1. System of linear equations:

Let us consider a system of two linear equations in three variables given below:
ax₁+bx₂+cx₃=r
dx₁+fx₂+gx₃=s … (i)
Note that in particular, if both r and s=0, then the system of linear equations (i) reduces to
ax₁+bx₂+cx₃=0
dx₁+fx₂+gx₃=0 … (ii)
This is called a homogenous system of two linear equations in three variables. It is well known that various physical situations are modelled in terms of systems of linear equations.

2. Coefficient matrix and augmented matrix:

$$\begin{aligned} &\hspace{-9.2cm} \text{The matrix } A = \begin{pmatrix} a & b & c \\\ d & e & f\end{pmatrix} \text{ and } B = \begin{pmatrix} r \\\ s \end {pmatrix} \end{aligned}$$

are obtained from the coefficients of the equations given in the above section. The matrix A is called coefficient matrix and the matrix [A|B] is known as augmented matrix and is denoted by A⁺.

3. Related matrix equation:

Let A be a matrix of order m×n and B be a column matrix of order m×1. Then AX=B is called a matrix equation. The matrix equation related to the given system of linear equations in Section 1 can be written as AX=B, where

$$\begin{aligned} &\hspace{-9.2cm} A = \begin{pmatrix} a & b & c \\\ d & e & f\end{pmatrix} \text{ , } X = \begin{pmatrix} x1 \\\ x2 \\\ x3 \end {pmatrix} \text{ and } B = \begin{pmatrix} r \\\ s \end {pmatrix} \end{aligned}$$

Its solution provides a solution of the system of linear equations and vice-versa. It may be noted that system of linear equations is consistent if and only if related matrix equation is consistent.

4. Example:

Consider 2x₁+3x₂+4x₃=1 and x₁-2x₂+3x₃=2 be a system of two linear equations in three variables. Then the coefficient matrix

$$\begin{aligned} &\hspace{-9.2cm} A = \begin{pmatrix} a & b & c \\\ d & e & f\end{pmatrix} \end{aligned}$$

and the given system of linear equations can be written as AX=B, where

$$\begin{aligned} &\hspace{-9.2cm} X = \begin{pmatrix} x1 \\\ x2 \\\ x3 \end {pmatrix} \text{ and } B = \begin{pmatrix} 1 \\\ 2 \end {pmatrix} \end{aligned}$$

Clearly

$$\begin{aligned} A^+ = \begin{pmatrix} 2 & 3 & 4 \vert & 1 \\ 1 & -2 & 3 \vert & 2 \end{pmatrix} \end{aligned}$$

is the augmented matrix.

5. Related linear map equation and its equivalence:

We discuss below the linear map equation associated with a matrix equation.

5.1. Related linear map and its range:

Let AX=B be a matrix equation. Recall from Experiment 5 that there is a linear map T:R³→R² associated with matrix A w.r.t. the standard basis of R³ and R² . By definition, Range(T) ={T(x): x∈R³}. That is, y∈Range(T) if and only if there exist x∈R³ such that T(x)=y. Note that Range(T) is generated by {T(e₁), T(e₂), T(e₃)}.

5.2 Linear map equation and its consistency:

The given system of linear equations represented by the matrix equation AX=B and leads to a linear map equation T(x)=b, where x=(x₁, x₂, x₃)R³ and b=(r, s)R². It is consistent, that is it has a solution, that is there exist x∈R³ such that T(x)=b, if b∈Range(T), as discussed in Experiment 7.

5.3 Equivalence:

The matrix equation AX=B is consistent if and only if linear map equation T(x)=b is consistent. Furthermore if X is a solution of AX=B, then X^T is a solution of T(x)=b and vice-versa.

6. Solution of matrix equation:

6.1. Consistency:

Proposition: (a.) Matrix equation AX=B is consistent if and only if RankA=RankA⁺.
(b.) Matrix equation AX=B is always consistent if B=0, that is AX=0 is always consistent.

