LINEAR ALGEBRA NOTES

Elementary, Identity & Inverse Matrices

Two Operations, Two Identities, Two Inverses

When working with matrices, we use 2 mathematical operations

addition and multiplication

and for each of these, there is an identity operator, and an inverse operator.

That is to say, there exists a "guy" such that

matrix + guy = matrix and matrix % guy = matrix for identity

and matrix + guy = identity and matrix % guy = identity for inverse.

The same, of course, is true for real numbers.

Additive Identity

The additive identity for real numbers is zero (0)
because a + 0 = 0 + a = a -- we add something but
change nothing. We still have a.

For Matrices, the additive identity is the zero matrix. (all entries = 0)

It changes nothing in matrix A when we add it to A.

The additive inverse is the "guy" we add to A to get back to 0.
In the set of Real numbers and Matrices, that "guy" is – A since A + ( – A) = 0 in both sets.

Multiplicative Identities

The multiplicative identity for Real numbers is one (1)
because a(1) = (1)a = a -- we multiply by something and change nothing. We still have a.
Therefore the multiplicative
inverse is the "guy" we multiply by to get back to 1.
That is: A
% guy = 1
In the set of Real numbers, that "guy" is a – 1 = (1/a) -- the reciprocal -- a – 1(a) = 1.

But multiplying Matrices is very different from multiplying real numbers, so
some matrices have no inverse A – 1.

Identity Matrices

Definition: an identity matrix I n is a square matrix such that
every entry on the main diagonal = 1 and every other entry = 0.
n is the number of rows and columns in the square matrix In

The Identity matrix is the multiplicative identity (1) for matrices.

In (An % m) = (An % m)Im = An % m.

Notice that a square Identity matrix can multiply a non-square matrix.

Note: Any matrix A can be expressed as AIn

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Inverse of a Matrix

Definition: If A is a square matrix and there exists a matrix B such that AB = BA = In
then A is said to be invertible and B is the inverse of A. (B = A
– 1).

Theorem: The inverse of an invertible matrix is unique.

Theorem: If A and B are invertible matrices of the same size then
AB is invertible and
(AB)– 1 = B – 1 A – 1.
That is: a product of invertible matrices is always invertible and the inverse
is the product of the inverses in the reverse order.

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Other Properties of the Inverse Matrix

An = AAA....A (n times) A0 = I A – n = A – 1A – 1.....A – 1 (n times)

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Theorem: If A – 1 exists then:

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Algorithm for the Inverse of a 2 % 2 Matrix

If A is a 2 % 2 matrix, and A is invertible, we can find A – 1 like this:

If det A is not = 0, then A – 1 exists and is equal to

***Note: This algorithm only works for 2 % 2 matrices.

Example 1

Find the inverse of

Det A = 9 – 6 = 3, therefore A – 1 exists. So,

So, the inverse of a 2 % 2 invertible matrix is
1 over det A, times the matrix we get from:
1) switching the main diagonal entries of A
and
2) negating the minor diagonal entries of A.

To prove that one matrix is the inverse of another, we show that their product = I
no matter which side we multiply on. That is, we show that AB = I = BA.

Let's show that the answer we got in example 1 really is the inverse matrix.

Example 2: Show that the 2 Matrices above are inverses.

To prove it properly, we must show that AB = I = BA

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ELEMENTARY MATRICES

Definition: A square matrix E is elementary if it is In altered by one row operation.

Example 3:

What row operation was performed on In to get these elementary matrices?

Solution:

In A, we switched row 1 & 2.
In B, we multiplied row 2 by – 6.
In C, we added 7 times row 1 to row 3.

Theorem: When we multiply matrix A on the left by E, we change A
with the
same elementary row operation that changed I into E.
That is, if we got E by switching row 1 and row 3 in I,
EA will switch rows 1 and 3 in matrix A.

Example 4:

Find the products AM, BN and CN using matrices A, B and C from example 3.

Solution:

A switches the rows so or M with its rows switched.

B multiplies row 2 by -6 so or B with -6(row 2).

C adds 7(row 1) to row 3 so or C with 7(row 1) + row 3.

Theorem: Every elementary matrix is invertible and its inverse is an elementary matrix.

We simply undo the operation that changed I into E.
In Example 3, A is its own inverse since it will undo the row switch.
The inverse of B multiplies row 2 by ( – 1/6) -- the inverse operation of multiplying by – 6.
The inverse of C subtracts 7(row 1) from row 3.

Definition: Matrices that can be obtained from one another by a finite number of elementary row operations are row equivalent.

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ELEMENTARY MATRIX METHOD OF FINDING A-1

If A is a square matrix, we set up a "side by side" matrix as shown below,
then use row operations to reduce A to
In on the left, and find A – 1 on the right.

Example 5: Use row reduction on matrix A to find A – 1 if

note: oprow is the operator row

A-1 is the 3 % 3 matrix right of the divider line.

Theorem: If A – 1 exists, then the system of equations AX=B has a unique solution X = A – 1B.

Here, A is the coefficient matrix, X is the matrix of variables, and B is the constant column.

We see quite easily that multiplying both sides on the left by A – 1 will give the result.

FUNDAMENTAL PROBLEM OF CONSISTENCY:

We're given AX = B. We need the conditions on B that make the system consistent.
We do row reduction. If A is invertible then there is a solution if the b's are from R.
If A is not invertible, then we will get at least one row of zeros.
So, the condition that renders the system consistent is that the constants
in the zero rows must also be zero.

example: if A, the augmented coefficient matrix reduces to:

then from row 3, b3 – b2 – b1 must = 0
The condition for consistency then is
b3 = b2 + b1

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Practice

1)

For the 4 elementary matrices shown,

a) describe the row operation applied to I to get the matrix.

b) write the inverse of each matrix.

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2)

a) Find E1 , E2 , E3 , such that (E3) (E2) (E1) B = I

b) Find E1' , E2' and E3' such that (E1') (E2' ) (E3') = B

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3) C =

a)

b) Find E1', E2' , E3' E4' such that (E1') (E2' ) (E3') (E4') = C

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4) and

Find E1 and E2 so that B = (E2)(E1)A

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5) ,

Find E1 and E2 so that E1A = E2B

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6) using row reduction, find A-1 if it exists.

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7) Find conditions the b's must satisfy for the system to be consistent.

a) b)
6x – 4y = b1 x – 2y + 5z = b1
3x – 2y = b2 4x + 5y + 8z = b2
  – 3x + 3y –3z = b3

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Solutions

1)

a)

Matrix A : row 2 –5 row 1 Matrix B : 2 times row 1
Matrix C : row 2 + 9 row 3 Matrix D : switch row 1 and row 4

b) inverses shown above.

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2) a)

b)

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3) a)

b)

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4)

5)

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6)

Row operations to find A – 1:

1) switch row 1 and row 3

2) row 2 minus 4(row 1), and row 3 minus 3(row 1)

3) switch row 2 and row 3

4) row 3 minus row 2

5) row 1 minus row 3, and row 2 minus row 3.

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7) a) obviously, b1 = 2 b2

b) after row reduction we find b1 = b2 + b3

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