If \( A \), \( B \), and \( \left( \text{adj}(A^{-1}) + \text{adj}(B^{-1}) \right) \) are non-singular matrices of the same order, then the inverse of \[ A \left( \text{adj}(A^{-1}) + \text{adj}(B^{-1}) \right) B \] is equal to:
To find the inverse of the matrix \(C = A \left( \text{adj}(A^{-1}) + \text{adj}(B^{-1}) \right) B\), we need to manipulate the expression using properties of adjugate and inverse matrices. Let's break it down:
The adjugate of an inverse matrix can be expressed in terms of the original matrix:
\[\text{adj}(X^{-1}) = |X|X\] Hence, \(\text{adj}(A^{-1}) = |A|A\) and \(\text{adj}(B^{-1}) = |B|B\).
Substitute these into the given expression:
\[\text{adj}(A^{-1}) + \text{adj}(B^{-1}) = |A|A + |B|B\]
The matrix \(C\) becomes:
\[C = A(|A|A + |B|B)B\]
Distribute the multiplication:
\[C = |A|A^2B + |B|AB^2\]
We need the inverse of this matrix \(C\). Using properties of inverses:
\[C^{-1} = \left(|A|A^2B + |B|AB^2\right)^{-1} = \frac{1}{|C|}\left(\text{adj}\left(|A|A^2B + |B|AB^2\right)\right)\]
Assuming:\(|C| = |A||B|\), \[C^{-1} =\frac{1}{|A||B|}\left(\text{adj}(|A|A^2B + |B|AB^2)\right)\]
Since adjugates are linear over matrix addition:
\[\text{adj}(|A|A^2B + |B|AB^2) = \text{adj}(|A|A^2B) + \text{adj}(|B|AB^2)\]
Applying properties of matrices:
\[\text{adj}(|A|A^2B) = |A|\text{adj}(A^2B)= |A|\text{adj}(A^2)\text{adj}(B)\]\[\text{adj}(|B|AB^2) = |B|\text{adj}(A)\text{adj}(B^2)\]
Thus, the inverse of \(C\) becomes:
\[C^{-1} = \frac{1}{|A||B|} \left( \text{adj}(B) + \text{adj}(A) \right)\]
This matches the answer:
\(\frac{1}{|A||B|} \left( \text{adj}(B) + \text{adj}(A) \right)\)
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: