Question 1
Question 2
Question 3
Question 4
Consider the matrix A with attributes {X1, X2, X3}
1
2
0
A = 2
4
0
3
6
1 then,
Number of columns in A = 3
R1 and R3 are linearly independent.
The rank of the matrix A which is the number of
non-zero rows in its echelon form are 2. we have,
AB = 0
Then we get, b1
+ 2*b2 = 0 b3 =
0
The null vector we can get is
The number of parameter in the general solution is the dimension
of the null space (which is 1 in this example). Thus, the sum of
the rank and the nullity of A is 2 + 1 which is equal to the
number of columns of A.
Your matrix
Sign
A1
A2
A3
A4
1
7
49
11
4
2
-6
-56
9
-7
3
12
28
-48
-36
4
-5
-35
8
3
+
Divide the 1st row by 7
Sign
+
A1
A2
A3
A4
1
1
7
11/7
4/7
2
-6
-56
9
-7
3
12
28
-48
-36
4
-5
-35
8
3
Multiply the 1st row by -6
Sign
A1
A2
A3
A4
1
-6
-42
-66/7
-24/7
2
-6
-56
9
-7
3
12
28
-48
-36
4
-5
-35
8
3
+
Subtract the 1st row from the 2nd row and restore it
Sign
A1
A2
A3
A4
1
1
7
11/7
4/7
2
0
-14
129/7
-25/7
3
12
28
-48
-36
4
-5
-35
8
3
+
Multiply the 1st row by 12
Sign
A1
1
12
A2
A3
A4
84
132/7
48/7
2
0
-14
129/7
-25/7
3
12
28
-48
-36
4
-5
-35
8
3
+
Subtract the 1st row from the 3rd row and restore it
Sign
+
1
A1
A2
A3
A4
1
7
11/7
4/7
2
3
4
0
-14
129/7
-25/7
0
-56
-468/7
-300/7
-5
-35
8
3
Multiply the 1st row by -5
Sign
A1
A2
A3
A4
1
-5
-35
-55/7
-20/7
2
0
-14
129/7
-25/7
3
0
-56
-468/7
-300/7
4
-5
-35
8
3
+
Subtract the 1st row from the 4th row and restore it
Sign
A1
A2
A3
A4
1
1
7
11/7
4/7
2
0
-14
129/7
-25/7
3
0
-56
-468/7
-300/7
4
0
0
111/7
41/7
+
Restore the 1st row to the original view
Sign
+
A1
A2
A3
A4
1
7
49
11
4
2
0
-14
129/7
-25/7
3
0
-56
-468/7
-300/7
4
0
0
111/7
41/7
Divide the 2nd row by -14
Sign
A1
A2
A3
A4
1
7
49
11
4
2
0
1
-129/98
25/98
3
0
-56
-468/7
-300/7
4
0
0
111/7
41/7
+
Multiply the 2nd row by -56
Sign
A1
A2
A3
A4
1
7
49
11
4
2
0
-56
516/7
-100/7
3
0
-56
-468/7
-300/7
4
0
0
111/7
41/7
+
Subtract the 2nd row from the 3rd row and restore it
Sign
+
A1
A2
A3
A4
1
7
49
11
4
2
0
1
-129/98
25/98
3
0
0
-984/7
-200/7
4
0
0
111/7
41/7
Restore the 2nd row to the original view
Sign
+
1
A1
A2
A3
A4
7
49
11
4
2
0
3
4
-14
129/7
-25/7
0
0
-984/7
-200/7
0
0
111/7
41/7
Divide the 3rd row by -984/7
Sign
A1
A2
A3
A4
1
7
49
11
4
2
0
-14
129/7
-25/7
3
0
0
1
25/123
4
0
0
111/7
41/7
+
Multiply the 3rd row by 111/7
Sign
A1
A2
A3
A4
1
7
49
11
4
2
0
-14
129/7
-25/7
3
0
0
111/7
925/287
4
0
0
111/7
41/7
+
Subtract the 3rd row from the 4th row and restore it
Sign
+
A1
A2
A3
A4
1
7
49
11
4
2
0
-14
129/7
-25/7
3
0
0
1
25/123
0
4
0
0
108/41
Restore the 3rd row to the original view
Sign
A1
A2
A3
A4
1
7
49
11
4
2
0
-14
129/7
-25/7
3
0
0
-984/7
-200/7
4
0
0
0
108/41
+
Multiply the main diagonal elements
Sign
A1
A2
A3
A4
1
7
49
11
4
2
0
-14
129/7
-25/7
3
0
0
-984/7
-200/7
4
0
0
0
108/41
+
Δ = 7 x (-14) x (-984/7) x 108/41 = 36288
Question 5
Solution of Determinant when Matrix’s second column is divided by 7
Your matrix
Sign
+
A1
A2
A3
A4
1
7
-7
11
4
2
-6
8
9
-7
3
12
-4
-48
-36
-5
4
5
8
3
Divide the 1st row by 7
Sign
A1
A2
A3
A4
1
1
-1
11/7
4/7
2
-6
8
9
-7
3
12
-4
-48
-36
4
-5
5
8
3
+
Multiply the 1st row by -6
Sign
A1
1
-6
A2
6
A3
A4
-66/7
-24/7
2
-6
8
9
-7
3
12
-4
-48
-36
4
-5
5
8
