# Finding the determinant using row operations.

So I have to find the determinant of $\begin{bmatrix}3&2&2\\2&2&1\\1&1&1\end{bmatrix}$ using row operations. From what I've learned, the row operations that change the determinate are things like swaping rows makes the determinant negative and dividing a row by a value means you have to multiply it by that value. Once you have an upper triangular matrix then you just multiply the diagonal and you should have the determinant. So how come I can't get the right answer? Here's what I did. I start by dividing the first row by 3. $3\begin{bmatrix}1&2/3&2/3\\2&2&1\\1&1&1\end{bmatrix}$. Then I subtract the second row by 2 times the first row $3\begin{bmatrix}1&2/3&2/3\\0&2/3&-1/3\\1&1&1\end{bmatrix}$. Then I subtract the third and first row. $3\begin{bmatrix}1&2/3&2/3\\0&2/3&-1/3\\0&1/3&1/3\end{bmatrix}$. Then I multiply the third row by 2 and subtract it from the second row. $3\begin{bmatrix}1&2/3&2/3\\0&2/3&-1/3\\0&0&1\end{bmatrix}$. So now I have an upper triangular matrix so now I just do $(3)(1)(2/3)(1)$ and I get 2. The answer to this question however is not 2 but 1. So where did I go wrong? If the things I read were true then I should've got the right answer but I didn't which tells me that there's another row operation rule that no one told me about.

• Your last step seems off. You should multiply the second row by $1/2$ and subtract it from the third row. The resulting matrix is not the one you have written down. Double check that. – user332239 Dec 6 '16 at 20:27
• Ok that works. So then if I want to make the third row 0, can I not multiply something by the third row and then subtract it from the second row? I have to multiply something by the second row to get rid of the third row? – david mah Dec 6 '16 at 20:31
• Just a side not (not a problem), but I would not start off by dividing the first row by 3. It just makes the arithmetic a little messy when dealing with fractions. – MathGuy Dec 6 '16 at 20:37
• @Dietrich Burde But why?! WHY do I have to do that? Why is what I did not correct. Tell me! – david mah Dec 6 '16 at 20:55
• So.....are you going to tell me why? – david mah Dec 6 '16 at 20:58

Why divide, when simple addition and subtraction will do? Subtract twice the bottom row from the middle and top rows and you are left with

$$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 1 & 1 & 1\end{vmatrix} = 1-0+0=1$$

by evaluating by minors.

Even easier is if you take that matrix and add row 2 to row 3 and subtract row 1 to get

$$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0\end{vmatrix} = 1-0+0=1$$

or take this and switch the last two rows to get

$$-\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{vmatrix} = -(-1)=1$$

Your last step doesn't do what you said though, if you take twice the third row from the second you get

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

which is what you expect, since you divided the original row by $3$.

• I don't care about my specific question. I want to know in the general case why did I get that wrong. I've done tons of examples like this but I keep getting them wrong. So why do I keep getting them wrong? – david mah Dec 6 '16 at 20:52
• @davidmah there you go, edited. You just didn't do the last step like you should have. – Adam Hughes Dec 6 '16 at 21:09
• @davidmah Perhaps you keep confusing the effects of two similar-seeming row operations, as you did here. $R_n\gets R_n+kR_m$ doesn’t change the determinant. On the other hand, $R_n\gets kR_n+R_m$ does, since it’s really two elementary operations done one after another: multiplying a row by $k$ ($R_n\gets kR_n$), which changes the determinant, and then adding one row to another ($R_n\gets R_n+R_m$), which doesn’t. – amd Dec 6 '16 at 21:50

I find it a bit easier to avoid fractions, therefore I would avoid dividing by 3 first. Instead, let's take 2R3-R2=R2 to obtain:

$\begin{bmatrix}3 & 2 & 2\\0 & 0 & 1\\1 & 1 & 1\end{bmatrix}$

Now, take R1-3R3=R3 to obtain

$\begin{bmatrix}3 & 2 & 2\\0 & 0 & 1\\0 & -1 & -1\end{bmatrix}$

Because we took 3R3, this will change the value of the determinant. We need to compensate by multiplying by 1/3. Leaving us with:

1/3$\begin{bmatrix}3 & 2 & 2\\0 & 0 & 1\\0 & -1 & -1\end{bmatrix}$

Now, interchange R2 and R3 to obtain

1/3$\begin{bmatrix}3 & 2 & 2\\0 & -1 & -1\\0 & 0 & 1\end{bmatrix}$

This will change the value of the determinant. We need to add a negative sign to compensate for the interchange. Leaving us with:

-1/3$\begin{bmatrix}3 & 2 & 2\\0 & -1 & -1\\0 & 0 & 1\end{bmatrix}$

Now, multiply the diagonal and then multiply by -1/3 leaving you with 1.

Your error is when you multiply the third row by $2$ and subtract the second row; this introduces a factor $2$ that you have to remove.

Use a more systematic way: \begin{align} \begin{bmatrix} 3&2&2\\ 2&2&1\\ 1&1&1 \end{bmatrix} &\to \begin{bmatrix} 1&2/3&2/3\\ 2&2&1\\ 1&1&1 \end{bmatrix} &&R_1\gets \tfrac{1}{3}R_1 &&\color{red}{3} \\[4px]&\to \begin{bmatrix} 1&2/3&2/3\\ 0&2/3&-1/3\\ 0&1/3&1/3 \end{bmatrix} && \begin{aligned}R_2&\gets R_2-2R_1\\R_3&\gets R_3-R_1\end{aligned} \\[4px]&\to \begin{bmatrix} 1&2/3&2/3\\ 0&1&-1/2\\ 0&1/3&1/3 \end{bmatrix} && R_2\gets\tfrac{3}{2}R_2 && \color{red}{\tfrac{2}{3}} \\[4px]&\to \begin{bmatrix} 1&2/3&2/3\\ 0&1&-1/2\\ 0&0&1/2 \end{bmatrix} &&R_3\gets R_3-\tfrac{1}{3}R_2 \\[4px]&\to \begin{bmatrix} 1&2/3&2/3\\ 0&1&-1/2\\ 0&0&1 \end{bmatrix} &&R_3\gets 2R_3 &&\color{red}{\tfrac{1}{2}} \end{align} The determinant is $$3\cdot\frac{2}{3}\cdot\frac{1}{2}=1$$