How Do You Simplify the Determinant of This 4x4 Matrix?

[erodger]

Homework Statement

Hello, I am stuck on this particular question for my homework.

It is a 4x4 Matrix that consists of

a b b b

b a b b

b b a b

b b b b

The Attempt at a Solution

My approach has been to factor out the a to give the first row of 1 b b b and then use that to make the first column consist of

1
0
0
0

and so i can then do the cofactor expansion and reduce it to a 3x3 matrix. but after that step, it gets extremely tedious and i believe that the approach must be wrong.

Can anyone point out a simpler approach to this question or was I on the right track and just have to endure the tedious algebra?

 

[pwsnafu]

My approach has been to factor out the a to give the first row of 1 b b b

How did you do that? If you factor out a, you get [itex]1 \quad \frac{b}{a} \quad \frac{b}{a} \quad \frac{b}{a}[/itex]

Can anyone point out a simpler approach to this question

Hint: What do you know about block matrices? Specifically, if each block is 2x2 and you have 2x2 of them?

 

[ehild]

It is a 4x4 Matrix that consists of

a b b b

b a b b

b b a b

b b b b




Can anyone point out a simpler approach to this question or was I on the right track and just have to endure the tedious algebra?

Why not subtracting the fourth row from all others?


ehild

 

[erodger]

yeah that fourth row subtraction may have been the best... oh well i solved it by converting that into 4 3x3 matrices and then solving all of those 3x3 matrices.

thanks for the replies though.

 

[ehild]

Subtracting the fourth raw from all other rows, you get the determinant

(a-b) 0 0 0
0 (a-b) 0 0
0 0 (a-b) 0
b b b b

Expand with respect to the fourth column.

ehild

 

[HallsofIvy]

In fact, because that is a "lower triangular" matrix, its determinant is just the product of the numbers on the main diagonal.

In any case, your suggestion that he subtract the last row from each of the other rows was excellent.

 

[ehild]

In fact, because that is a "lower triangular" matrix, its determinant is just the product of the numbers on the main diagonal.
.

I knew it but I could not tell the complete solution.

ehild

 

[HallsofIvy]

Well, he still has to do the multiplication!