00a0 = ( \\(x^2 + xy + y^2\\))(x – y) – (\\( x^2 – xy + y^2\\))(x + y)<\/p>\n
= (\\(x^3 – y^3\\)) – (\\(x^3 + y^3\\)) = \\(-2y^3\\)<\/p>\n
Example 5<\/span><\/strong> : find the determinant of \\(\\begin{vmatrix} 1 & log_ba \\\\ log_ab & 1 \\end{vmatrix}\\).<\/p>\n
Solution <\/span><\/strong>: Let | A | = \\(\\begin{vmatrix} 1 & log_ab \\\\ log_ab & 1 \\end{vmatrix}\\)<\/p>\n
By definition, we obtain<\/p>\n
| A |\u00a0 = 1 – ( \\(log_ab \\times log_ba\\)) = 1 – 1 = 0<\/p>\n\n\n