# Thread: [SOLVED] Finding determinant of 4x4 matrix

1. ## [SOLVED] Finding determinant of 4x4 matrix

So I have the matrix

1 0 1 0
0 1 0 1
-2 1 1 0
0 0 1 1

I chose row 3 column 2 and will take it's minor. The minor is

1 1 0
0 0 1
0 1 1

Row two has the most zeros, and so I will select row 2. This leaves me to calculate only row 2 column 3. Which is

1 1
0 1

If I'm not mistaken, the determinant should be 1.

From the 4x4 matrix I have (-1)^3+2 * 1 = 1
From the 3x3 I have only (-1)^2+3 * 1 = 1

All of this just ends up as 1. Am I correct?

2. The determinant should be 4.

3. The correct answer is definatly 4, I used two programs to check
sorry, I can't seem to find where you're going wrong

4. You just forgot to compute some steps...you still have to go back to your original matrix and repeat the same process.

I decided to choose the first row for the first step instead of the 2nd row like you chose cause the first row has the most zeros. I just wrote it by hand cause ... well, it was simpler that way. Tell me if this makes sense!

(Also, those should be -1's, not 1's, although it doesn't make a difference cause they are raised to even powers and become even anyway)

5. Originally Posted by thekrown
So I have the matrix

1 0 1 0
0 1 0 1
-2 1 1 0
0 0 1 1

I chose row 3 column 2 and will take it's minor. The minor is

1 1 0
0 0 1
0 1 1

Row two has the most zeros, and so I will select row 2. This leaves me to calculate only row 2 column 3. Which is

1 1
0 1

If I'm not mistaken, the determinant should be 1.

From the 4x4 matrix I have (-1)^3+2 * 1 = 1
From the 3x3 I have only (-1)^2+3 * 1 = 1

All of this just ends up as 1. Am I correct?
It is not at all clear what you are doing! Since you say "I chose row 3 column 2", you are expanding on either the third row or the second column but what you do after that is confusing. You may be just finding minors for a single entry. you can't do that- you expand on an entire row or column. Since the third row has three non-zero entries and the second column only two, it is easier to expand on the second column:

$\displaystyle \left|\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ -2 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{array}\right|$$\displaystyle = 1\left|\begin{array}{ccc}1 & 1 & 0 \\ -2 & 1 & 0 \\ 0 & 1 & 1\end{array}\right|$$\displaystyle - 1\left|\begin{array}{ccc}1 & 1 & 0 \\ 0 & 0 & 1\\ 0 & 1 & 1\end{array}\right|$

Strictly speaking there would two more minors but they would be multiplied by the "0"s in those positions.

To evaluate those two determinants, expand the first one on the last column:
$\displaystyle \left|\begin{array}{ccc}1 & 1 & 0 \\ -2 & 1 & 0 \\ 0 & 1 & 1\end{array}\right|= 1\left|\begin{array}{cc}1 & 1 \\- 2 & 1\end{array}\right|= 1(1+ 2)= 3$
Expand the second on the second row:
$\displaystyle \left|\begin{array}{ccc}1 & 1 & 0 \\ 0 & 0 & 1\\ 0 & 1 & 1\end{array}\right|$$\displaystyle = -1\left|\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right|= -1(1- 0)= -1$.

The reason for the choice of the last column and second row was that there was only one non-zero entry so we needed only one minor.

The original determinant is 3- (-1)= 4.