Activity 2.3
II. FINDING THE INVERSE OF A MATRIX
2.
Determinant is not zero, therefore inverse matrix exists. SOLUTION:
Write the augmented matrix
A1
A2
B1
B2
1
1
-2
1
0
2
2
3
0
1
Find the pivot in the 1st column in the 1st row
A1
A2
B1
B2
1
1
-2
1
0
2
2
3
0
1
Multiply the 1st row by 2
A1
A2
B1
B2
1
2
-4
2
0
2
2
3
0
1
Subtract the 1st row from the 2nd row and restore it
A1
A2
B1
B2
1
1
-2
1
0
2
0
7
-2
1
Make the pivot in the 2nd column by dividing the 2nd row by 7
1
A1
A2
B1
B2
1
-2
1
0
2
0
1
-2/7
1/7
Multiply the 2nd row by -2
A1
A2
B1
B2
1
1
-2
1
0
2
0
-2
4/7
-2/7
Subtract the 2nd row from the 1st row and restore it
A1
A2
B1
B2
1
1
0
3/7
2/7
2
0
1
-2/7
1/7
There is the inverse matrix on the right
A1
A2
B1
B2
1
1
0
3/7
2/7
2
0
1
-2/7
1/7
3.Determinant is not zero, therefore inverse matrix exists. SOLUTION:
1.
Write the augmented matrix
A1
A2
A3
B1
B2
B3
1
1
1
2
1
0
0
2
3
1
0
0
1
0
3
-2
0
3
0
0
1
2.
Find the pivot in the 1st column in the 1st row
A1
A2
A3
B1
B2
B3
1
1
1
2
1
0
0
2
3
1
0
0
1
0
3
-2
0
3
0
0
1
3.
4.
Multiply the 1st row by 3
A1
A2
A3
B1
B2
B3
1
3
3
6
3
0
0
2
3
1
0
0
1
0
3
-2
0
3
0
0
1
Subtract the 1st row from the 2nd row and restore it
A1
A2
A3
B1
B2
B3
1
1
1
2
1
0
0
2
0
-2
-6
-3
1
0
3
-2
0
3
0
0
1
5.
Multiply the 1st row by -2
A1
A2
A3
B1
B2
B3
1
-2
-2
-4
-2
0
0
2
0
-2
-6
-3
1
0
3
-2
0
3
0
0
1
6.
Subtract the 1st row from the 3rd row and restore it
A1
A2
A3
B1
B2
B3
1
1
1
2
1
0
0
2
0
-2
-6
-3
1
0
3
0
2
7
2
0
1
7.
Make the pivot in the 2nd column by dividing the 2nd row by -2
A1
A2
A3
B1
B2
B3
1
1
1
2
1
0
0
2
0
1
3
3/2
-1/2
0
3
0
2
7
2
0
1
8.
Subtract the 2nd row from the 1st row1-ой
A1
A2
A3
B1
B2
B3
1
1
0
-1
-1/2
1/2
0
2
0
1
3
3/2
-1/2
0
3
0
2
7
2
0
1
9.
Multiply the 2nd row by 2
A1
A2
A3
B1
B2
B3
1
1
0
-1
-1/2
1/2
0
2
0
2
6
3
-1
0
3
0
2
7
2
0
1
10. Subtract the 2nd row from the 3rd row and restore it
A1
A2
A3
B1
B2
B3
1
1
0
-1
-1/2
1/2
0
2
0
1
3
3/2
-1/2
0
3
0
0
1
-1
1
1
11. Find the pivot in the 3rd column in the 3rd row
1
A1
A2
A3
B1
B2
B3
1
0
-1
-1/2
1/2
0
2
0
1
3
3/2
-1/2
0
3
0
0
1
-1
1
1
12. Multiply the 3rd row by -1
A1
A2
A3
B1
B2
B3
1
1
0
-1
-1/2
1/2
0
2
0
1
3
3/2
-1/2
0
3
0
0
-1
1
-1
-1
13. Subtract the 3rd row from the 1st row and restore it
A1
A2
A3
B1
B2
B3
1
1
0
0
-3/2
3/2
1
2
0
1
3
3/2
-1/2
0
3
0
0
1
-1
1
1
14. Multiply the 3rd row by 3
A1
A2
A3
B1
B2
B3
1
1
0
0
-3/2
3/2
1
2
0
1
3
3/2
-1/2
0
3
0
0
3
-3
3
3
15. Subtract the 3rd row from the 2nd row and restore it
A1
A2
A3
B1
B2
B3
1
1
0
0
-3/2
3/2
1
2
0
1
0
9/2
-7/2
-3
3
0
0
1
-1
1
1
16. There is the inverse matrix on the right
A1
A2
A3
B1
B2
B3
1
1
0
0
-3/2
3/2
1
2
0
1
0
9/2
-7/2
-3
3
0
0
1
-1
1
1
ASSIGNMENT 2.4
10. Proof. Indeed, it is easy to see that for every elementary operation q if we apply q and then q-1 then we return
to the original matrix. In particular, if we apply q and then q-1 to I, we return to I. But by the theorem about
elementary matrices, application of a row operation to a matrix is equivalent to multiplying this matrix by the
corresponding elementary matrix. Thus let E be the elementary matrix corresponding to the operation q, let F be
the elementary matrix corresponding to the operation q-1. Then
I=F*E*I=F*E,
I=E*F*I=E*F
Thus E is invertible and F is the inverse of E.
