Compute the determinant of the 4x4 matrix with rows (18,5,7,8), (18,3,4,4), (9,1,1,1) and (9,2,3,4).

Step-by-step solution

1. 1

   Reduce before expanding. A blind cofactor expansion of a $4\times 4$ needs four $3\times 3$ determinants. Adding a multiple of one row to another leaves the determinant unchanged, so it is far cheaper to first create a column with a single nonzero entry.

2. 2

   Clear the first column using row 3. Row $3$ starts with $9$, and the other first-column entries are $18$, $18$ and $9$, so

   $R_1\leftarrow R_1-2R_3,\qquad R_2\leftarrow R_2-2R_3,\qquad R_4\leftarrow R_4-R_3$

   gives

   $\begin{vmatrix}0&3&5&6\\0&1&2&2\\9&1&1&1\\0&1&2&3\end{vmatrix}$

3. 3

   Expand along the first column. Only the entry $9$ in position $(3,1)$ survives, and its cofactor sign is $(-1)^{3+1}=+1$:

   $\det=9\begin{vmatrix}3&5&6\\1&2&2\\1&2&3\end{vmatrix}$

   (The minor keeps rows $1,2,4$ and columns $2,3,4$ of the reduced matrix.)

4. 4

   Evaluate the 3×3 minor. Subtract row $2$ from row $3$ to get $(0,0,1)$, then expand along that row:

   $\begin{vmatrix}3&5&6\\1&2&2\\0&0&1\end{vmatrix}=1\cdot\begin{vmatrix}3&5\\1&2\end{vmatrix}=3(2)-5(1)=1$

5. 5

   Multiply back.

   $\det=9\times 1=9$

   $\boxed{9}$

6. 6

   Check by direct cofactor expansion along the third row. Expanding the original matrix along $(9,1,1,1)$ gives $9M_{31}-1\cdot M_{32}+1\cdot M_{33}-1\cdot M_{34}$ with the appropriate signs; carrying that out also returns $9$. The agreement of two independent routes is the check that no row operation was mis-signed.