Question

If A is a non-singular matrix of order n then which of the following is / are correct?

• A. adj ($A^T$) = (adj A)$^T$
• B. adj ($A^{-1}$) = (adj A)$^{-1}$
• C. adj ($A^2$) = (adj A)$^2$
• D. If n = 3 then adj(2A) = 4adj A

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

The problem asks us to identify the correct properties of the adjoint of a non-singular matrix A of order n. We will check each option using the fundamental properties of determinants, inverses, and adjoints.

Key properties of adjoint matrix:

1. For any square matrix A, $A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A| I_n$, where $|A|$ is the determinant of A and $I_n$ is the identity matrix of order n.
2. If A is non-singular (i.e., $|A| \neq 0$), then $A^{-1} = \frac{1}{|A|} \text{adj}(A)$. From this, we can also write $\text{adj}(A) = |A| A^{-1}$.

Let's analyze each option:

(A) adj ($A^T$) = (adj A)$^T$ We use the property $\text{adj}(X) = |X| X^{-1}$. LHS: $\text{adj}(A^T) = |A^T| (A^T)^{-1}$. Since $|A^T| = |A|$ and $(A^T)^{-1} = (A^{-1})^T$, $\text{adj}(A^T) = |A| (A^{-1})^T$.

RHS: $(\text{adj A})^T = (|A| A^{-1})^T$. Using the property $(kM)^T = kM^T$ for a scalar k and matrix M, $(\text{adj A})^T = |A| (A^{-1})^T$.

Since LHS = RHS, option (A) is correct.

(B) adj ($A^{-1}$) = (adj A)$^{-1}$ We use the property $\text{adj}(X) = |X| X^{-1}$. LHS: $\text{adj}(A^{-1}) = |A^{-1}| (A^{-1})^{-1}$. Since $|A^{-1}| = \frac{1}{|A|}$ and $(A^{-1})^{-1} = A$, $\text{adj}(A^{-1}) = \frac{1}{|A|} A$.

RHS: $(\text{adj A})^{-1} = (|A| A^{-1})^{-1}$. Using the property $(kM)^{-1} = \frac{1}{k} M^{-1}$ for a scalar k and matrix M, $(\text{adj A})^{-1} = \frac{1}{|A|} (A^{-1})^{-1} = \frac{1}{|A|} A$.

Since LHS = RHS, option (B) is correct.

(C) adj ($A^2$) = (adj A)$^2$ We use the property $\text{adj}(X) = |X| X^{-1}$. LHS: $\text{adj}(A^2) = |A^2| (A^2)^{-1}$. Since $|A^2| = |A \cdot A| = |A||A| = |A|^2$ and $(A^2)^{-1} = (A \cdot A)^{-1} = A^{-1} A^{-1} = (A^{-1})^2$, $\text{adj}(A^2) = |A|^2 (A^{-1})^2$.

RHS: $(\text{adj A})^2 = (|A| A^{-1})^2$. $(|A| A^{-1})^2 = (|A| A^{-1}) (|A| A^{-1}) = |A|^2 (A^{-1} A^{-1}) = |A|^2 (A^{-1})^2$.

Since LHS = RHS, option (C) is correct. Alternatively, we can use the property $\text{adj}(XY) = \text{adj}(Y)\text{adj}(X)$. For $X=A$ and $Y=A$, we get $\text{adj}(A \cdot A) = \text{adj}(A)\text{adj}(A)$, which means $\text{adj}(A^2) = (\text{adj A})^2$.

(D) If n = 3 then adj(2A) = 4adj A We use the general property $\text{adj}(kA) = k^{n-1} \text{adj}(A)$, where n is the order of the matrix. In this option, $k=2$ and $n=3$. So, $\text{adj}(2A) = 2^{3-1} \text{adj}(A) = 2^2 \text{adj}(A) = 4 \text{adj}(A)$.

This property is correct.

Since all four options (A), (B), (C), and (D) are correct, this is a multiple correct options question.