11. F - The product would result in two elementary row operations, when an elementary matrix consists of only
one elementary row operation.
12. T
13. F
- Just like an expression of the matrix A as a product of elementary matrices is also not unique.
14. T
- supported on page 74 of the book (REMARKS section)
15. F - Multiplication of a matrix by a scalar is not a single elementary row operation so it cannot be represented
by a corresponding elementary matrix.
16. T
- supported by Theorem 2.13 of the book
17. F - The elementary matrix is obtained from the Identity matrix by using any elementary row (column)
operations. We never get the zero matrix from identity matrix by using any row (column) operations. Therefore,
the given statement “The zero matrix is an elementary matrix” is false.
18. F - All elementary matrix are square but not all square matrices are elementary matrices.
19. T - A square matrix is nonsingular when it can be written as the product of elementary matricies.
20. T - S suppose A has only trivial solution ,i.e nullity (A)=0;now rank -nullity theorem says that
rank(A)+nullityA)=n;where A is n*n matrix. Hence, it is true.
ACTIVITY 3.1
43. a. - F. The determinant is the difference of the products of the two diagonals of the matrix.
b. - T. The determinant of a matrix of order 1 is simply the entry of the matrix. For example, if A = [−2], then
det(A) = −2.
c. - F. If A is a square matrix, then the minor Mij of the entry aij is the determinant of the matrix obtained by
deleting the ith row and jth column of A. The cofactor Cij of the entry aij is Cij = (−1)i+ jMij.
ACTIVITY 3.2
26.
Row
Operation
1:
1
Row
Operation
2:
1
Row
Operation
3:
1
1
1
2 -1 -2
1 -2 -1
1
1
0 -3 -4
1 -2 -1
1
add -2 times the 1st row to the 2nd row
the determinant does not change
1
0 -3 -4
0 -3 -2
1
0 -3 -4
1 -2 -1
1
add -1 times the 1st row to the 3rd row
the determinant does not change
1
1
0 -3 -4
0 -3 -2
1
1
1
1
1
4
multiply the 2nd row by -1/3
the determinant is multiplied by -1/3
0
1
3
0 -3 -2
Row
Operation
4:
1
1
0
1
1
4
3
1 1 1
add 3 times the 2nd row to the 3rd row
the determinant does not change
0 -3 -2
Row
Operation
5:
0 1
3
3
0 0 2
1 1 1
4
4
0 1
1 1 1
multiply the 3rd row by 1/2
the determinant is multiplied by 1/2
4
0 1
3
0 0 2
0 0 1
Step 2: Calculate the determinant
1
Let us denote the original matrix: A =
1
1
2 -1 -2
1 -2 -1
1 1 1
4
and the row echelon form obtained above: B =
0 1
3
0 0 1
We then have the following relationship:
det(A) (-1/3) (1/2) = det(B).
Since det(B) = 1 , we have,
det(A) = -6.
37.
a. T – Theorem 3.3 Part 1 stated that when B is obtained from A by interchanging two rows of A, det (B) =
- det (A)
b. T – Theorem 3.3 Part 3 stated that when B is obtained from A by multiplying a row of A by nonzero
constant c, det (B) = c det (A)
c. T – Theorem 3.4 Part 2 states that two rows (or columns) are equal if A is a square matrix and the first
condition stated is also true, then det(A) = 0.
38.
a. F – by Theorem 3.3 – Part 2. We know that adding a multiple of a row of a matrix to another does not
change the determinant
b. T – by definition of column equivalent matrices
c. T – by theorem 3.4