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.

"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 $$\det(A)=-\dfrac{1\cdot 3\cdot (6\cdot 21)\cdot (-12)}{-21\cdot 6}=-36.$$

• 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

"I then did −21⋅r4+6⋅r3 to replace r4 and got"..

Whenever we are doing Row operations on a particular row, whatever the co-efficients that we are multiplying on the same row should be taken as a divisor with sign outside of determinant. For example, You have a matrix A and it's determinant is |A|

If we are doing the below operation, R3 -> 3 R2 - 5R3 The operation shout be processed by taking (-1/5) outside. Concept is, we are indirectly multiplying the Row 3 by (-5) through this operation. We no need to worry about the multiplier 3 with R2, as it will not affect the determinant value(we are altering the Row 3, hence remaining Row's co-efficients will not affect |A|).

In your case, Till the final step, the Row operations didn't have co-efficients for the particular Row alterations. In final step you did,
R4 -> -21 R4 + 6 R3 So, you should take (-1/21) outside. After taking outside, the determinant calculation will be like:

|A| = - (-1/21) (1) (3) (21) (-12) = -36