考虑一个3x3的矩阵,例如:a1 a2 a3 b1 b2 b3 c1 c2 c3 其行列式的计算公式为:a1(b2c3-b3c2) - a2(b1c3-b3c1) + a3(b1c2-b2c1)。这表示第一行第一个元素a1乘以其余子式(b2c3-b3c2)的值,减去第二行第一个元素a2乘以其余子式(b1c3-b3c1)的值,再加上第三行第一个元素a3乘...
Determinant of a 3x3 matrix Let us now tackle the case of a matrix . There are six possible permutations of the set of the first three natural numbers. We report them below, together with the number of inversions and their sign (deriving them is left as an exercise - revise the lecture ...
Learn to write the determinant of a 3x3 matrix. Using a 3x3 determinant formula and the shortcut method, understand how to find the determinant of...
矩阵的行列式是线性代数中的一个基本概念,对于一个3x3的矩阵,其行列式的计算可以通过以下步骤进行:假设给定矩阵为:```a1 a2 a3 b1 b2 b3 c1 c2 c3 ```其中,a、b、c 都是实数。该矩阵的行列式(determinant),记作 `|A|` 或 `det(A)`,可以通过以下公式计算:```det(A) =...
Even gpt got this result by running matrix via python. So i believe it is just NumPy library bug. Here is code. Got it when i was playing with random generated matrixes created by numpy.random.randint in scale from 1 to 9. Matrix size is 3x3 and was meeting a few of this type ...
假如矩阵为 a1 a2 a3 b1 b2 b3 c1 c2 c3 (a, b, c 均为实数)则该矩阵的行列式等于:a1(b2c3-b3c2) - a2(b1c3-b3c1) + a3(b1c2-b2c1)即 a1*( b2b3c2c3的行列式 ) - a2*( b1b3c1c3的行列式 ) + a3*( b1b2c1c2的行列式 )
1matrix determinant 矩阵的行列式3x3的矩阵,怎么求determinant呢?我要一般形式的答案。比如A B CD E FG H I的determinant是多少?我知道A BC D的= AD-BC还有。若2 -1 43 0 5 = x 44 1 6 5 x求x。我的理解是这个等号意味着两个式子的determinant一样,对么?右边的determinant为 x^2-20 请告诉我左边...
假如矩阵为 a1 a2 a3 b1 b2 b3 c1 c2 c3 (a, b, c 均为实数) 则该矩阵的行列式等于: a1(b2c3-b3c2) - a2(b1c3-b3c1) + a3(b1c2-b2c1) 即 a1*( b2b3c2c3的行列式 ) - a2*( b1b3c1c3的行列式 ) + a3*( b1b2c1c2的行列式 )...
楼主说的是三阶行列式的求法吧,我截个图给你看看 而关于n阶的去看《线性代数》里n阶行列式的求法
To find the determinant of a 2x2 matrix, use the formula |A| = (ad - bc), where A is the matrix: [a b] [c d] How do I find the determinant of a 3x3 matrix? To find the determinant of a 3x3 matrix, use the formula |A| = a(ei - fh) - b(di - fg) + c(dh - eg...