Matrix Properties Analysis: Non-Singular Matrix A

The question asks us to evaluate the correctness of three statements regarding a non-singular matrix \(A\) of order \(n\). A non-singular matrix is a square matrix that has an inverse (i.e., its determinant is non-zero).

Evaluating Statement 1: \(A(\text{adj}A^T) = A(\text{adj}A)^T\)

Let's break down this statement:

• Key Property 1: For any square matrix \(M\), the adjoint of its transpose is the transpose of its adjoint. That is, \(\text{adj}(M^T) = (\text{adj}M)^T\). Applying this to matrix \(A\), we have \(\text{adj}(A^T) = (\text{adj}A)^T\).
• Substituting Property 1: If we substitute \((\text{adj}A)^T\) for \(\text{adj}(A^T)\) in the left side of the statement, we get \(A(\text{adj}A^T) = A((\text{adj}A)^T)\).
• Comparing Sides: The statement essentially asks if \(A((\text{adj}A)^T) = A(\text{adj}A)^T\). This equality is true by definition after the substitution.
• Alternative Verification: We also know that for any non-singular matrix \(A\), \(A(\text{adj}A) = |A|I_n\), where \(|A|\) is the determinant of \(A\) and \(I_n\) is the identity matrix of order \(n\).
  • Left Hand Side (LHS): \(A(\text{adj}A^T)\). Taking the transpose of the LHS: \((A(\text{adj}A^T))^T = (\text{adj}A^T)^T (A^T)^T = (\text{adj}A^T)^T A\). Using Property 1, this becomes \(((\text{adj}A)^T)^T A = (\text{adj}A) A\). Since \((\text{adj}A) A = |A|I_n\), we have \((A(\text{adj}A^T))^T = |A|I_n\). Taking the transpose again, we get \(A(\text{adj}A^T) = (|A|I_n)^T = |A|I_n\). So, LHS = \(|A|I_n\).
  • Right Hand Side (RHS): \(A(\text{adj}A)^T\). Let \(X = A(\text{adj}A)\). We know \(X = |A|I_n\). Then the RHS is \(X^T\). Since \(X = |A|I_n\), \(X^T = (|A|I_n)^T = |A|I_n^T = |A|I_n\). So, RHS = \(|A|I_n\).
  Since LHS = RHS, Statement 1 is correct.

Evaluating Statement 2: If \(A^2 = A\), then \(A\) is the identity matrix of order \(n\).

Let's analyze the condition \(A^2 = A\) for a non-singular matrix \(A\).

• Rewrite the equation: \(A^2 - A = 0\).
• Factor out \(A\): \(A(A - I) = 0\), where \(I\) is the identity matrix.
• Since \(A\) is non-singular, its inverse \(A^{-1}\) exists. Multiply both sides by \(A^{-1}\) on the left:

  \(A^{-1}A(A - I) = A^{-1}0\)

• Simplify using \(A^{-1}A = I\) and \(A^{-1}0 = 0\):

  \(I(A - I) = 0\)

  \(A - I = 0\)

• Solve for \(A\): \(A = I\).

This derivation shows that if \(A\) is non-singular and \(A^2 = A\), then \(A\) must indeed be the identity matrix. Therefore, Statement 2 is correct.

Evaluating Statement 3: If \(A^3 = A\), then \(A\) is the identity matrix of order \(n\).

Let's examine the condition \(A^3 = A\) for a non-singular matrix \(A\).

• Rewrite the equation: \(A^3 - A = 0\).
• Factor out \(A\): \(A(A^2 - I) = 0\).
• Multiply by \(A^{-1}\) on the left (since \(A\) is non-singular):

  \(A^{-1}A(A^2 - I) = A^{-1}0\)

  \(I(A^2 - I) = 0\)

  \(A^2 - I = 0\)

  \(A^2 = I\)

• The condition \(A^3 = A\) for a non-singular matrix implies \(A^2 = I\). However, \(A^2 = I\) does not necessarily mean \(A = I\).
• Counterexample: Consider the matrix \(A = -I\).
  • Is \(A\) non-singular? The determinant is \(|A| = |-I| = (-1)^n\). This is either 1 or -1, and never 0. So, \(A=-I\) is always non-singular.
  • Does \(A^3 = A\) hold? \(A^3 = (-I)^3 = (-1)^3 I^3 = -I\). Since \(A = -I\), the condition \(A^3 = A\) is satisfied.
  • Is \(A=I\)? \(A = -I\). This equals \(I\) only if \(-I = I\), which implies \(2I = 0\). This is impossible for matrices over real or complex numbers unless \(I\) is the zero matrix (which isn't the identity matrix). Thus, \(A = -I\) is not the identity matrix.

Since we found a non-singular matrix (\(A=-I\)) for which \(A^3=A\) but \(A \neq I\), Statement 3 is incorrect.

Final Conclusion

Based on the analysis of each statement:

• Statement 1 is correct.
• Statement 2 is correct.
• Statement 3 is incorrect.

Therefore, the correct statements are 1 and 2.