Linear Algebra Lecture 2

Linear Algebra Lecture 2

1. Elimination

2. Back-substitution

3. Elimination matrices

4. Multiply matrices

Example 1 :

To solve :⎧⎩⎨x+2y+z=2
4000
3x+8y+z=124y+z=2

Here is a system of equations, three equations and three unknowns. Ax = b

Change the equations into matrix form:

A=⎡⎣⎢130284111⎤⎦⎥

Elimination steps :

1.Accept the first equation is okay, multiply that equation by the right number, and going to subtract it from the second equation. With the purpose to knock out the x part of equation two, to eliminate x.

2.Find the first pivot(主元), won’t change the first row, because it’s the pivot row. Then find the multiplier, 3 times first equation will knock out that 3 in the second equation.

⎡⎣⎢130284111⎤⎦⎥→⎡⎣⎢1002241−21⎤⎦⎥

3.We have eliminated x, we are down to two equations in y and z, find the second pivot 2, then eliminate y.

⎡⎣⎢1002241−21⎤⎦⎥→⎡⎣⎢1002201−25⎤⎦⎥

4.Then the matrix is an upper triangular matrix(上三角矩阵), call U. The whole purpose of elimination was to get from A to U.

How could the elimination have failed?

If the first number was 0, there were no x in first equation. Then need switch rows, if there is a zero in the pivot position, need to exchange rows.

How we identify failure?

There’s temporary failure when we can do a row exchange and get out of it or there’s complete failure when we get a 0 and below.

There wouldn’t have been all pivots. The matrix would have not been invertible(矩阵不可逆).

Back substitution

Now bring the right-hand side as an extra column.

⎡⎣⎢1302841112122⎤⎦⎥

This is called augmented matrix(增广矩阵). And we should do the same step to the right-hand side column.

⎡⎣⎢1302841112122⎤⎦⎥→⎡⎣⎢1002241−21262⎤⎦⎥→⎡⎣⎢1002201−2526−10⎤⎦⎥

Then we need to do back substitution.

⎧⎩⎨x+2y+z=22y−2z=65z=−10

Then easy to solve the equations. Here is the back substitution(回代).

⎧⎩⎨x=2y=1z=−2

Back substitution is the simple step solving the equations in reverse order because the system is triangular now.

Elimination matrices

Row operations

Matrix times a columns is a column.

A row times a matrix is a row.

When as we do matrix multiplication, we got a linear combination of rows.

Elimination matrices(消元矩阵)

What’s the matrix that does the first step, subtracts 3 of equation one from equation two and leave the other rows the same?

[⋮⋮⋮]⎡⎣⎢130284111⎤⎦⎥=⎡⎣⎢1002241−21⎤⎦⎥

We are just doing this step changes row two, so actually, row one is not changing.

The first row of the matrix will be [100], because that’s just the right thing that takes one of the first row and none of the other rows. And the last row is [001], takes one of the third row and none of the other rows.

The second row is a matrix [−310], means that takes minus 3 of row one plus the row two.

⎡⎣⎢1−30010001⎤⎦⎥⎡⎣⎢130284111⎤⎦⎥=⎡⎣⎢1002241−21⎤⎦⎥

What matrix does nothing?

⎡⎣⎢100010001⎤⎦⎥

The identity matrix(单位矩阵), it just multiplies everything and leaves it where it is, like the number 1.

Elementary matrix(初等矩阵)

E21 is the matrix we needed to fix the (2,1) position to be 0.

Step 2 is subtract 2 times row 2 from row 3

E32→⎡⎣⎢10001−2001⎤⎦⎥⎡⎣⎢1002241−21⎤⎦⎥=⎡⎣⎢1002201−25⎤⎦⎥

The elimination matrices that take each step.Put those together into a matrix that does it all.

E32(E21A)=U→(E32E21)A=U

What matrix takes me from A to U? E

Permutation

There’s another type of elementary matrix, the matrix that exchanges two rows, it’s called a permutation matrix(置换矩阵).

Permutation = Exchange rows 1 and 2

P→[0110][acbd]=[cadb]

If need to do column exchanges,

[acbd][0110]=[bdac]

In short, to do column operations, the permutation matrix multiplies on the right, to do row operations , it multiplies on the left.

Inverse Matrix

Inverses

⎡⎣⎢130010001⎤⎦⎥⎡⎣⎢1−30010001⎤⎦⎥=⎡⎣⎢100010001⎤⎦⎥

E−1E=I