Question

If \(A\) is a square matrix of order 3 such that \(|\mathrm{A}|=2\), then \(\left|\left(\operatorname{adj} \mathrm{A}^{-1}\right)^{-1}\right|\) is (1) \(\frac{1}{4}\) (2) 2 (3) 4 (4) 8

Short answer

The value is 4 (option 3).

Step-by-step solution

Understand the problem

Given a 3x3 matrix A where the determinant \(\text{det}(\text{A}) \) is 2, the task is to find the value of \(\big|\big(\text{adj} \left(\text{A}^{-1}\right)^{-1}\big)\big| \).

Recall properties of determinants

Recall that for any square matrix A, \(\text{adj}(\text{A}) = \text{det}(\text{A}) \cdot \text{A}^{-1}\).

Find the determinant of the adjoint and inverse

The determinant of an inverse matrix follows the property \(\text{det}(A^{-1}) = \frac{1}{\text{det}(A)}\). Thus, if \(\text{det}(\text{A}) = 2\), we have \(\text{det}(\text{A}^{-1}) = \frac{1}{2}\). Using the property of adjoints, we get \(\text{det}(\text{adj}(\text{A})) = (\text{det}(\text{A}))^{n-1}\) where \(n\) is the order of the matrix.

Calculate determinant of adjugate and its inverse

Since A is a 3x3 matrix, \(\text{det}(\text{adj}(\text{A}^{-1})) = (\text{det}(\text{A}^{-1}))^2\). Substitute \(\text{det}(\text{A}^{-1}) = \frac{1}{2}\):\[ \text{det}(\text{adj}(\text{A}^{-1})) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\]

Inverse of adjugate determinant

For any matrix B, \(\text{det}(B^{-1}) = \frac{1}{\text{det}(B)}\), so: \[\text{det} \big((\text{adj}(\text{A}^{-1}))^{-1}\big) = \frac{1}{\text{det}(\text{adj}(\text{A}^{-1}))} = \frac{1}{\frac{1}{4}} = 4\]

Key concepts

Matrix Inverse

To understand the matrix inverse, let's first recall what it means for a matrix to be invertible. A square matrix \(\text{A}\) is said to be invertible if there exists another matrix \(\text{A}^{-1}\) such that:
\[\text{A} \times \text{A}^{-1} = \text{I}\]
where \(\text{I}\) is the identity matrix. The identity matrix is a special kind of matrix that doesn't change another matrix when multiplied by it, like multiplying by 1 for numbers.
If \(\text{A}\) is a 3x3 matrix and its determinant \(\text{det}(\text{A}) = 2\), then its inverse has a determinant given by:
\[\text{det}(\text{A}^{-1}) = \frac{1}{\text{det}(\text{A})} = \frac{1}{2}\]
This is important because the determinant helps determine whether a matrix is invertible. If \(\text{det}(\text{A})\) were 0, \(\text{A}\) wouldn't be invertible.

Adjugate Matrix

The adjugate matrix, sometimes called the adjoint matrix, is crucial in finding the inverse of a matrix. For any given square matrix \(\text{A}\), the adjugate matrix \(\text{adj}(\text{A})\) is obtained by taking the transpose of the cofactor matrix.
Mathematically, if \(\text{A}\) is a 3x3 matrix, then \(\text{adj}(\text{A})\) is:\[\text{adj}(\text{A}) = \text{det}(\text{A}) \times \text{A}^{-1}\]
This means each element of the adjugate matrix is related to the inverses of the original matrix \(\text{A}\).
Thus, given our example \(\text{A}\) with \(\text{det}(\text{A}) = 2\), we can infer properties about \(\text{adj}(\text{A}^{-1})\). Specifically, knowing the determinant of the inverse matrix helps us find:\[\text{det}(\text{adj}(\text{A}^{-1})) = \text{det}((\text{det}(\text{A})) \times \text{A}^{-1})^{n-1} = \big(\text{det}(\text{A}^{-1})\big)^2 = \big(\frac{1}{2}\big)^2 = \frac{1}{4}\]

Determinant Calculations

Determinant calculations for matrices often puzzle students, but understanding the basics simplifies the process. The determinant of a 3x3 matrix \(\text{A}\) is calculated through the formula involving minors and cofactors.
For our problem, we start with \(\text{det}(\text{A}) = 2\). A crucial property is: \[\text{det}(\text{A}^{-1}) = \frac{1}{\text{det}(\text{A})} = \frac{1}{2}\]
This directly helps in determining \(\text{adj}(\text{A}^{-1})\). The determinant of the adjugate of a matrix's inverse is:\[\text{det}(\text{adj}(\text{A})) = (\text{det}(\text{A}))^{n-1}\]
where \(\text{A}\) is an \(\text{n} \times \text{n}\) matrix. For \(\text{n} = 3\), \(\text{adj}(\text{A}^{-1})\) gives:\[\text{det}(\text{adj}(\text{A}^{-1})) = \big(\frac{1}{2}\big)^2 = \frac{1}{4}\]
Finally, applying the inverse property to determinants:\[\text{det} \big((\text{adj}(\text{A}^{-1}))^{-1}\big) = \frac{1}{\text{det}(\text{adj}(\text{A}^{-1}))} = \frac{1}{\frac{1}{4}} = 4\]
So, the value of \(\big|\big(\text{adj} \big(\text{A}^{-1}\big)^{-1}\big)\big|\) is 4.