# Det of a 3x3 Matrix

1. Aug 1, 2007

### Cmunro

The question is as follows:

Solve for a: |(a-1) ( 1) (0) |
.................|(-10) (a+1) (a^2) | =0
.................|(2a) ( 2) (-1) |

(Sorry that is my attempt at the determinant of a 3x3 matrix - the brackets are there to show which bit goes with which as they seem to group together)

My calculations: a-1(-a-1-2a^2) + (2a^3 -10)
-2a^2-2a-a^2-a-a-1-2a^3-10=0
2a^3-3a^2-4a-11=0

Since we have not covered binomial expansion yet..I can only assume that I have made a mistake in the calculations here. Any suggestions?

Thank you!

Last edited: Aug 1, 2007
2. Aug 1, 2007

### mjsd

you should get
a^2 -9 =0

error in det expansion.

3. Aug 1, 2007

### Cmunro

You're right.. I lost a ^2 in my calculations. I tried to work that out.. but I got this:

a-1(-a-1-2a^2) +(2a^3 -10)
(-a(a-1) -1(a-1)-2a^2(a-1) +(2a^3 -10)
-a^2+a-a+1-2a^3+2a+2a^3-10
-a^2+2a-9=0

Is this right or did I go wrong again?

4. Aug 1, 2007

### HallsofIvy

Expanding by the first row,
(a-1)((a+1)(-1)- (2)(a^2))- (1)((-10)(-1)- (2a)(a^2))
(a-1)(-a-1) + (2a^3- 10)
I don't see where you got the "(-a-1-2a^2)" in the first term.

5. Aug 1, 2007

### EugP

$$\begin{array}{|l cr| }a - 1&1&0\\-10&a + 1&a^2\\2a&2&-1\end{array}$$

First row expansion:

$$(a - 1)[(a + 1)(-1) - 2a^2] + [(-10)(-1) - 2a^3] - 0[(-10)(2) - 2a(a + 1)] = 0$$

$$(a - 1)[-a - 1) - 2a^2] + [10 - 2a^3] = 0$$

$$a^2 - a + a + 1 - 2a^3 + 2a^2 + 10 - 2a^3 = 0$$

$$-4a^3 + 3a^2 + 11 = 0$$

This is what I get, you should be able to solve it now.

6. Aug 1, 2007

### Dick

Hmmm. Another wrong solution. You messed up the alternating signs as you expanded across the row. mjsd's solution in the first response is correct.

7. Aug 1, 2007

### EugP

Oh, heh sorry and thanks for pointing it out. What is the proper method? It is:$$\Delta_1 + \Delta_2 - \Delta__3$$, right?

8. Aug 1, 2007

### Dick

$$\Delta_1 - \Delta_2 + \Delta__3$$. The sign on each is (-1)^(i+j) where i is the row number and j is the column number.

9. Aug 1, 2007

For the determinant of a 3x3 matrix, it's just what Dick wrote; some find it helpful to use this scheme:

+ - +
- + -
+ - +

10. Aug 1, 2007

### EugP

Ooo, thanks so much. I've been doing this wrong by hand the whole time! Good thing I use my calculator for quick results.

11. Aug 8, 2007

### Cmunro

Sorry I have been away and did not have internet access.

Thank you all so much! (I'm sorry for the delay in thanking you)

It all gets very complicated with the pluses and minuses everywhere, and I find myself losing numbers here and there which is hardly useful. Anyway, thank you, now I see how to get the a^2 -9, so a =3.

Oh and Hallsofivy : no idea where I got "(-a-1-2a^2)" from!