Determinants: Conceptual Doubts Cleared (4)

Question

If $A$ is a $3 \times 3$ matrix with $\det(A) = 4$, find $\det(2A)$, $\det(A^{-1})$, and $\det(A^T)$.

Solution — Step by Step

Final answers: $\det(2A) = 32$, $\det(A^{-1}) = 1/4$, $\det(A^T) = 4$.

Why This Works

Each property has a clean algebraic reason:

• $\det(kA) = k^n \det(A)$ because multiplying a matrix by $k$ scales every row by $k$, and pulling each row’s $k$ out of the determinant gives $k^n$.
• $\det(A^{-1}) = 1/\det(A)$ because $A A^{-1} = I$ implies $\det(A) \det(A^{-1}) = \det(I) = 1$.
• $\det(A^T) = \det(A)$ comes from the cofactor expansion working identically along rows or columns.

Alternative Method

Pick a specific $3 \times 3$ matrix with $\det = 4$, e.g., $A = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$. Compute each quantity directly. $\det(2A) = 8 \cdot 1 \cdot 1 \cdot 4 = 32$. $\det(A^{-1}) = \det(\text{diag}(1/4, 1, 1)) = 1/4$. $\det(A^T) = 4$. Confirms our properties.

Common Mistake

Writing $\det(2A) = 2 \det(A)$ for a $3 \times 3$ matrix. The factor is $2^n$, not 2. For a $2 \times 2$ matrix it would be $4 \det A$; for $3 \times 3$, $8 \det A$.