Homework Due Class 13

1. Compute the determinant of the following matrix using cofactor expansion along the third row:

[ 2	-3	4	1 ]
[ 4	-2	3	2 ]
[ a	b	c	d ]
[ 3	-1	4	3 ]

Using cofactor expansion, we take each term from the third row multiplied by the determinant of the matrix that does not contain that term's row or column and by (-1)^3+j where j is the number of the term. So we obtain:

a*8 - b*(-15) + c*12 - d*19 = 8a + 15b + 12c - 19d

2. Compute the determinant of the following n by n matrix:

[ x	y	0	0	...	0	0 ]
[ 0	x	y	0	...	0	0 ]
[ 0	0	x	y	...	0	0 ]
......................
[ 0	0	0	0	...	x	y ]
[ y	0	0	0	...	0	x ]

Using cofactor expansion along the last row gives us:

(-1)n+1 * y *

det> [ y 0 0 ... 0 0 ] [ x y 0 ... 0 0 ] [ 0 x y ... 0 0 ] ...................... [ 0 0 0 ... x y ]

+ x *

det [ x y 0 0 ... 0 ] [ 0 x y 0 ... 0 ] [ 0 0 x y ... 0 ] ...................... [ 0 0 0 0 ... x ]

Looking at these two matrices that we need the determinants of, we see that the first is a lower-triangular matrix, while the second is upper-triangular. So the first determinant is y^n-1, while the second determinant is x^n-1. So our equation now looks like:

= (-1)ⁿ⁺¹*y*y(n-1) + x*x(n-1)

= (-1)^n-1*yⁿ + x

So if n was even, the solution is xⁿ - yⁿ, while if n is odd, the solution is xⁿ + yⁿ.

3. Compute the determinant of the following n by n matrix by using row operations:

[ 1	2	3	...	n ]
[ -1	0	3	...	n ]
[ -1	-2	0	...	n ]
...................
[ -1	-2	-3	...	0 ]

We can form an upper-triangular matrix from this by adding the first row into every other row. By our theorems on determinants, this will have no effect on the determinant of the matrix. So after these additions, the new matrix looks like:

[ 1	2	3	...	n ]
[ 0	2	6	...	2n ]
[ 0	0	3	...	2n ]
...................
[ 0	0	0	...	n ]

As the resulting matrix is upper-triangular, we can compute the determinant by multiplying the terms along the diagonal. So the determinant is 1*2*3*4* ... *n, which is n!.