Find the determine for the given matrix as a two days by usi

Find the determine for the given matrix as a two days, by using cofactor expression (a) along the raw of your choosing and (b) along the column of your choosing use the determinant to decide if T(x) = Ax is invertible A = [4 2 1 0 1 0 3 -1 1 2 0 -1 0 0 1 0 1 2 0 3 0 0 1 0 1]

Solution

Let the given matrix be denoted by A.

(a). The 3^rd and the 5^th rows of the given 5x5 matrix A have 3 zeros each. Thus, it will be relatively easy to compute the value of det(A) if we select the 3^rd or the 5^th row. Let us select the 5^th row. The cofactor of the 1^st 1 in the 5^th row(3^rd column) is

4

2

0

1

0

3

1

2

0

-1

0

1

0

1

0

3

which is equal to 4 multiplied by

3

1

2

-1

0

1

1

0

3

as the other entries in the 1^st column of the above determinant are all 0 and the cofactors of 0 do not matter. Thus, the cofactor of the 1^st 1 in the 5^th row(3^rd column) is 4[ 3(0*3-1*0) -1(-1*3-1*1)+2(-1*0)-0*1)] = 4[0+4+0) = 16.

The cofactor of the 2^nd 1 in the 5^th row(5^th column) is

4

2

1

0

0

3

-1

1

0

-1

0

0

0

1

2

0

which is equal to 4 multiplied by

3

-1

1

-1

0

0

1

2

0

= 4[ 3(0*0 -0*2)-(-1)(-1*0-0*1)+1(-1*2-0*1)] = 4[0+0-2] = -8

Therefore det(A) = 16-8 = 8.

(b). We can select the 1^st or the 4^th column of A for cofactor expansion as these have the maximum number of zeros (4 each). Let us select the 1^st column. Here, the cofactor of 4 only matters as the other entries in rthe 1^st column are all 0. Then det(A) is 4 multiplied by

3

-1

1

2

-1

0

0

1

1

2

0

3

0

1

0

1

On choosing the 3^rd column of this determinant for further cofactor expansion ( as it has 3 zeros), det(A)= 4 multiplied by

-1

0

1

1

2

3

0

1

1

= 4[ -1(2*1-3*1)-0(1*1-3*0)+1(1*1-2*0)] = 4[ 1+0+1] = 4*2 = 8.

Thus det(A) = 8.

Now, since TX = AX and since A is invertible ( as det(A) = 8 0), therefore T is also invertible.

4	2	0	1
0	3	1	2
0	-1	0	1
0	1	0	3