Determinant by nullifying

Question

I am supposed to calculate the value for the determinant of this matrix. I didn't know what to do, so I looked up for the sample solution, which I don't understand.

$$\left|\begin{array}{ccc} 18 & 4 & 6\\ 9 & 2 & 3\\ 11 & 5 & 27\end{array}\right| = \left|\begin{array}{ccc} 2\cdot 9 & 2\cdot 2 & 2\cdot 3\\ 9 & 2 & 3\\ 11 & 5 & 27\end{array}\right| = 2\cdot\left|\begin{array}{ccc} 9 & 2 & 3\\ 9 & 2 & 3\\ 11 & 5 & 27\end{array}\right| = 2\cdot 0 = 0.$$

Why does the multiplication with $2$ only apply to the first row in the third matrix?

Answer 1

The second equality uses the fact that if you multiply a row of a square matrix $A$ by a scalar $k$ to obtain a new matrix $\hat{A}$, then $\det(\hat{A}) = k\det(A)$.

The third equality uses the fact that a square matrix with two identical rows has determinant zero.

Answer 2

One was of seeing this is to note that if $A,B$ are matrices, then $\det (AB) = \det A \det B$.

In the following, let $A$ be the leftmost matrix.

If you treat the determinant as a function of $n$ rows (or columns), then it is linear in each row separately. In the above, the top row is twice the middle row, so you can use linearity to 'pull' the 2 out.

Explicitly: If we let $B=\begin{bmatrix} {1 \over 2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$, then $\det B = {1 \over 2}$ and so we get $\det A = {1 \over \det B} \det (AB)$, since ${1 \over \det B} = 2$ and post multiplying by $B$ divides the first row by $2$ we get the result.

If you swap two rows the determinant switches sign. So, if two rows are equal, the determinant must be zero.

In this case, if we let $C=\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$, see that post multiplying by $C$ swaps the top two rows and $\det C = -1$. Let $D$ be the matrix in the middle, then since we have $D = DC$, so $\det D = -\det D$, so we must have $\det D = 0$.