Consider the homogeneous linear equation Ax = 0, (when the right-hand-side vector is the zero vector, the equation is called homogeneous.) If A has linearly independent columns, then the solution of Ax = 0 can only be the zero vector. Can you see why? It is because Ax = [a₁, a2,, an] 21 x₁α₁ + + Cnn is essentially a linear combination of the columns of A, where the scaling coefficient of the i-th columna is exactly S Xn the i-th component x; in the solution. When the LC sums up to the RHS zero vector 0, the linear independence of the columns of a1, a2,, an would force the coefficients #₁,...,n to be all zeros! (Otherwise it means the columns of A are linearly dependent.) Now try the above theoretical reasoning to the following equation with a concrete A 1 0 0 1 1 0 1 1 1 1 0 1 Ax = x1 x2 0 Pay attention to the dimension matches here (explicitly written out for you already). Multiply Ax out using standard matrix-vector product, then use LC of column to express Ax, you should get the same product! Then make it equal to the RHS zero vector, then find the solution: 1 type your answer... 2 type your answer... 3 type your answer... Spend some time to convince yourself that this is the only solution you can get. Can any other solution for the above homogeneous linear equation Ax = 0 exist? type your answer... (input only yes or no)
Consider the homogeneous linear equation Ax = 0, (when the right-hand-side vector is the zero vector, the equation is called homogeneous.) If A has linearly independent columns, then the solution of Ax = 0 can only be the zero vector. Can you see why? It is because Ax = [a₁, a2,, an] 21 x₁α₁ + + Cnn is essentially a linear combination of the columns of A, where the scaling coefficient of the i-th columna is exactly S Xn the i-th component x; in the solution. When the LC sums up to the RHS zero vector 0, the linear independence of the columns of a1, a2,, an would force the coefficients #₁,...,n to be all zeros! (Otherwise it means the columns of A are linearly dependent.) Now try the above theoretical reasoning to the following equation with a concrete A 1 0 0 1 1 0 1 1 1 1 0 1 Ax = x1 x2 0 Pay attention to the dimension matches here (explicitly written out for you already). Multiply Ax out using standard matrix-vector product, then use LC of column to express Ax, you should get the same product! Then make it equal to the RHS zero vector, then find the solution: 1 type your answer... 2 type your answer... 3 type your answer... Spend some time to convince yourself that this is the only solution you can get. Can any other solution for the above homogeneous linear equation Ax = 0 exist? type your answer... (input only yes or no)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Consider the homogeneous linear equation Ax = 0, (when the right-hand-side vector is the zero vector, the equation is called homogeneous.)
If A has linearly independent columns, then the solution of Ax = 0 can only be the zero vector. Can you see why? It is because
21
-0-
Xn
Ax = [a₁, a2,, an]
x₁α₁ + + Cnn is essentially a linear combination of the columns of A, where the scaling coefficient of the i-th columna is exactly
the i-th component x; in the solution. When the LC sums up to the RHS zero vector 0, the linear independence of the columns of a1, a2,, an would force the
coefficients #₁,...,n to be all zeros! (Otherwise it means the columns of A are linearly dependent.)
Now try the above theoretical reasoning to the following equation with a concrete A
1
0 0
1
1
0
1 1
1
1
0 1
Ax =
x1
x2
0
Pay attention to the dimension matches here (explicitly written out for you already).
Multiply Ax out using standard matrix-vector product, then use LC of column to express Ax, you should get the same product!
Then make it equal to the RHS zero vector, then find the solution:
1
type your answer...
2
type your answer...
3
type your answer...
Spend some time to convince yourself that this is the only solution you can get.
Can any other solution for the above homogeneous linear equation Ax = 0 exist?
type your answer...
(input only yes or no)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc0bb319e-1598-41b9-bb4c-4a8a1d4b3ce7%2F1311ea63-4513-4c20-aff9-81a4fa3ed4ea%2Fkhmwmvd_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the homogeneous linear equation Ax = 0, (when the right-hand-side vector is the zero vector, the equation is called homogeneous.)
If A has linearly independent columns, then the solution of Ax = 0 can only be the zero vector. Can you see why? It is because
21
-0-
Xn
Ax = [a₁, a2,, an]
x₁α₁ + + Cnn is essentially a linear combination of the columns of A, where the scaling coefficient of the i-th columna is exactly
the i-th component x; in the solution. When the LC sums up to the RHS zero vector 0, the linear independence of the columns of a1, a2,, an would force the
coefficients #₁,...,n to be all zeros! (Otherwise it means the columns of A are linearly dependent.)
Now try the above theoretical reasoning to the following equation with a concrete A
1
0 0
1
1
0
1 1
1
1
0 1
Ax =
x1
x2
0
Pay attention to the dimension matches here (explicitly written out for you already).
Multiply Ax out using standard matrix-vector product, then use LC of column to express Ax, you should get the same product!
Then make it equal to the RHS zero vector, then find the solution:
1
type your answer...
2
type your answer...
3
type your answer...
Spend some time to convince yourself that this is the only solution you can get.
Can any other solution for the above homogeneous linear equation Ax = 0 exist?
type your answer...
(input only yes or no)
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