Proving a Determinant Identity for Beginners

[Vishalrox]

1. The problem

Prove that

| (a+b-c) (-c+a-b) (a+b+c) |
| (a-c) (c-a) (b-a) | = (a+b-c)(-c+a-b)(a-c)
| (a-b) (a-c) (a+b) |

using properties of determinants without expanding a determinant

2. The attempt at a solution

I tried a lot of ways like with following steps C(3) -> C(3) + C(2) , R(1) -> R(1) + R(2) ,
R(1) -> R(1) + R(2) - R(3)...but i could get nothing...its not my homework sum...i am just a 7th grade maths enthusiast...i saw this sum somewhere...i tried to solve this but couldn't get...can anyone help me up to solve this...or atleast guide me...where C(1) means column 1 and R(1) means Row 1 and similar things to that

 

[HallsofIvy]

(a+b-c) (-c+a-b) (a+b+c) |
| (a-c) (c-a) (b-a) | = (a+b-c)(-c+a-b)(a-c)
| (a-b) (a-c) (a+b) |
[tex]\left|\begin{array}{ccc}a+b- c & -c+a-b & a+b+c \\ a-c & c-a & b-a \\ a-b & a-c & a+b\end{array}\right|[/tex]

"C(3) -> C(3) + C(2)". Okay, that gives
[tex]\left|\begin{array}{ccc}a+b- c & -c+a-b & 2a \\ a-c & c-a & b+c-2a \\ a-b & a-c & 2a+b-c\end{array}\right|[/tex]

But may I ask why you did that? How does it improve the situation?

Notice that the right side, (a+b-c)(-c+a-b)(a-c), is a product of three numbers, one of which is the upper left element of the determinant. The determinant of a diagonal or triangular matrix is just the product of the numbers on the main diagonal so I would recommend using row operations to change all numbers below the main diagonal to 0. For example, changing R2 to R2- (a-c)/(a+b-c)R1 will make the second row of the first column equal to 0.

 

[Vishalrox]

that was one of the steps i did...i tried with many properties meddled with this a lot...but couldn't get a thing...instead of what you said i would just add R(2) and R(3) and get a 0..(R2 -> R2 + R3)...so that's aint the problem...u will get zeros...bu the problem gets more complicated i u proceed...so what i want is...a stock algorithm to get the answer...

 

[Ray Vickson]

1. The problem

Prove that

| (a+b-c) (-c+a-b) (a+b+c) |
| (a-c) (c-a) (b-a) | = (a+b-c)(-c+a-b)(a-c)
| (a-b) (a-c) (a+b) |

using properties of determinants without expanding a determinant

2. The attempt at a solution

I tried a lot of ways like with following steps C(3) -> C(3) + C(2) , R(1) -> R(1) + R(2) ,
R(1) -> R(1) + R(2) - R(3)...but i could get nothing...its not my homework sum...i am just a 7th grade maths enthusiast...i saw this sum somewhere...i tried to solve this but couldn't get...can anyone help me up to solve this...or atleast guide me...where C(1) means column 1 and R(1) means Row 1 and similar things to that

The claimed result is false. Let D denote the determinant and R the expression on the right. When a=1, b=2, c=3 we have D = 16 but R = 0. When a=-1, b=0, c=2 we have D = 24 but R = -27.

There must be something wrong with the expressions you wrote.

RGV

 

[HallsofIvy]

Thanks, Ray. I didn't check.

 

[Vishalrox]

Oh thanks...