Use the method of elimination to evaluate the determinants

Question

So this is my process, but I used calculator to check if I got them right or not. and it seems like i got both of them wrong. number 6 supposed to be 36 and 1 supposed to be 135. can anyone please explain where did i do wrong?

Answer 1

The operation $\frac{1}{3}R_3$ changes the determinant, dividing it by $3$; similarly, the operation $R_2+2R_3\to R_3$ multiplies the determinant by $2$.

Further, the last operation leaves $1$ in position $(3,3)$. As a consequence, the determinant is $$ 1\cdot(-2)\cdot1\cdot 3\cdot\frac{1}{2}=-3 $$ which the computer confirms:

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parisize = 8000000, primelimit = 500000
? a=[1,-3,-3;-1,1,2;2,-3,-3]
%1 = 
[ 1 -3 -3]

[-1  1  2]

[ 2 -3 -3]

? matdet(a)
%2 = -3

The same program confirms that the second determinant is $135$; indeed, the last operation you do multiplies the determinant by $-29$; the others don't modify it.

Answer 2

For the first determinant, there is a final error: the last determinant should be $$\begin{vmatrix} 1&-3&-3 \\ 0&-2&-1 \\ 0&\phantom{-}0&\phantom{-}\color{red}1 \end{vmatrix}, $$ so the final determinant is $-2$.

There are also two conceptual errors:

• The step $\frac13 R_3\to R_3$ multiplies the determinant by $\frac13$;
• the step $R_2+2R_3\to R_3$ multiplies the determinant by $2$.

So to get the original determinant, you have to multiply the final determinant by $\frac32$, which yields $\color{red}{-3}$ as the sought determinant.