Compute the determinant of the 3x3 matrix with rows (9,2,2), (27,7,7) and (18,5,4).

Step-by-step solution

1. 1

   Look for a shortcut first. The first two rows both have equal second and third entries ($2,2$ and $7,7$), which hints that subtracting column $3$ from column $2$ will create zeros. But with only three rows, straight cofactor expansion along the top row is already short, so we use that and keep the structure in reserve as a check.

2. 2

   Write the cofactor expansion with the alternating signs. Along the first row the signs are $+,-,+$:

   $\det=9\begin{vmatrix}7&7\\5&4\end{vmatrix}-2\begin{vmatrix}27&7\\18&4\end{vmatrix}+2\begin{vmatrix}27&7\\18&5\end{vmatrix}$

3. 3

   Evaluate the three 2×2 minors.

   $\begin{vmatrix}7&7\\5&4\end{vmatrix}=28-35=-7,\quad \begin{vmatrix}27&7\\18&4\end{vmatrix}=108-126=-18,\quad \begin{vmatrix}27&7\\18&5\end{vmatrix}=135-126=9$

4. 4

   Combine.

   $\det=9(-7)-2(-18)+2(9)=-63+36+18=-9$

   $\boxed{-9}$

5. 5

   Check with the column shortcut. Replacing $C_2$ by $C_2-C_3$ (which leaves the determinant unchanged) gives

   $\begin{vmatrix}9&0&2\\27&0&7\\18&1&4\end{vmatrix}$

   Expanding along the new second column, only the entry $1$ at position $(3,2)$ survives, with sign $(-1)^{3+2}=-1$:

   $\det=-1\begin{vmatrix}9&2\\27&7\end{vmatrix}=-(63-54)=-9$

   The two methods agree.