Matrices: Tricky Questions Solved (3)

Question

Let $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$. Find $A^{-1}$, verify by computing $A \cdot A^{-1}$, and use it to solve the system $x + 2y = 5$, $3x + 4y = 11$. Finally, explain why this method fails for some matrices and how to recognise that case.

Solution — Step by Step

Why This Works

The matrix-inverse method turns a linear system into a product. When $A$ is invertible (i.e., $\det A \neq 0$), there’s a unique solution $\vec{x} = A^{-1}\vec{b}$. This is the cleanest way to solve $2\times 2$ and $3\times 3$ systems for board exams.

The method fails when $\det A = 0$ — the matrix is singular, no inverse exists. Geometrically, the two equations represent either parallel lines (no solution) or the same line (infinite solutions). You must then use elimination or Cramer’s rule with care.

Alternative Method

Cramer’s rule:

$x = \frac{\det A_x}{\det A}, \quad y = \frac{\det A_y}{\det A}$

where $A_x$ replaces column 1 with $\vec{b}$ and $A_y$ replaces column 2.

$\det A_x = \begin{vmatrix} 5 & 2 \\ 11 & 4\end{vmatrix} = 20 - 22 = -2 \implies x = -2/-2 = 1$

$\det A_y = \begin{vmatrix} 1 & 5 \\ 3 & 11\end{vmatrix} = 11 - 15 = -4 \implies y = -4/-2 = 2$

Same answer.

Common Mistake

Final answer: $A^{-1} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}$, $x = 1$, $y = 2$.