2019年4月21日 · Given: A and B be (3 × 3) matrices with det A = 4 and det B = 3. As we know that, if A and B are square matrices of order n, then det (m × AB) = m n × det (A) × det (B) where m ∈ R is a scalar. ⇒ det (2AB) = 2 3 × det (A) × det (B) = 8 × 4 × 3 = 96.

For large matrices, the determinant can be calculated using a method called expansion by minors. This involves expanding the determinant along one of the rows or columns and using the …

2022年9月17日 · Let A A be an n × n n × n matrix and let B B be a matrix which results from switching two rows of A. A. Then det(B) = − det(A). det (B) = − det (A). When we switch two rows of a matrix, the determinant is multiplied by −1 − 1. Consider the following example. Let A = [1 3 2 4] A = [1 2 3 4] and let B = [3 1 4 2] B = [3 4 1 2].

对于方阵A和B，有 det(AB) = det(A) * det(B) 。 即方阵乘积的行列式等于各个矩阵行列式的乘积。 零元素： 如果方阵A中存在一行（或一列）全为零，则det(A) = 0。即方阵的行列式为零，当且仅当矩阵中存在一行（或一列）全为零。

determinant inequality $\det(A^2+AB+B^2)\geq\det(AB-BA)$ 4 Prove $\det \left[\begin{smallmatrix} A&B\\\\C&D\\ \end{smallmatrix}\right] =\det(AD-BC)$ for $A,B,C,D$ upper triangular complex matrices

维空间长成的平行六面体(n-dimensional parallelepiped) 的有向体积。 定理1. 如果A 存在一个列向量是0 ,则det(A) = 0。 如果A 存在两列向量相同,则det(A) = 0。 如果A 存在一列是其他列的倍数,则det(A) = 0。 定理2. 如果rank(A) < n, 则det(A) = 0。 定理3....

The sign of $\det(A)$ tells you whether $A$ preserves or reverses orientation. Examples: Let $n=2$ so we are dealing with areas in the plane. If $A$ is a rotation matrix, then its effect on the plane is a rotation. $\det(A)$ is positive 1 because $A$ actually preserves all areas (so absolute value 1) and preserves orientation (so positive).

If A is an n × n matrix, then $det(uA)$ = $u^ndetA$ for any number u. So for this case if i was given the det(uA) would i just divide the 2, so the answer is 3? linear-algebra

公式中，det(AB)代表矩阵A与矩阵B相乘后的结果矩阵的行列式。 简而言之，此公式指出：两个矩阵相乘后的行列式等于这两个矩阵各自行列式的乘积。 这一结论是线性代数中一个重要的性质，其直观解释在于矩阵的线性变换性质。

Let A and B be a 3x3 matrices with det(A) = 4 and det (B) = 5. Find the Value of (a) det(AB) (b) det(3A) (c) det(2AB) (d) (A^-1 * B) There are 3 steps to solve this one.