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 $= -24$ .
Best Answer
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 $-24$.