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

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:

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

## Expert Answer

**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**.