\(|A| A\)
For a non singular matrix \(A\), the adjugate of the inverse of \(A\), denoted as \(\text{Adj}(A^{-1})\), can be derived from the relationship: \[ A \cdot \text{Adj}(A) = |A| I \] where \(I\) is the identity matrix and \(|A|\) is the determinant of \(A\). By applying the properties of the adjugate and inverse matrices, and using the formula for the inverse of a product, we have: \[ \text{Adj}(A^{-1}) = \text{Adj}((\text{Adj}(A) \cdot |A|^{-1} I)^{-1}) = (\text{Adj}(A) \cdot |A|^{-1})^{-1} \] Since \(\text{Adj}(A) \cdot |A|^{-1}\) yields \(A^{-1}\), taking the inverse of this expression gives: \[ (\text{Adj}(A) \cdot |A|^{-1})^{-1} = (\text{Adj}(A))^{-1} \cdot |A| I = (\text{Adj}(A))^{-1} \] This leads us to conclude that \(\text{Adj}(A^{-1}) = (\text{Adj} A)^{-1}\), confirming the correct option.
A, B, C, D are square matrices such that A + B is symmetric, A - B is skew-symmetric, and D is the transpose of C.
If
\[ A = \begin{bmatrix} -1 & 2 & 3 \\ 4 & 3 & -2 \\ 3 & -4 & 5 \end{bmatrix} \]
and
\[ C = \begin{bmatrix} 0 & 1 & -2 \\ 2 & -1 & 0 \\ 0 & 2 & 1 \end{bmatrix} \]
then the matrix \( B + D \) is:
Given matrices \( A \) and \( B \) where:
and the condition:
If matrix \( C \) is defined as:
then the trace of \( C \) is:
Matrix Inverse Sum Calculation
Given the matrix:
A = | 1 2 2 | | 3 2 3 | | 1 1 2 |
The inverse matrix is represented as:
A-1 = | a11 a12 a13 | | a21 a22 a23 | | a31 a32 a33 |
The sum of all elements in A-1 is:
Match the following: