# Determine the value of h such that the matrix is the augmented matrix of a consistent linear system.

$\boldsymbol{ \left[ \begin{array}{ c c | c } 1 & 3 & -8 \\ -4 & h & 1 \end{array} \right] }$

The aim of this question is to understand the solution of the system of linear equations using the row operations and row echelon form.

Any matrix is said to be in the row echelon form if it fulfills three requirements. First, the first non-zero number in every row must be 1 (called the leading 1). Second, each leading 1 must be on the right of the leading 1 in the previous row. Third, all non-zero rows must precede the zero rows. For example:

$\left[ \begin{array}{ c c c | c } 1 & x & x & x \\ 0 & 0 & 1 & x \\ 0 & 0 & 0 & 0 \end{array} \right]$

Where x can have any value.

The row echelon form can be used to solve a system of linear equations. We simply write the augmented matrix and then convert it to the row echelon form. Then we convert it back to the equation form and find the solutions by back substitution.

The linear system of equations represented by an augmented matrix will have a unique solution (consistency) if the following condition is satisfied:

$\text{ no. of non-zero rows } \ = \ \text{ no. of unknown variables }$

Given:

$\left[ \begin{array}{ c c | c } 1 & 3 & -8 \\ -4 & h & 1 \end{array} \right]$

Reducing to row echelon form:

$R_2 \ + \ 4R_1 \rightarrow \left[ \begin{array}{ c c | c } 1 & 3 & -8 \\ 0 & h-12 & -31 \end{array} \right]$

It can be deduced from the above matrix that the system of linear equations formed by these coefficients will have a unique solution on all possible values of $R^n$ except when h = 12 (because this nullifies the 2nd equation and the system reduces to a single equation describing two variables).

## Numerical Result

$h$ can have all possible values of $R^n$ excluding $h = 12$.

## Example

Find all the possible values of $y$ such that the following augmented matrix represents a consistent system of linear equations:

$\boldsymbol{ \left[ \begin{array}{ c c | c } 9 & 18 & 0 \\ 5 & y & 1 \end{array} \right] }$

Reducing the given matrix to row echelon form via row operations:

$\dfrac{ 1 }{ 9 } R_1 \rightarrow \left[ \begin{array}{ c c | c } 1 & 2 & 0 \\ 5 & y & 1 \end{array} \right]$

$R_2 – 5 R_1 \rightarrow \left[ \begin{array}{ c c | c } 1 & 2 & 0 \\ 0 & y-10 & 1 \end{array} \right]$

It can be deduced from the above matrix that the system of linear equations formed by these coefficients will have a unique solution on all possible values of $R^n$ except when y = 10.