Solve 3x3 Matrix Determinant: a-1(-a-1-2a^2)+2a^3-10

[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!

 

[mjsd]

you should get
a^2 -9 =0

error in det expansion.

 

[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?

 

[HallsofIvy]

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?

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.

 

[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.

 

[Dick]

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

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.

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.

 

[EugP]

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.

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

 

[Dick]

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

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

 

[radou]

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

+ - +
- + -
+ - +

 

[EugP]

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

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

 

[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!