Let matrix B be the adjoint of a square matrix A, l be the identify matrix of same order as A. If k (≠ 0) is the determinate of the matrix A, then what is AB equal to?
kl
The question asks for the value of the product of a square matrix A and its adjoint matrix B, given that the determinant of A is k and l is the identity matrix of the same order as A.
There is a fundamental property relating a square matrix A, its adjoint adj(A), its determinant \(\det(A)\), and the identity matrix I of the same order. This property is given by the equation:
\[ A \cdot \text{adj}(A) = \det(A) \cdot I \]
We are given:
Substituting these given values into the property \(A \cdot \text{adj}(A) = \det(A) \cdot I\), we get:
\[ A \cdot B = k \cdot l \]
Thus, the product AB is equal to kl.
Based on the fundamental property of matrices, the product of a square matrix A and its adjoint B (where B = adj(A)) is equal to the determinant of A multiplied by the identity matrix of the same order. Given \(\det(A) = k\) and the identity matrix is l, we find that \(AB = kl\).
Comparing our result \(kl\) with the given options:
Our result matches option 2.
| Property | Description | Formula |
|---|---|---|
| Matrix times its Adjoint | The product of a square matrix and its adjoint is the determinant times the identity matrix. | \(A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = \det(A) \cdot I\) |
| Determinant of Adjoint | The determinant of the adjoint of an n x n matrix A. | \(\det(\text{adj}(A)) = (\det(A))^{n-1}\) |
| Adjoint of Adjoint | The adjoint of the adjoint of an n x n matrix A. | \(\text{adj}(\text{adj}(A)) = (\det(A))^{n-2} \cdot A\) |
| Inverse of a Matrix | If \(\det(A) \ne 0\), the inverse of A exists. | \(A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A)\) |
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