Solution of Non-homogenous Systems

Consider the system: $A_{n\times n}X_{n\times 1}=B_{n\times 1}$. $$

• $|A| \neq 0 \iff \text{Rank }A = \text{Rank }(A|B)=n \iff \text{unique solution exists}$
• $|A|=0 \implies \text{no solution} \lor \text{infinitely many solutions}$
• $\text{Rank }A <\text{Rank }(A|B) \implies \text{no solutions}$
• $\text{Rank }A =\text{Rank }(A|B) <n \implies \text{infinitely many solutions}$

Methods

Method 1: Direct approach

Used when coefficient matrix $A$ is invertible. It means the system has a unique set of solutions. $$

$$AX=B \implies X=A^{-1}B$$

Method 2: Cramer’s Rule

Let $AX=B$, where $A$ is the coefficient matrix and $X=(x_i)_{n\times 1}$.

$$x_i=\frac{\lvert{A_i}\rvert}{\lvert{A}\rvert}$$

Where $A_i$ is the matrix obtained by replacing $i$th column in matrix $A$ by $B$.

Method 3: Reducing to Echelon Form

Start with $(A|B)$. Convert the $\text{LHS}$ to echelon form. The solution can be found easily. If a contradiction is encountered while solving the equation, then the system has no solutions.