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equation is redundant. An analysis or rank (see below) will reveal which, but elimination can still take place. Irrelevant equations will be removed with rows of zeros appearing, while inconsistent systems will be revealed by impossible equations (a row with zeros except for the right-most column).

6.10 Inversion by Row operations

It is also possible to find the inverse of a matrix entirely via row-operations.

The theory is beyond the scope of this module, but the principle is that to invert A an augmented matrix is formed thus

(A|I)

with the appropriately sized identity matrix to the right of the line. Row operations are undertaken on this matrix until the identity matrix is produced on the left hand side. The matrix on the right will then be the inverse matrix.

I|A⁻¹

providing this is possible. As noted above, for singular matrices a row con-taining zeros left of the line will be produced, which cannot be made into the identity and will signify this.

6.11 Rank

The rank of a matrix A, denoted r(A) is the order of the largest square matrix contained within A that has a non-zero determinant. The submatrix can be formed out of any of the rows and columns of A.

The rank is essentially a measure of the amount of information in a matrix, or alternatively the number of redundant rows in the matrix. Clearly the rank of an m × n matrix cannot exceed the smaller of m and n, as no larger square matrix can be constituted.

6.11.1 Example

Find the rank of the following matrices.

(a)





0 0 0 0 0 0 0 0 0





6.11 Rank 118

(a) It should be clear that the determinant of this 3 × 3 matrix is zero.

Therefore the rank is not 3. One can construct 9 possible submatrices ⁶ that are 2×2, but these again all consist of zeros, and the determinant will be zero.

Therefore the rank cannot be 2. Finally there are 9 possible submatrices that are 1 × 1, which are all zero. Therefore the rank cannot be 1. So the rank in this case is zero.

Can you see that if exactly one number in the matrix was changed to 4 say then the rank would become 1?

(b) We take the determinant of this matrix to obtain 1 6= 0 and therefore the rank of this matrix is 3.

(c) It is left as an exercise for the reader to show that the determinant of this 3 × 3 matrix is 0. Therefore the rank of the matrix cannot be three. By eliminating any one of the rows and any one of the columns we can generate 9 matrices which are 2 × 2. For example, if we eliminate row 3 and column 2 we obtain so that we see the rank for this matrix is 2.

We can see looking at example (a) that no information is contained in this matrix and the rank is 0, (b) may surprise by having a rank of 3 compared to that in example (c) which is 2, but all the rows in (b) are genuinely different.

A close analysis of row 3 in example (c) will show it to be R3 = 2 ∗ R1 − R2, and therefore that row really doesn’t add anything to the discussion.

6It is important to notice this. The possible 2 × 2 matrices can be found by eliminating R1 and C1, R1 and C2, R1 and C3, R2 and C1, etc. and it is not simply a matter of testing the matrices found by eliminating the “outer rows and columns”.

6.11 Rank 119

6.11.2 Systems of equations

Matrix rank is particularly useful due to what it can tell you about the solu-tions presented by a system of equasolu-tions. In particular, a comparison between the rank of a co-efficient matrix A and the rank of the augmented matrix A_B is very useful. We look at some examples that may prove instructive.

6.11.3 Example

Find the rank of the co-efficient and augmented matrix for the following system of equations.

3x + 2y = 4 (i) 2x − y = 5 (ii) Solution

You may note that this is the very simple system we looked at before in 6.9.

This system gave rise to the co-efficient matrix and augmented matrix as follows.

A =

 3 2

2 −1



; A_B =

 3 2 4

2 −1 5



Now |A| = (3)(−1) − (2)(2) = −7 6= 0, and so r(A) = 2. To look at the rank of the augmented matrix there are two square 2×2 matrices that can be formed, but we already know what the left hand one (found by eliminating the third column) has a non zero determinant because it is the same as A.

Therefore

r(A) = r(A_B) = 2.

6.11.4 Example

Find the rank of the co-efficient and augmented matrix for the following system of equations.

3x + 2y = 1 (i) 6x + 4y = 2 (ii)

6.11 Rank 120

Solution

It is clear here that equation (ii) is simply double equation (i). Therefore there is actually only one equation’s worth of information here.

This system gives rise to the co-efficient matrix and augmented matrix as follows.

A =

 3 2 6 4



; A_B =

 3 2 1

6 4 2



Now |A| = (3)(4) − (2)(6) = 0, and so the rank of A cannot be two.

Taking any of the possible one by one submatrices we can clearly see that the determinant will not be zero. Therefore r(A) = 1.

For the augmented matrix we know that the submatrix formed by elim-inating the third column has a zero determinant, but we have not checked the submatrices formed by eliminating the first or middle column. However a brief inspection shows that both of these also have a zero determinant. Once again, any of the six possible 1 × 1 submatrices show that r(A_B = 1.

r(A) = r(A_B) = 1.

It should be clear that since both of these equations are the same, then by simply taking equation (i) we obtain that

3x + 2y = 1 ⇒ y = 1

2(1 − 3x)

and consequently since there is no contraint on the choice of x and a choice of x determines y there are infinitely many solutions.

6.11.5 Example

Find the rank of the co-efficient and augmented matrix for the following system of equations.

3x + 2y = 1 (i) 6x + 4y = 1 (ii) Solution

This is very similar to the previous example. However the second equation is not quite the exact double of the first anymore. In fact it should be obvious that this system of equations is inconsistent, since the two contradict each other.

6.11 Rank 121

This system gives rise to the co-efficient matrix and augmented matrix as follows.

A =

 3 2 6 4



; A_B =

 3 2 1

6 4 1



Now |A| = (3)(4) − (2)(6) = 0, and so the rank of A cannot be two.

Taking any of the possible one by one submatrices we can clearly see that the determinant will not be zero. Therefore r(A) = 1.

For the augmented matrix we know that the submatrix formed by elim-inating the third column has a zero determinant, but we have not checked the submatrices formed by eliminating the first or middle column. If we eliminate the first column the resulting submatrix has a determinant of (2)(1) − (1)(4) = −2 6= 0. Therefore the r(A_B) = 2.

r(A) = 1 < r(A_B) = 2.

Remember that this system of equations has no solutions.

6.11.6 Summary

Although the previous examples do not constitute a proof, they illustrate the following results.

With a system of equations in n unknowns such that the matrix of co-efficients is A and the augmented matrix is A_B, then if

• r(A) = r(A_B) = n then there exists a unique solution;

• r(A) = r(A_B) < n then there are infinitely many solutions;⁷

• r(A) < r(A_B) then there is no solution.

We have previously known that to obtain a unique solution for n un-knowns we require n equations. Rank can be thought of as a way of showing the the n equations are genuinely different and we have sufficient information to solve the system uniquely.

When the rank of the co-efficient and augmented matrix match but are less than n then we do not have enough information to uniquely determine a solution, but there is a consistency.

7Note that this implies that the matrix A has a zero determinant, and thus is singular (no inverse exists) and therefore other methods must be employed to find the solutions.