Step 1: Understanding the Concept:
We need to identify the incorrect statement among the given properties of invertible matrices. An invertible matrix is a square matrix that has a non-zero determinant.
Step 3: Detailed Explanation:
Let's analyze each statement.
1. \( \text{adj}(A) = |A|A^{-1} \):
The formula for the inverse of a matrix \( A \) is given by: \[ A^{-1} = \frac{1}{|A|} \text{adj}(A) \] Multiplying both sides by \( |A| \) (which is non-zero since \( A \) is invertible), we get: \[ |A|A^{-1} = \text{adj}(A) \] This statement is correct.
2. \( (A + B)^{-1} = A^{-1} + B^{-1} \):
This property states that the inverse of a sum is the sum of the inverses. This is generally not true for matrices. We can show this with a counterexample.
Let \( A = I \) and \( B = I \) (where \( I \) is the identity matrix). Both are invertible.
LHS = \( (A + B)^{-1} = (I + I)^{-1} = (2I)^{-1} = \frac{1}{2}I^{-1} = \frac{1}{2}I = \begin{bmatrix} 1/2 & 0 \\ 0 & 1/2 \end{bmatrix} \).
RHS = \( A^{-1} + B^{-1} = I^{-1} + I^{-1} = I + I = 2I = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \).
Since LHS \( \neq \) RHS, the statement is NOT correct.
3. \( |A^{-1}| = |A|^{-1} \):
We know that a matrix and its inverse satisfy the relation \( AA^{-1} = I \).
Taking the determinant of both sides: \[ |AA^{-1}| = |I| \] Using the property \( |XY| = |X||Y| \), we get: \[ |A||A^{-1}| = 1 \] Since \( A \) is invertible, \( |A| \neq 0 \). We can divide by \( |A| \): \[ |A^{-1}| = \frac{1}{|A|} = |A|^{-1} \] This statement is correct.
4. \( (AB)^{-1} = B^{-1}A^{-1} \):
This is the well-known reversal law for the inverse of a product of matrices. It is a standard property.
This statement is correct.
Step 4: Final Answer:
The statement that is not correct is \( (A + B)^{-1} = A^{-1} + B^{-1} \).
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) \(\begin{bmatrix} \tfrac{2}{15} & -\tfrac{1}{15} \\[6pt] \tfrac{1}{6} & \tfrac{1}{6} \end{bmatrix}\) |