Question

If $A$ is a square matrix of order 3 such that $|A|=5$ then $|(adjA^{-1})^{-1}|$ is

• A. 9
• B. 4
• C. 6
• D. 25

Answer 1

Answer: (D)

25

Answer 2

Let $A$ be a square matrix of order $n=3$. Given $|A|=5$. We need to find the value of $|(adjA^{-1})^{-1}|$.

We will use the following properties of matrices and determinants:

1. For any invertible matrix $X$, $(X^{-1})^{-1} = X$.

2. For an invertible matrix $X$, $adj(X^{-1}) = (adj X)^{-1}$.

3. For a square matrix $X$ of order $n$, $|adj X| = |X|^{n-1}$.

Now, let's simplify the given expression step-by-step:

• Step 1: Apply the property $adj(X^{-1}) = (adj X)^{-1}$ with $X=A$. So, $adj(A^{-1}) = (adj A)^{-1}$. The expression becomes $|(adj(A^{-1}))^{-1}| = |((adj A)^{-1})^{-1}|$.

• Step 2: Apply the property $(Y^{-1})^{-1} = Y$ with $Y = adj A$. So, $((adj A)^{-1})^{-1} = adj A$. The expression simplifies to $|adj A|$.

• Step 3: Apply the property $|adj A| = |A|^{n-1}$. Given that the order of matrix $A$ is $n=3$. So, $|adj A| = |A|^{3-1} = |A|^2$.

• Step 4: Substitute the given value of $|A|$. Given $|A|=5$. Therefore, $|adj A| = (5)^2 = 25$.

Thus, $|(adjA^{-1})^{-1}| = 25$.