Proof: (a.) Let AX=B is consistent. Then the matrix equation AX=B has a solution. Thus T(x)=b has a solution. Recall from Experiment 7 that T(x)=b has a solution implies that b∈Range(T). Recall from Experiment 5 that Range of T is generated by image of basis elements and image of each basis element leads to a column of A. Thus if b∈Range(T), then B is a linear combination of columns of A. This is equivalent to saying RankA=RankA⁺.

Conversely, let RankA=RankA⁺. Then it means B is a linear combination of columns of A.

As discussed in the preceding paragraph, Range(T) is generated by image of basis elements and image of each basis element leads to a column of A. Thus b∈Range(T) which means T(x)=b is consistent. This shows that the given matrix equation is consistent.

(b.) As RankA=RankA⁺ from the above proposition, the given homogenous system of linear equations is consistent.

6.2. Uniqueness:

Let A be a square matrix and let the matrix equation AX=B be consistent. Then AX=B has a unique solution if and only if A is invertible. In this case the solution is X=A^-1B.

Proof: Recall from Experiment 7 that T(x)=b has a unique solution, where T(x)=b is the linear map equation associated to the matrix equation, has a unique solution (if it exists) if and only if T is one-to-one. Thus a consistent system of linear equations represented by AX=B has a unique solution if and only if A is invertible since A is a square matrix.

6.3. Examples:

(i.) Consider the matrix equation AX=B, where

$$\begin{aligned} &\hspace{-9.2cm} A = \begin{pmatrix} 1 & 2 \\\ 2 & 6\end{pmatrix} \text{ and } B = \begin{pmatrix} 3 \\\ 1 \end {pmatrix} \end{aligned}$$

Consistency & uniqueness: Clearly

$$\begin{aligned} A^+ = \begin{pmatrix} 1 & 2 & \vert & 4 \\ 2 & 6 & \vert & 2 \end{pmatrix} \end{aligned}$$

The rank of A is 2 and the rank of A⁺ is also 2. Hence RankA=RankA⁺. Thus solution exists that is, the given matrix equation is consistent. As A is a square matrix and is invertible, by Section 6.2. (i), solution is unique.
(ii.) Consider the matrix equation AX=B, where

$$\begin{aligned} A = \begin{pmatrix} 1 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix} \text{ and } B = \begin{pmatrix} 2 \\ 4 \end{pmatrix}. \end{aligned}$$

Consistency: Clearly

$$\begin{aligned} A^+ = \begin{pmatrix} 1 & 1 & -1 \vert & 2 \\ 0 & 0 & 0 \vert & 4 \end{pmatrix} \end{aligned}$$

The rank of A is 1 and the rank of A⁺ is also 2. Hence RankA≠RankA⁺. Thus the given matrix equation is inconsistent.
(iii.) Consider the matrix equation AX=B, where

$$\begin{aligned} A = \begin{pmatrix} 1 & 1 \\ 1 & -1 \\ 1 & 2 \end{pmatrix} \text{ and } B = \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix}. \end{aligned}$$

Consistency & uniqueness: Clearly

$$\begin{aligned} A^+ = \begin{pmatrix} 1 & 1 \vert & 2 \\ 1 & -1 \vert & 0 \\ 1 & 2 \vert & 3 \end{pmatrix} \end{aligned}$$

The rank of A is 2 and the rank of A⁺ is also 2. Hence RankA=RankA⁺. Thus solution exists and the given matrix equation is consistent. As (1, 1) is the only solution which satisfy the above matrix equation, solution is not unique.
(iv.) Consider the matrix equation AX=B, where

$$\begin{aligned} A = \begin{pmatrix} 1 & 1 \\ 2 & 2 \\ 3 & 3 \end{pmatrix} \text{ and } B = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}. \end{aligned}$$

Consistency & uniqueness: Clearly

$$\begin{aligned} A^+ = \begin{pmatrix} 1 & 1 \vert & 2 \\ 2 & 2 \vert & 4 \\ 3 & 3 \vert & 6 \end{pmatrix} \end{aligned}$$

The rank of A is 1 and the rank of A⁺ is also 1. Hence RankA=RankA⁺. Thus solution exists and the given matrix equation is consistent. As (2, 0) and (0, 2) both satisfy the above matrix equation, solution is not unique.

