Compute the determinant of the 3x3 matrix with rows (9,2,2), (28,8,8) and (18,5,4).

Step-by-step solution

1. 1

   Note the common factor in row 2. Every entry of $(28,8,8)$ is divisible by $4$. Scaling a row by $k$ scales the determinant by $k$, so

   $\det=4\begin{vmatrix}9&2&2\\7&2&2\\18&5&4\end{vmatrix}$

   This keeps the arithmetic small — an important habit once the entries get large.

2. 2

   Simplify further with a row operation. Subtracting row $2$ from row $1$ (which does not change the determinant) leaves

   $4\begin{vmatrix}2&0&0\\7&2&2\\18&5&4\end{vmatrix}$

   since the second and third entries of both rows are $2,2$.

3. 3

   Expand along the new first row. Only the entry $2$ at position $(1,1)$ is nonzero, with cofactor sign $+$:

   $4\cdot 2\begin{vmatrix}2&2\\5&4\end{vmatrix}=8(8-10)=8(-2)=-16$

   $\boxed{-16}$

4. 4

   Verify by direct cofactor expansion on the original.

   $9(8\cdot 4-8\cdot 5)-2(28\cdot 4-8\cdot 18)+2(28\cdot 5-8\cdot 18)$

   $=9(32-40)-2(112-144)+2(140-144)=-72+64-8=-16$

   The two routes agree.

5. 5

   Compare with the neighbouring problem. Changing only the middle row from $(27,7,7)$ to $(28,8,8)$ moves the determinant from $-9$ to $-16$. Determinants are highly sensitive to small entry changes, which is exactly why exam sets often present such near-identical matrices — each must be computed from scratch.