3
+
Subtract the 1st row from the 2nd row and restore it
Sign
+
A1
A2
A3
A4
1
1
-1
11/7
4/7
2
0
2
129/7
-25/7
3
12
-4
-48
-36
4
-5
5
8
3
Multiply the 1st row by 12
Sign
A1
A2
A3
A4
1
12
-12
132/7
48/7
2
0
2
129/7
-25/7
3
12
-4
-48
-36
4
-5
5
8
3
+
Subtract the 1st row from the 3rd row and restore it
Sign
A1
A2
A3
A4
1
1
-1
11/7
4/7
2
0
2
129/7
-25/7
3
0
8
-468/7
4
-5
5
8
+
-300/7
3
Multiply the 1st row by -5
Sign
+
A1
A2
A3
A4
1
-5
5
-55/7
-20/7
2
0
2
129/7
-25/7
3
0
8
-468/7
-300/7
4
-5
5
8
3
Subtract the 1st row from the 4th row and restore it
Sign
+
A1
A2
A3
A4
1
1
-1
11/7
4/7
2
0
2
129/7
-25/7
3
0
8
-468/7
-300/7
4
0
0
111/7
41/7
Restore the 1st row to the original view
Sign
A1
1
7
A2
A3
-7
11
A4
4
2
0
2
129/7
-25/7
3
0
8
-468/7
-300/7
4
0
0
111/7
41/7
+
Divide the 2nd row by 2
Sign
A1
A2
A3
A4
1
7
-7
11
4
2
0
1
129/14
-25/14
3
0
8
-468/7
-300/7
4
0
0
111/7
41/7
+
Multiply the 2nd row by 8
Sign
A1
A2
A3
A4
1
7
-7
11
4
2
0
8
516/7
-100/7
3
0
8
-468/7
-300/7
4
0
0
111/7
41/7
+
Subtract the 2nd row from the 3rd row and restore it
Sign
A1
A2
A3
A4
1
7
-7
11
4
2
0
1
129/14
-25/14
3
0
0
-984/7
-200/7
4
0
0
111/7
41/7
+
Restore the 2nd row to the original view
Sign
A1
A2
A3
A4
1
7
-7
11
4
2
0
2
129/7
-25/7
3
0
0
-984/7
-200/7
4
0
0
111/7
41/7
+
Divide the 3rd row by -984/7
Sign
+
A1
A2
A3
A4
1
7
-7
11
4
2
0
2
129/7
-25/7
3
0
0
1
25/123
4
0
0
111/7
41/7
Multiply the 3rd row by 111/7
Sign
+
A1
A2
A3
A4
1
7
-7
11
4
2
0
2
129/7
-25/7
3
0
0
111/7
925/287
4
0
0
111/7
41/7
Subtract the 3rd row from the 4th row and restore it
Sign
A1
A2
A3
A4
1
7
-7
11
4
2
0
2
129/7
-25/7
3
0
0
1
25/123
4
0
0
0
108/41
+
Restore the 3rd row to the original view
Sign
+
A1
A2
A3
A4
1
7
-7
11
4
2
0
2
129/7
-25/7
3
0
0
-984/7
-200/7
4
0
0
0
108/41
Multiply the main diagonal elements
Sign
A1
A2
A3
A4
1
7
-7
11
4
2
0
2
129/7
-25/7
3
0
0
-984/7
-200/7
4
0
0
0
108/41
+
Δ = 7 x 2 x (-984/7) x 108/41 = -5184
Solutions to 6
Solution to 6
In linear algebra, an orthogonal matrix, or orthonormal matrix, is a real square matrix whose
columns and rows are orthonormal vectors. One way to express this is: QTQ = QQT = I
where QT is the transpose of Q and I is the identity matrix.
This leads to the equivalent characterization: a matrix Q is orthogonal if its transpose is equal to
its inverse:
where Q−1 is the inverse of Q.
We say that 2 vectors are orthogonal if they are perpendicular to each other. i.e. the
dot product of the two vectors is zero. ... A set of vectors S is orthonormal if every
vector in S has magnitude 1 and the set of vectors are mutually orthogonal.
Orthonormal Complex
Question 7
If V
is a
vector space over a field K and if W is a subset of V, then W is a linear subspace of V if
under the operations of V, W is a vector space over K. ... These are called the trivial
subspaces of the vector space. Question 7B
The constant polynomial
whose coefficients are all equal to 0. The corresponding polynomial function is
the constant function with value 0, also called the zero map. The zero polynomial is the additive identity of the
additive group of polynomials.
The degree of the zero polynomial is undefined, but many authors conventionally set it equal to
or
.