6.4. Understanding the condition for consistency and uniqueness of matrix equation (through an example):

Example: Consider the matrix equation AX=B, where

$$\begin{aligned} A = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & 1 \end{pmatrix} \text{ and } B = \begin{pmatrix} 4 \\ 6 \end{pmatrix}. \end{aligned}$$

Linear map: We first find the linear transformation associated with A w.r.t. the standard basis B₁={e₁, e₂, e₃} and B₂={f₁, f₁} of R³ and R² respectively. Here e₁=(1, 0, 0), e₂=(0, 1, 0), e₃=(0, 0,1), f₁=(1, 0) and f₂=(0, 1). Define T(1, 0, 0)=1(1, 0)+0(0, 1)=(1, 0), T(0, 1, 0)=-1(1, 0)+1(0, 1)=(-1, 1) and T(0, 0, 1)=0(1, 0)+1(0, 1)=(0, 1). If (x, y, z)=a(1, 0, 0)+b(0, 1, 0)+c(0, 0, 1); then a=x, b=y and c=z. Thus define T:R³→R² by T(x, y, z)= aT(1, 0, 0)+bT(0, 1, 0)+cT(0, 0, 1)=a(1, 0)+b(-1, 1)+c(0, 1)=(a-b, b+c)=(x-y, y+z), where x, y, z∈R. Thus the linear transformation T:R³→R² associated with the matrix A w.r.t. the basis B₁ and B₂ is T(x, y, z)=(x-y, y+z), where x, y, z∈R.

Linear map equation: T(x, y, z)=b, where b=(4, 2)=B^T is the linear map equation equivalent to the given matrix equation.

Consistency of linear map equation: The linear map equation is T(x, y, z)=(4, 2) is consistent as b=(4, 2)∈Range(T).

Uniqueness of linear map equation: Solution of linear map equation T(x, y, z)=(4, 2) is not unique as both (4, 0, 2) and (6, 2, 0) map to (4, 2). Notice that T is not one-to-one.

Understanding: As b≡(4, 2)∈Range(T), (4, 2) which is the last column vector of A⁺ is a linear combination of column vectors of A. This implies that RankA=RankA⁺.

6.5. Remark:

i.Let A be a square matrix and the matrix equation AX=B is consistent. Then it has a unique solution if and only if A is invertible.
ii.Let A be a matrix of order 2×3 and the matrix equation AX=B is consistent. Then it does not have a unique solution.
iii.Let A be a matrix of order 3×2 and the matrix equation AX=B is consistent. Then it may or may not have a unique solution. It has a unique solution if and only if RankA=2.

7. Consistency (existence of solution) of a system of linear equations:

7.1. Consistency:

Consider a system of two linear equations in three variables as given by equation (i). Then it is consistent if and only if related matrix equation AX=B is consistent if and only if RankA =RankA⁺.
In particular, a homogenous system of two linear equations in three variables as given in equation (ii) is consistent if and only if matrix equation AX=0, where

$$\begin{aligned} 0 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{aligned}$$

is consistent. In this case, clearly since A=A⁺, a homogenous system of two linear equations in three variables is always consistent. It is evident that (0, 0, 0) is a solution.

7.2. Uniqueness:

Let solution of a system of two linear equations in three variables as given in equation (i) exists. Then it has a unique solution if and only if matrix equation AX=B has a unique solution.

7.3. Example:

(i.) Consider the system of three linear equations in two variables given as follows x₁+x₂=2 ... (i),
2x₁+2x₂=4 …(ii) and
3x₁+3x₂=6 …(iii)
Then the related matrix equation is AX=B, where A=, X= and B=. Its solution exists which is shown in the Example 6.4. (iv) above. Hence the given system of linear equations is consistent and the solution is not unique.

8. Conclusion:

Following are equivalent:
(a) System of linear equations is consistent.
(b) Matrix equation is consistent.
(c) Linear map equation is consistent.
Therefore we study the consistency of a system of linear equations through the consistency of related matrix equation, i.e. which is consistent if and only if RankA=RankA⁺. The condition for consistency and uniqueness of matrix equation is understood through related linear map equation.