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:
\(\frac{1}{|A||B|} \left( \text{adj}(B) + \text{adj}(A) \right)\)
Match List-I with List-II
List-I (Matrix) | List-II (Inverse of the Matrix) |
---|---|
(A) \(\begin{bmatrix} 1 & 7 \\ 4 & -2 \end{bmatrix}\) | (I) \(\begin{bmatrix} \tfrac{2}{15} & \tfrac{1}{10} \\[6pt] -\tfrac{1}{15} & \tfrac{1}{5} \end{bmatrix}\) |
(B) \(\begin{bmatrix} 6 & -3 \\ 2 & 4 \end{bmatrix}\) | (II) \(\begin{bmatrix} \tfrac{1}{5} & -\tfrac{2}{15} \\[6pt] -\tfrac{1}{10} & \tfrac{7}{30} \end{bmatrix}\) |
(C) \(\begin{bmatrix} 5 & 2 \\ -5 & 4 \end{bmatrix}\) | (III) \(\begin{bmatrix} \tfrac{1}{15} & \tfrac{7}{30} \\[6pt] \tfrac{2}{15} & -\tfrac{1}{30} \end{bmatrix}\) |
(D) \(\begin{bmatrix} 7 & 4 \\ 3 & 6 \end{bmatrix}\) | (IV) \(\begin{bmatrix} \tfrac{2}{15} & -\tfrac{1}{15} \\[6pt] \tfrac{1}{6} & \tfrac{1}{6} \end{bmatrix}\) |
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
Resonance in X$_2$Y can be represented as
The enthalpy of formation of X$_2$Y is 80 kJ mol$^{-1}$, and the magnitude of resonance energy of X$_2$Y is: