Evaluating the determinant of adjugate matrices and scalar multiples

Key concept: The identity det(adj A) = det(A)^{n-1} is the central tool for the matrices A, B = adj(A), and C = 5A.

Step by step

Step 1: Use the determinant identity for adjugates

For any 3 x 3 matrix A,

$|\operatorname{adj}A| = |A|^{3-1} = |A|^2$

So the first job is to find $|A|$.

For

$A=\begin{bmatrix}1&-1&1\\0&2&-3\\2&1&0\end{bmatrix},$

expand along the first row:

$|A| = 1\begin{vmatrix}2&-3\\1&0\end{vmatrix} -(-1)\begin{vmatrix}0&-3\\2&0\end{vmatrix} +1\begin{vmatrix}0&2\\2&1\end{vmatrix}$

$= 1(3) + 1(6) + 1(-4) = 5$

Thus

$|B| = |\operatorname{adj}A| = 5^2 = 25$

Step 2: Find |adj B|

Since B is also 3 x 3,

$|\operatorname{adj}B| = |B|^2 = 25^2 = 625$

This follows from the same identity applied again.

Step 3: Find |C|

Given $C=5A$, and A is 3 x 3,

$|C| = |5A| = 5^3|A| = 125\cdot 5 = 625$

Now the ratio becomes

$\frac{|\operatorname{adj}B|}{|C|} = \frac{625}{625} = 1$

Step 4: Match the answer choice

The correct value is

$\boxed{1}$

The computation is short once the determinant rules for adjugates and scalar multiples are used correctly.

Pitfall alert

A common error with adjugate determinants is to use the exponent n instead of n minus 1. For a 3 x 3 matrix, det(adj A) is not det(A)^3; it is det(A)^2. Another mistake is to treat det(5A) as 5 times det(A), but scalar multiplication in a 3 x 3 determinant contributes 5^3, not 5. In this problem those two exponent rules interact, and mixing them up leads to choices 5 or 25 very quickly. It also helps to compute det(A) first, because once det(A)=5 is known, the rest collapses cleanly.

Try different conditions

If C were changed from 5A to 2A, the same method would give a different ratio. The matrix A is unchanged, so det(A)=5 still implies det(B)=25 and det(adj B)=625. But now det(C)=det(2A)=2^3 times det(A)=8 times 5, which is 40. The new ratio would be 625 divided by 40, or 125 over 8. If instead B were changed from adj A to adj of 2A, the adjugate would scale by 2^2 because A is 3 x 3, and that would change the entire determinant chain. These variants all hinge on knowing how adjugates and scalar multiples behave under determinants.

Further reading

adjugate determinant identity, scalar multiple determinant, 3x3 matrix determinant