Q.Solve the system of equations $2x + 5y = 1$, $3x + 2y = 7$.

Concept: Matrix Equation Solving – Represent the system as $A\mathbf{x} = \mathbf{b}$ and solve by finding $A^{-1}$.

Write the system in matrix form:

The inverse of $A = \begin{pmatrix} 2 & 5 \\ 3 & 2 \end{pmatrix}$ is

Multiply both sides by $A^{-1}$:

Why the matrix approach?

A system like $2x + 5y = 1$, $3x + 2y = 7$ is just a compact way of asking: what pair $(x, y)$ makes both equations true at the same time? Instead of elimination or substitution, we can think of it as a single matrix equation:

If we call the coefficient matrix $A$, the variable column $X$, and the constant column $B$, then $AX = B$. The neat idea: if $A$ has an inverse $A^{-1}$, multiply both sides on the left by $A^{-1}$ to get $X = A^{-1}B$. That gives the solution directly — no guessing, no back-substitution.

1. Write the system in matrix form

So $AX = B$.

2. Check if $A$ is invertible — compute its determinant

Since $\det(A) \neq 0$, $A^{-1}$ exists.

3. Find $A^{-1}$ using the formula for a $2 \times 2$ matrix

   For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the inverse is $\frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.

4. Multiply $A^{-1}$ by $B$ to get $X$

Compute each entry:

• For $x$: $(-\frac{2}{11})(1) + (\frac{5}{11})(7) = -\frac{2}{11} + \frac{35}{11} = \frac{33}{11} = 3$
• For $y$: $(\frac{3}{11})(1) + (-\frac{2}{11})(7) = \frac{3}{11} - \frac{14}{11} = -\frac{11}{11} = -1$

So $x = 3$, $y = -1$.

5. Verify by plugging back into the original equations

   • $2(3) + 5(-1) = 6 - 5 = 1$ ✓
   • $3(3) + 2(-1) = 9 - 2 = 7$ ✓

Method: Solving a 2-Variable Linear System by the Matrix (Inverse) Method

This method solves any system of two linear equations in two unknowns by packaging it as a single matrix equation $AX=B$ and solving via $X=A^{-1}B$.

Steps

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

Collect the coefficients into a $2\times2$ matrix $A$, the unknowns into a column $X$, and the constants into a column $B$:

Step 2: Compute $\det(A)$ and check it's nonzero

If this is zero, $A^{-1}$ doesn't exist and the matrix method can't be used directly — the system needs a different treatment (inconsistent or infinitely many solutions).

Step 3: Find $A^{-1}$

Step 4: Multiply on the left: $X=A^{-1}B$

Carry out the $2\times2$-by-$2\times1$ matrix multiplication carefully, term by term.

Step 5: Verify by substituting back

Plug the found $x,y$ into both original equations to confirm — this is quick and catches an arithmetic slip before it's submitted as the final answer.

This is the base case (two unknowns) of the general matrix method — the same $X=A^{-1}B$ idea scales directly to three or more unknowns, just with a $3\times3$ (or larger) inverse via the adjoint instead of the $2\times2$ shortcut.