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# 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-1A = 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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