# Linear Equations and Regular Matrices

**Theorem 5** Let A be a square matrix. Then the column vectors of A are

linearly independent if and only if det(A) ≠ 0.

A square matrix A with det(A) ≠ 0 is called regular; a square matrix A

with det(A) = 0 is called singular.

When the number n of rows and columns of A is large, the formula for

computing det(A) is quite complicated, and calculating det(A) is best left to

the computer. You will only be responsible for computing det(A) by hand

only if A has dimensions 2 × 2, that is, if

In this case the formula for the determinant becomes:

**Exercise 6** Consider the matrices

(a) Find det(M) and det(N).

(b) Use the result of (a) to determine whether the column vectors of M are

linearly independent ; similarly for the column vectors of N.

(c) Try to derive the results you obtained in point (b) directly from the

definition of linear independence.

Now let A be a square matrix. An inverse matrix of a
square matrix A

is a matrix A^{-1} such that A^{-1}A = I. Recall that the identity matrix I acts

with respect to matrix multiplication like the number 1 for multiplication

of numbers, that is, BI = B = IB whenever these products are defined .

Thus A^{-1} can be thought of as a kind of reciprocal of a matrix A. This

analogy can be spun a little further: Not every real number has a reciprocal

(0 being the exception here) and similarly, not every square matrix has an

inverse. However, if A^{-1} exists, then A^{-1} is uniquely defined (thus we can

speak about the inverse matrix of A), and we also have AA^{-1} = I. A square

matrix that has an inverse matrix A^{-1} is called invertible.

Finding inverse matrices requires quite extensive computations and we

will leave these to the computer. However, if we want to verify that a given

matrix B is the inverse matrix of a given square matrix A, all we need to

do is to compute the matrix product BA (or AB) and check whether the

product evaluates to I.

**Exercise 7** Let

Verify that B is the inverse matrix of A.

Now let us look at the system (1) in yet another way. Let

**Exercise 8** By writing out the definition of matrix
multiplication, convince

yourself that is a solution of system (1) if and only if

Equation (19) is called the matrix version of system (1);
it says the same

as (1), only in matrix notation. Note that Equation (19) looks very similar

to the linear equation

where a; b are given real numbers and x is an unknown real
number. If

a ≠ 0, we can solve Equation (20) by multiplying both sides of it by a^{-1}

and we get:

Now suppose A is an invertible square matrix. Let us
multiply both

sides of (19) from the right by A^{-1}. We get:

Thus as long as the coefficient matrix of (19) is an
invertble square

matrix, we can solve the system by multiplying the right-hand side of (19)

by the inverse of the coefficient matrix.

**Exercise 9** Show that the inverse matrix of the coefficient matrix of the

system (4) of Exercise 2 is given by

and use this fact to solve system (4).

You will have noticed that the solution of (4) that you found in Exercise 9

is unique. It has to be unique if A has an inverse matrix, since both the

inverse matrix and the product of matrix multiplication on the right-hand

side of (22) are uniquely determined. Thus we have noticed an important

connection: If the coefficient matrix A of (1) is invertible, then (1) has a

unique solution. It can be shown that this also works the other way around:

If a linear system (1) with a square coefficient matrix A has a unique solution,

then this coefficient matrix A must be invertible.

So, when, exactly, is a square matrix invertible? Let us take a square

matrix A and consider the homogeneous system

As we just saw, A is invertible if and only if (24) has a
unique solution. As

you showed in Exercise 1, the zero vector is always a solution of (24),

and we found earlier in this note that is the unique solution of (24) if and

only if the column vectors of A are linearly independent. By Theorem 5 in

turn, this is the case if and only if det(A) ≠ 0. Thus a square matrix is

invertible if and only if it is regular.

We have considered a lot of properties of square matrices
in this note.

These properties are stated in different terminology , but in the end, they

all say the same thing about the matrix. Mathematicians refer to this kind

of situation by saying that the properties are equivalent. Let us summarize

what we have learned in a single theorem:

**Theorem 10** Let A be a square matrix. Then the following conditions are

equivalent:

1. A^{-1} exists, that is, A is invertible.

2. The column vectors of A are linearly independent.

3. det(A) ≠ 0, that is A is regular.

4. Every system has a unique solution.

5. The zero vector is the unique solution of the homogeneous system

For convenience and completeness, let us also state a version of Theo-

rem 10 for noninvertible matrices:

**Theorem 11** Let A be a square matrix. Then the following conditions are

equivalent:

1. A^{-1} does not exist, that is, A is not invertible.

2. The column vectors of A are linearly independent.

3. det(A) = 0, that is, A is singular.

4. Every system is either inconsistent or underdetermined.

5. The homogeneous system has a nonzero solution
.

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