{"id":5061,"date":"2021-09-05T16:04:02","date_gmt":"2021-09-05T10:34:02","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5061"},"modified":"2021-11-21T16:18:13","modified_gmt":"2021-11-21T10:48:13","slug":"determinants-of-matrix-4x4","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/determinants-of-matrix-4x4\/","title":{"rendered":"Determinants of Matrix 4×4 with Examples"},"content":{"rendered":"
Her you will learn how to find determinants of matrix 4×4 with example.<\/p>\n
Let’s begin –<\/p>\n
To evaluate the determinant of a square matrix of order 4 we follow the same procedure as discussed in previous post in evaluating the determinant of a square matrix of order 3.<\/p>\n
If A = \\(\\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\\\ a_{21} & a_{22} & a_{23} & a_{24} \\\\ a_{31} & a_{32} & a_{33} & a_{34} \\\\ a_{41} & a_{42} & a_{43} & a_{44} \\end{bmatrix}\\) is a square matrix of order 4,<\/p>\n
then | A | = \\(a_{11}\\begin{vmatrix} a_{22} & a_{23} & a_{24} \\\\ a_{32} & a_{33} & a_{34} \\\\\u00a0 a_{42} & a_{43} & a_{44} \\end{vmatrix}\\) – \\(a_{12}\\begin{vmatrix} a_{21} & a_{23} & a_{24} \\\\ a_{31} & a_{33} & a_{34} \\\\\u00a0 a_{41} & a_{43} & a_{44} \\end{vmatrix}\\) + \\(a_{13}\\begin{vmatrix} a_{21} & a_{22} & a_{24} \\\\ a_{31} & a_{32} & a_{34} \\\\\u00a0 a_{41} & a_{42} & a_{44} \\end{vmatrix}\\) – \\(a_{14}\\begin{vmatrix} a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \\\\\u00a0 a_{41} & a_{42} & a_{43} \\end{vmatrix}\\)<\/p>\n
Find the determinant of A = \\(\\begin{bmatrix} 1 & 2 & -1\u00a0 &\u00a0 3 \\\\ 2 & 1 & -2\u00a0 & 3\\\\ 3 & 1 & 2 & 1 \\\\ 1 & -1 & 0 & 2 \\end{bmatrix}\\).<\/p>\n
Solution<\/strong><\/span> : | A | = \\(\\begin{vmatrix} 1 & 2 & -1\u00a0 &\u00a0 3 \\\\ 2 & 1 & -2\u00a0 & 3\\\\ 3 & 1 & 2 & 1 \\\\ 1 & -1 & 0 & 2 \\end{vmatrix}\\)<\/p>\n \\(\\implies\\) | A | = \\(1\\begin{vmatrix} 1 & -2 & 3 \\\\ 1 & 2 & 1 \\\\\u00a0 -1 & 0 & 2 \\end{vmatrix}\\) – \\(2\\begin{vmatrix} 2 & -2 & 3 \\\\ 3 & 2 & 1 \\\\\u00a0 1 & 0 & 2 \\end{vmatrix}\\) + \\((-1)\\begin{vmatrix} 2 & 1 & 3 \\\\ 3 & 1 & 1 \\\\\u00a0 1 & -1 & 2 \\end{vmatrix}\\) – \\(3\\begin{vmatrix} 2 & 1 & -2 \\\\ 3 & 1 & 2 \\\\\u00a0 1 & -1 & 0 \\end{vmatrix}\\)<\/p>\n | A | =\u00a0 (1){\\((1)\\begin{vmatrix} 2 & 1 \\\\\u00a0 0 & 2 \\end{vmatrix}\\) – \\((-2)\\begin{vmatrix} 1 & 1 \\\\ -1 & 2 \\end{vmatrix}\\) + \\((3)\\begin{vmatrix} 1 & 2 \\\\ -1 & 0 \\end{vmatrix}\\)}<\/p>\n – (2){\\((2)\\begin{vmatrix} 2 & 1 \\\\\u00a0 0 & 2 \\end{vmatrix}\\) – \\((-2)\\begin{vmatrix} 3 & 1 \\\\ 1 & 2 \\end{vmatrix}\\) + \\((3)\\begin{vmatrix} 3 & 2 \\\\ 1 & 0 \\end{vmatrix}\\)}<\/p>\n + (-1){\\((2)\\begin{vmatrix} 1 & 1 \\\\\u00a0 -1 & 2 \\end{vmatrix}\\) – \\((1)\\begin{vmatrix} 3 & 1 \\\\ 1 & 2 \\end{vmatrix}\\) + \\((3)\\begin{vmatrix} 3 & 1 \\\\ 1 & -1 \\end{vmatrix}\\)}<\/p>\n – (3){\\((2)\\begin{vmatrix} 1 & 2 \\\\\u00a0 -1 & 0 \\end{vmatrix}\\) – \\((1)\\begin{vmatrix} 3 & 2 \\\\ 1 & 0 \\end{vmatrix}\\) + \\((-2)\\begin{vmatrix} 3 & 1 \\\\ 1 & -1 \\end{vmatrix}\\)}<\/p>\n \\(\\implies\\) | A | = 1{(1)(4 – 0) – (-2)(2 + 1) + (3)(0 + 2)} – 2{(2)(4 – 0) – (-2)(6 – 1) + (3)(0 – 2)} – (-1){(2)(2 + 1) – (1)(6 – 1) + (3)(-3 – 1)} – 3{(2)(0 + 2) – (1)(0 – 2) + (-2)(-3 – 1)}<\/p>\n \\(\\implies\\) | A | = 1(16) – 2(12) + (-1)(-11) – 3(14) = -39<\/p>\n\n\n