Finding the $2\times 2$ determinant using row operations and upper triangular matrix.

Question

Calculate $\begin{vmatrix} 3 &-5 \\ 2&6 \end{vmatrix}$.

I want to use row operations, and so what I did first was interchange the rows, which multiplies the determinant by $-1$. Then I wanted the $2$ to be a leading one, so I multiplied the first row by $\frac{1}{2}$, which multiplies the determinant by $\frac{1}{2}$. Then I eliminate the $3$ in the second row. I get the resulting determinant:

$-\frac{1}{2}\begin{vmatrix} 1&3 \\ 0&-14 \end{vmatrix}$

So then the determinant is $7$. However, I know this is wrong. Could someone please tell me what I did wrong to get an incorrect determinant?

Answer 1

It's important to know what types of operations change the determinant. Replacing a row after you've scaled it will scale the determinant. So you should replace a row you didn't scale. For example:$$\underbrace{\begin{bmatrix} 3&-5\\ 2&6 \end{bmatrix}\sim\begin{bmatrix} 0&-14\\ 2&6 \end{bmatrix}}_{R_1-\frac{3}{2}R_2\implies R_1}$$ Which has determinant $28.$

Now, you swapped two rows, so you are right to change the sign. But when you multiplied by $\frac{1}{2}$ you then need to multiply by 2 to get the correct determinant in the end.

Answer 2

You took the $2$ out by taking $1/2$ out, i.e., you actually used $(2a) = 1/2*a$.

Answer 3

Along those lines, what you should have done was:\begin{align}\begin{vmatrix}3&-5\\2&6\end{vmatrix}&=-\begin{vmatrix}2&6\\3&-5\end{vmatrix}\\&=-2\begin{vmatrix}1&3\\3&-5\end{vmatrix}\\&=-2\begin{vmatrix}1&3\\0&-14\end{vmatrix}\\&=28.\end{align}