For a $2 \times 2$ matrix, if one row is a multiple of the other, the determinant is zero. Use this fact and the method of minors and cofactors to show that the determinant of a $3 \times 3$ matrix is zero if one row is a multiple of another.

Solution

Suppose we have a $3 \times 3$ matrix in which the first and second rows are multiples of each other. Our task is to show that the determinant of this matrix is zero.

Let's write the matrix in a general form:

$$A = \begin{pmatrix}a_1 & a_2 & a_3 \\ k a_1 & k a_2 & k a_3 \\ b_1 & b_2 & b_3\end{pmatrix}$$ for some constant $k$.

We want to express the determinant of $A$ by a minor and cofactor expansion.

Because there is no assumed relationship between the $a$'s and $b$'s, we can not directly see that $\text{ det } A = 0$ without directly computing the $2 \times 2$ determinants.

We notice that each of these $2 \times 2$ matrices has a row that is a multiple of another row. Hence all these $2 \times 2$ determinants are zero.

We conclude that $\text{det }A = 0$.

In this solution, we assumed that the first two rows of $A$ were the ones that were related by a constant factor. If, instead, the related rows were the first and third or the second and third rows, the same procedure would work by taking the minor and cofactor expansion along the unrelated row.