Q.Examine the consistency of the following system of equations: $x + 2y = 2$ $2x + 3y = 3$

Concept: Matrix Equation Solving — we write the system as $AX = B$ and check if $\text{rank}(A) = \text{rank}([A|B])$.

Step 1: Matrix form

Step 2: Augmented matrix

Step 3: Row reduce

$R_2 \to R_2 - 2R_1$ gives

Both $A$ and $[A|B]$ have rank 2 (no zero rows), and rank equals number of variables (2).

Why This Approach Works

When we ask whether a system of equations is consistent, we are really asking: Does there exist at least one pair $(x, y)$ that satisfies both equations at the same time? For a system of two linear equations in two variables, there are three possibilities:

• Unique solution — the lines intersect at exactly one point.
• Infinitely many solutions — the lines coincide (same line).
• No solution — the lines are parallel and distinct.

The fastest way to decide which case we have is to examine the coefficient matrix and its determinant. If the determinant is non-zero, the matrix is invertible, and a unique solution exists — the system is automatically consistent. If the determinant is zero, we must check further (the equations might be dependent or contradictory).

Here, the equations are:

Let’s work through it.

1. Write the system in matrix form. The coefficient matrix $A$ and the constant vector $\mathbf{b}$ are:

The system is $A \mathbf{x} = \mathbf{b}$, where $\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}$.

2. Compute the determinant of $A$. For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is $ad - bc$.

Since $\det(A) = -1 \neq 0$, the matrix is invertible. This immediately tells us that the system has a unique solution — and therefore is consistent.

3. Find the solution (optional, but confirms consistency).

   We can solve using the inverse of $A$, or by elimination. Let’s use elimination for clarity.

   From the first equation: $x = 2 - 2y$.

   Substitute into the second:

Then $x = 2 - 2(1) = 0$.

So the unique solution is $(x, y) = (0, 1)$.

4. Interpret geometrically. The two lines $y = -\frac{1}{2}x + 1$ and $y = -\frac{2}{3}x + 1$ have different slopes ($-\frac{1}{2}$ and $-\frac{2}{3}$), so they intersect at exactly one point — which we found to be $(0, 1)$. This confirms consistency.

Method: Testing Consistency via the Coefficient Determinant

This method decides whether a system of linear equations has a solution at all, and how many, often without solving the system in full.

Steps

Step 1: Write the system as $AX = B$

Collect the coefficients into a square matrix $A$, the unknowns into a column $X$, and the constants into a column $B$:

Step 2: Compute $\det(A)$

For a $2\times2$ matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, $\det(A)=ad-bc$. This single number is the fastest possible consistency test — compute it before attempting anything else.

Step 3: Read the consistency straight off the determinant

If instead $\det(A)=0$, $A^{-1}$ does not exist and you cannot stop here — you must examine the equations directly to decide between "no solution" and "infinitely many."

Step 4: Applying this to a pure consistency question

When a question only asks you to examine consistency (not to find $x$ and $y$), stating '$\det(A)\neq0$, so the system has a unique solution and is consistent' is already a complete, correct answer. Solving for $x,y$ afterward — by elimination or by computing $A^{-1}B$ — is a useful confirmation, not a requirement.

Common Mistakes

Mistake 1: Computing the determinant as $bc-ad$ instead of $ad-bc$

Why it's wrong: swapping which diagonal product is subtracted flips the sign of $\det(A)$. Here $\det(A)=(1)(3)-(2)(2)=-1$; computing it the wrong way round gives $+1$. In this particular case the sign flip doesn't change the consistency verdict (still non-zero), but if you go on to compute $A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A)$ with the wrong-sign determinant, every entry of the final solution flips sign, producing a wrong $(x,y)$. Correct approach: always compute $ad-bc$ (top-left times bottom-right, minus top-right times bottom-left) and write out both products before subtracting.

Mistake 2: Thinking a negative determinant signals inconsistency

Why it's wrong: the sign of $\det(A)$ carries no information about consistency — only whether it is zero or non-zero matters. A determinant of $-1$ is just as 'invertible' as $+1$. Correct approach: check only $\det(A)=0$ vs. $\det(A)\neq0$; ignore the sign when deciding consistency.