The system of the equations

x + y + 2z = a

x + z = b

2x + y + 3z = c

has a solution if

Concept:

Consider the system of m linear equations

a₁₁ x₁ + a₁₂ x₂ + … + a_1n x_n = b₁

a₂₁ x₁ + a₂₂ x₂ + … + a_2n x_n = b₂

…

a_m1 x₁ + a_m2 x₂ + … + a_mn x_n = b_m

The above equations containing the n unknowns x₁, x₂, …, x_n. To determine whether the above system of equations is consistent or not, we need to find the rank of following matrices.

A = \(\rm \begin{bmatrix} a_{11} & a_{12} & ... & a_{1n} \\ a_{21} & a_{22} & ... & a_{2n} \\ ... & ... & ... & ... \\ a_{m1} & a_{m2} & ... & a_{mn} \end{bmatrix}\) and [A|B] = \(\rm \begin{bmatrix} a_{11} & a_{12} & ... & a_{1n}& b_1\\ a_{21} & a_{22} & ... & a_{2n} &b_2\\ ... & ... & ... & ...& ...\\ a_{m1} & a_{m2} & ... & a_{mn}&b_m \end{bmatrix}\)

A is the coefficient matrix and [A|B] is called an augmented matrix of the given system of equations.

We can find the consistency of the given system of equations as follows:

(i) If the rank of matrix A is equal to the rank of an augmented matrix and it is equal to the number of unknowns, then the system is consistent and there is a unique solution.

Rank of A = Rank of augmented matrix = n

(ii) If the rank of matrix A is equal to the rank of an augmented matrix and it is less than the number of unknowns, then the system is consistent and there are an infinite number of solutions.

Rank of A = Rank of augmented matrix < n

(iii) If the rank of matrix A is not equal to the rank of the augmented matrix, then the system is inconsistent, and it has no solution.

Rank of A ≠ Rank of the augmented matrix

Calculation:

From the given equations:

x + y + 2z = a

x + z = b

2x + y + 3z = c

The coefficient matrix is:

A = \(\rm \begin{bmatrix} 1 & 1 & 2 \\ 1 & 0 & 1 \\ 2 & 1 & 3 \end{bmatrix}\) and the augmented matrix, [A|B] = \(\rm \begin{bmatrix} 1 & 1 & 2 & a \\ 1 & 0 & 1 & b \\ 2 & 1 & 3 & c \end{bmatrix}\) 

The determinant of the matrix [A] = \(\begin{vmatrix} 1&1 & 2\\ 1& 0 & 1\\ 2& 1 & 3 \end{vmatrix}\) 

R₃ = R₃ - R₁ - R₂ 

⇒ |A| = \(\begin{vmatrix} 1&1 & 2\\ 1& 0 & 1\\ 0& 0 & 0 \end{vmatrix}\) = 0

∴ The rank of matrix is 2

So, the solution if the rank of matrix A is equal to rank of augmented matrix A|B = 2

⇒ [A|B] = \(\rm \begin{bmatrix} 1 & 1 & 2 & a \\ 1 & 0 & 1 & b \\ 2 & 1 & 3 & c \end{bmatrix}\) 

R3 = R3 - R1 - R2 

[A|B] = \(\rm \begin{bmatrix} 1 & 1 & 2 & a \\ 1 & 0 & 1 & b \\ 0 & 0 & 0 & c - a-b \end{bmatrix}\) 

For rank of [A|B] to be 2

c - a - b = 0

c = a + b