# Find the triangular matrix and determinant.

I have a 4x4 matrix and I want to find the triangular matrix (lower half entries are zero).

$$A= \begin{bmatrix} 2 & -8 & 6 & 8\\ 3 & -9 & 5 & 10\\ -3 & 0 & 1 & -2\\ 1 & -4 & 0 & 6 \end{bmatrix}$$

Here are the elementary row operations I performed to get it into triangular form.

row swap rows 1 and row 4

$r_2 - 3\cdot r_1$ replacing $r_2$

$r_3 + 3\cdot r_1$ replacing $r_3$

$r_4 - 2\cdot r_1$ replacing $r_4$

I get this matrix

$$A= - \begin{bmatrix} 1 & -4 & 0 & 6\\ 0 & 3 & 5 & -8\\ 0 & -12 & 1 & 16\\ 0 & 0 & 6 & -4 \end{bmatrix}$$

I then did $4\cdot r_2 + r_3$ to replace $r_3$ and got

$$A= - \begin{bmatrix} 1 & -4 & 0 & 6\\ 0 & 3 & 5 & -8\\ 0 & 0 & 21 & -16\\ 0 & 0 & 6 & -4 \end{bmatrix}$$

I then did $-21\cdot r_4 + 6\cdot r_3$ to replace $r_4$ and got

$$A= - \begin{bmatrix} 1 & -4 & 0 & 6\\ 0 & 3 & 5 & -8\\ 0 & 0 & 21 & -16\\ 0 & 0 & 0 & -12 \end{bmatrix}$$

I am not sure if I did this correctly but the determinant of the matrix should be -36. When I multiply the diagonal entries it isn't -36. I can't figure out what I am doing wrong.

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"I then did -21*row 4 + 6*row 3 to replace row 4 and got"

This is a determinant altering operation and not an elementary operation.

Don't write that $A$ equals something which isn't $A$.

Picking up where you errored and using the same idea you had one gets:

\begin{align} \begin{bmatrix} 1 & -4 & 0 & 6\\ 0 & 3 & 5 & -8\\ 0 & 0 & 21 & -16\\ 0 & 0 & 6 & -4 \end{bmatrix}&\leadsto \begin{bmatrix} 1 & -4 & 0 & 6\\ 0 & 3 & 5 & -8\\ 0 & 0 & 6\cdot 21 & -6\cdot 16\\ 0 & 0 & -21\cdot 6 & (-21)\cdot (-4) \end{bmatrix}\\ &\leadsto \begin{bmatrix} 1 & -4 & 0 & 6\\ 0 & 3 & 5 & -8\\ 0 & 0 & 6\cdot 21 & -16\\ 0 & 0 & 0 & -12 \end{bmatrix}_.\end{align}

Making the proper compensation yields $$\det(A)=-\dfrac{1\cdot 3\cdot (6\cdot 21)\cdot (-12)}{-21\cdot 6}=-36.$$

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A part of the theorem in my book says that If a row of A is multiplied by k to produce a matrix B, then det B is k det A. I think that is what you used but I cannot seem to understand it. Could you explain that to me? – Kot Jul 20 '14 at 1:39
Let $X$ be the first matrix in my answer and $Y$ the second. In the first $\leadsto$ one gets $\det(Y)=6\cdot (-21)\cdot \det(X)$, therefore $\det(X)=\dfrac{\det(Y)}{-21\cdot 6}$. – Git Gud Jul 20 '14 at 1:43
Ahh I see, you multiplied row 3 and 4 by constants as a separate step then added them together. – Kot Jul 20 '14 at 1:47
@Kot Yes, my first $\leadsto$ is actually two elementary operations combined. – Git Gud Jul 20 '14 at 1:54