How would I go about showing this DETERMINANT??

Question

If a=
0 x y z
-x 0 1 -1
-y -1 0 1
-z 1 -1 0

show that det(A)=(x+y+z)^2.

How do I do this? You guys don't have to do it for me, just get me started.

Thanks!!

note: spacing is off but it is a 4x4 matrix

Answer 1

This uses properties of determinants. I will use { and } to denote determinants

det(A) = {0 x y z
-x 0 1 -1
-y -1 0 1
-z 1 -1 0}

interchanging C3 and C1
-{y x 0 z
1 0 -x -1
0 -1 -y 1
-1 1 -z 0}

interchanging R1 and R2
{1 0 -x -1
y x 0 z
0 -1 -y 1
-1 1 -z 0}

using R3 = x * R1 + R3 and R4 = R1 + R4
{1 0 0 0
y x xy y+z
0 -1 -y 1
-1 1 -x-z -1}

Simplifying, the other terms are 0 * {a-determinant} so ignored
1 * {x xy y+z
-1 -y 1
1 -x-z -1}

Expanding
x(y + x + z) - xy(1 - 1) + (y + z)(x + z + y)

= x(y + x + z) + (y + z)(x + z + y)

taking (x + z + y) out
= (y + x + z)(x + y + z)
= (x + y + z)^2

Hope it helps :).

Answer 2

Expand the matrix by using the simplification of a 4 x 4 matrix
|a b c d|
|e f g h|
|j k n o|
|p q r s|
= aR - bS + cT - dU

where
R is
|f g h|
|k n o|
|q r s|

S is
|e g h|
|j n o|
|p r s|

T is
|e f h|
|j k o|
|p q s|

U is
|e f g|
|j k n|
|p q r|




After that, use the simplification of the 3 x 3 matrix
|a b c|
|d e f|
|g h k|
= aR - bS + cT

where
R is
|e f|
|h k|

S is
|d f|
|g k|

T is
|d e|
|g h|




Finally, simplify the 2 x 2 matrix
|a b|
|c d|
= ad - bc

^_^

Answer 3

YOu should start to work on your math and school home work .