Question

For any square matrix $A$, $A \cdot \text{adj}(A) =$

• A. \( A \)
• B. \( |A| I \)
• C. \( I \)
• D. \( |A|^2 A \)

Hint:

This property is exceptionally useful for solving determinant operations on adjugates. Taking the determinant of both sides reveals that: \(|A \cdot \text{adj}(A)| = ||A|I| \implies |A| \cdot |\text{adj}(A)| = |A|^n \implies |\text{adj}(A)| = |A|^{n-1}\).

Answer 1

The Correct Option is B

Concept:

For any square matrix \(A\) of order \(n\), a fundamental identity connects the matrix, its adjugate, and its determinant:

\[ A \cdot \operatorname{adj}(A) = \operatorname{adj}(A) \cdot A = |A| I_n \]

Here, \(|A|\) represents the determinant of matrix \(A\), and \(I_n\) is the identity matrix of order \(n\).

This identity is also used to derive the inverse of a matrix:

\[ A^{-1} = \frac{\operatorname{adj}(A)}{|A|} \] 
 

Step 1: Verification Using Cofactor Properties

Consider the multiplication of matrix \(A\) with its adjugate matrix. When the elements of row \(i\) of \(A\) are multiplied with the cofactors of the same row, the result becomes the determinant:

\[ \sum_{k=1}^{n} a_{ik} C_{ik} = |A| \]

If the elements of row \(i\) are multiplied with cofactors corresponding to a different row \(j\) where \(i \neq j\), then the sum becomes zero:

\[ \sum_{k=1}^{n} a_{ik} C_{jk} = 0 \] 
 

Step 2: Formation of the Final Matrix

Thus, after performing row-by-column multiplication, the resulting matrix contains determinant values along the diagonal and zeros elsewhere:

\[ A \cdot \operatorname{adj}(A) = \begin{bmatrix} |A| & 0 & \cdots & 0 \\ 0 & |A| & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & |A| \end{bmatrix} \]

Taking \(|A|\) common:

\[ A \cdot \operatorname{adj}(A) = |A| \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix} = |A| I_n \]