# Using row operations to compute the following 3x3 determinant

Use row operations to compute the following determinant $\begin{bmatrix}3&3&-3\\3&4&-4\\2&-3&-5\end{bmatrix}$

I know how to easily compute the determinant using $i - j + k$ method... The problem is I have put the matrix in LTF (Lower Triangular Form) and then used the product of the diagonal. So I did:

$R_2 \leftarrow R_1 + (-1)R_2$

$R_3 \leftarrow 2R_1 + (-3)R_3$

$R_3 \leftarrow 15R_2 + R_3$

to get the matrix in LTF $\begin{bmatrix}3&3&-3\\0&-1&1\\0&0&23\end{bmatrix} \implies \det(A) = (3)(-1)(23) = -69$ I know that since I used these operations I changed the determinant, how exactly can I fix it back? I also know that the $\det(A)$ should be $= -24$ .

• You should get $24$ (not $23$) in the bottom right corner. – Oscar Cunningham Jul 12 '16 at 23:09
• Every time you replace a row or a column by itself multiplied by a constant, you have to divide by the same constant if you do not want to change your determinant. So for the first computation you would get -1 for the second one $-\tfrac{1}{3}$ and the last is fine. So your result, which seems to -72 has to be multiplied with -1 and $-\tfrac{1}{3}$ to get the value of the determinant. – Alex Jul 12 '16 at 23:14
• This is exactly the same mistake that the OP made in this question. – amd Jul 13 '16 at 7:09

The final matrix should be $$\begin{bmatrix} 3 & 3 & -3 \\ 0 & -1 & 1 \\ 0 & 0 & 24 \end{bmatrix}$$ However, you have multiplied the determinant by $-1$ with the first operation and by $-3$ with the second one, so you get $$\frac{3\cdot(-1)\cdot24}{(-1)\cdot(-3)}=-24$$

I use a different method, reducing the pivots to $1$: \begin{align} \begin{bmatrix} 3 & 3 & -3 \\ 3 & 4 & -4 \\ 2 & -3 & -5 \end{bmatrix} &\to \begin{bmatrix} 1 & 1 & -1 \\ 3 & 4 & -4 \\ 2 & -3 & -5 \end{bmatrix} && R_1\gets\color{red}{\frac{1}{3}}R_1 \\[6px] &\to \begin{bmatrix} 1 & 1 & -1 \\ 0 & 1 & -1 \\ 2 & -3 & -5 \end{bmatrix} && R_2\gets R_2-R_1 \\[6px] &\to \begin{bmatrix} 1 & 1 & -1 \\ 0 & 1 & -1 \\ 0 & -5 & -3 \end{bmatrix} && R_3\gets R_3-2R_1 \\[6px] &\to \begin{bmatrix} 1 & 1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & -8 \end{bmatrix} && R_3\gets R_3+5R_2 \\[6px] &\to \begin{bmatrix} 1 & 1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix} && R_3\gets \color{red}{-\frac{1}{8}}R_3 \end{align} The red numbers tell that the determinant has been multiplied by $$-\frac{1}{24}$$ so it is $-24$.

Multiplying any row by $\lambda$ multiplies the determinant by $\lambda$.

Adding any multiple of a row to another row doesn't change the determinant.

So

$R_2 \leftarrow R_1 + (-1)R_2$

multiplies the determinant by $-1$ (because multiplying $R_2$ by $-1$ multiplies the determinant by $-1$ and adding $R_1$ to $R_2$ does nothing to the determinant).

$R_3 \leftarrow 2R_1 + (-3)R_3$

multiplies the determinant by $-3$.

$R_3 \leftarrow 15R_2 + R_3$

multiplies the determinant by $1$ (i.e. does nothing).

Overall the determinant has been multiplied by a factor of $-1\times-3\times1=3$. So dividing the new determinant by $3$ will give the original determinant.

The problem is that the operations you did were not elementary row operations, but rather compound operations that involved multiplying the individual rows before performing a row operation.

When you are doing the operations, you can swap two rows (that changes the determinant by a factor of $-1$), and you can replace a row $R_i$ with $R_i+aR_j$ for some different $R_j$, but you cannot replace $R_i$ with $aR_i+bR_j$ for any $a\neq 1$, because that is the equivalent to multiplying $R_i$ by $a$ first, which multiplies the determinant by $a$.

Of course, if you are careful to keep track of all the ways you are multiplying, you can keep track of what you're doing to the determinant, and then compensate at the end.