The Identity Matrix , Matrix Inverses, and Matrix Equations

The identity matrix , I, is an nxn (square of size n) matrix. If a matrix A can be multiplied on the left by I, that is if A has n rows, then IA=A.

The 2x2 identity is . the 3x3 identity is

These are the only ones we will use.

If A is square (nxn) and EA=I for some matrix E, then E is called the inverse of A, E= A.

Not all square matrices have inverses. A matrix that does not have an inverse is called singular. We will use the calculator to find inverse matrices. The calculator will show

"error singular matrix" if there is no inverse.

When a system of n equations in n variables (nxn system) has a unique solution the matrix of coefficients has an inverse and we can solve the system as follows:

Let A be the matrix of coefficients, not augmented by the constant column.

The system can be written as the matrix equation AX = B where B is the column of constants and X is the column of variables, X= if n=3.

Multiplying the matrix equation, AX = B, on the left by the inverse of A we have:

AAX = AB. Then since A A= I and IX=X

X= AB.

Example: Solve the matrix equation

The solution is

Enter A= in the calculator. Enter B=. Using the key find AB which is .

Compare this to rref(

So what is the advantage?

Sometimes we can solve other matrix equations and using an inverse matrix is more convenient.

Solve the matrix equation AX+3X=B for a given nxn matrix, A, and nx1 column, B.

1) Factor X out of AX+3X. With numbers, the expression ax+3x is (a+3)x.

But A+3 does not make sense in matrices. We cannot add a number to a matrix.

But we can say, AX+3X=AX+3IX = (A+3I)X

Now the equation looks like (A+3I)X = B

2) If A+3I has an inverse then X=(A+3I)B

For example: A= and B = and AX+3X = B we have

X = The calculator instructions for finding X follow:

Enter matrices A and B.

Type (A+3matrix>math5 3))B matrix>math5 thells the calculator we want the identity matrix and The 3(following matrix>math5 ) tells the calculator we want the 3x3 identity.

Summary problems:

Solve the system of equations x + z = s, 2x – y = 1, 3y + z = -2.

First write the 3x3 matrix of coefficients and enter it into matrix A in your calculator.

Find the inverse to A.

By hand, multiply the inverse of A by the column of constants. You will have s in the answer.

For the same A as above, solve for X, the matrix equation 4AX – 2X = B where B is a column 30, -60, 120.

(4A – 2I(3))^(-1) B should give x= -1, y=8.67, z=