hw31

1. Evaluate the determinants of the following matrices using only the definition of determinants :
1.

[ 1	2	0	0 ]
[ 0	1	0	0 ]
[ 1	0	0	1 ]
[ 0	0	1	0 ]

[ 1	2	3	4 ]
[ 0	1	2	3 ]
[ 0	0	1	2 ]
[ 1	0	0	1 ]

We evaluate the determinant by taking the sum of

(-1)^p A(p₁,1) A(p₂,2)... A(p_n,n)

as we evaluate all the permutations of the matrix. In each term, we will have one entry from each row and one from each column. To make things simpler, we can realize that any term containing a zero will be zero. To make sure we don't miss any permutations, we lets have maple generate those for us. We will use an arbitrary matrix, with zero's substituted in at the appropriate positions:

> A:=matrix([[a11,a12,0,0],[0,a22,0,0],[a31,0,0,a34],[0,0,a43,0]]);

A :=	[ a₁₁	a₁₂	0	0 ]
	[ 0	a₂₂	0	0 ]
	[ a₃₁	0	0	a₃₄ ]
	[ 0	0	a₄₃	0 ]

> D:=det(A);

D := - a₁₁ a₂₂ a₃₄ a₄₃

As maple makes clear, there is only one term that does not contain any zero values. We can see why this is true by looking at the original matrix. There is only one non-zero entry in each of the second and fourth rows. Since we use only one entry from each row and column, from the first and third rows we need non-zero entries in the first and fourth columns. Since the first row can only provide a non-zero value in the first column, the only non-zero permutation we have is (1,1)(2,2)(4,3)(3,4) (arranged to have the column values in sequential order). As we can easily see, it requires one transposition to return the row values to their proper order, so the sign of this term is (-1)¹, or -1.

Hence, the value of det(A) is -(1*1*1*1), or -1.

We approach the second matrix the same way: this time the original matrix is

[ 1	2	3	4 ]
[ 0	1	2	3 ]
[ 0	0	1	2 ]
[ 1	0	0	1 ]

> A:=matrix([[a11,a12,a13,a14],[0,a22,a23,a24],[0,0,a33,a34],[a41,0,0,a44]]);

A :=	[ a₁₁	a₁₂	a₁₃	a₁₄ ]
	[ 0	a₂₂	a₂₃	a₂₄ ]
	[ 0	0	a₃₃	a₃₄ ]
	[ a₄₁	0	0	a₄₄ ]

> D:=det(A);

D := a₁₁ a₂₂ a₃₃ a₄₄ - a₄₁ a₁₂ a₂₃ a₃₄ + a₄₁ a₁₂ a₂₄ a₃₃ + a₄₁ a₂₂ a₁₃ a₃₄ - a₄₁ a₂₂ a₁₄ a₃₃

We can find these permutations manually without too much difficulty. Beginning with the first column, we see that we have two choices for an entry; a₁₁ or a₄₁. If we start with a₁₁, then the only available entry in the fourth row is a₄₄, which leaves us with a33 in the third row and a22 in the second. Hence A(1,1)A(2,2)A(3,3)A(4,4) is our only permutation begininng with a₁₁. Beginning with a₄₁, we then have two choices in the second column: a₁₂ or a₂₂. Each of these will give us two choices in the third column, and leaving us with our last entry from the necessary position in the fourth column. So A(4,1)A(1,2)A(2,3)A(3,4), A(4,1)A(1,2)A(2,3)A(3,4), A(4,1)A(2,2)A(1,3)A(3,4), and A(4,1)A(2,2)A(3,3)A(1,4) our our remaining non-zero permutations.

Now we need to find signs on all of these. The first permutation A(1,1)A(2,2)A(3,3)A(4,4) is already in trivial order, and so has a positive sign. A(4,1)A(1,2)A(3,3)A(2,4) requires three transpositions, and so has a negative sign. A(4,1)A(1,2)A(2,3)A(3,4) requires four transpositions, and so has a positive sign. A(4,1)A(2,2)A(1,3)A(3,4) requires four transpositions, and so has a positive sign. A(4,1)A(2,2)A(3,3)A(1,4) requires five transpositions, and so has a negative sign. So substituting values from the original matrix, we find the determinant to be:
+(1*1*1*1) -(1*2*2*2) +(1*2*1*3) +(1*1*3*2) -(1*1*1*4)
= 1 - 8 + 6 + 6 - 4
= 1

2. If the determinant of this matrix

[ a	b	c	d ]
[ k	l	m	n ]
[ p	q	r	s ]
[ x	y	z	t ]

is 5. What is this determinant of this matrix

[ 3a	3b	3c	3d ]
[ k-x	l-y	m-z	n-t ]
[ p	q	r	s ]
[ x	y	z	t ]

In order to find the determinant of this matrix, we must determine what steps were taken to derive the second matrix from the first.

1) First we multiply the first row by 3

[ 3a	3b	3c	3d ]
[ k	l	m	n ]
[ p	q	r	s ]
[ x	y	z	t ]

2)Then we added the third row to the first row

[ 3a+p	3b+q	3c+r	3d+s ]
[ k	l	m	n ]
[ p	q	r	s ]
[ x	y	z	t ]

3) Then the fourth row is multiplied by -1 and added to the second row

[ 3a+p	3b+q	3c+r	3d+s ]
[ k-x	l-y	m-z	n-t ]
[ p	q	r	s ]
[ x	y	z	t ]

4) Then the fourth and the third row were swaped

[ 3a+p	3b+q	3c+r	3d+s ]
[ k-x	l-y	m-z	n-t ]
[ x	y	z	t ]
[ p	q	r	s ]

Now, we look at the steps taken to derive this matrix and apply the theorems of determinants from the notes to them.

1) If we multiply a row of a matrix by a number, the determinant will be multiplied by the same number. Therefore, the determinant of 5 is changed to 15 by this row operation.

2) If we add a row to another row, the determinant does not change. Therefore, the determinant stays at 15 through this row operation.

3) If we add a row multiplied by a scalar k to another row, the determinant will not change. Therefore, the determinant stays at 15

4)If we swap two rows of a matrix, the determinant will change its sign. Therefore, the determinant is changed to -15 by this row operation.

The determinant of this final matrix is therefore -15.