Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

If $A = \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}$ then using $A^{-1}$, solve the following system of equations: $x - 2y = -1, 2x + y = 2$.

Options:

$x = 3, y = 4$

$x = \frac{3}{5}, y = \frac{4}{5}$

$x = 1, y = 1$

$x = -1, y = 2$

Correct Answer:

$x = \frac{3}{5}, y = \frac{4}{5}$

Explanation:

The correct answer is Option (2) → $x = \frac{3}{5}, y = \frac{4}{5}$ ##

$|A| = 5$

$\text{adj. } A = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}$

$A^{-1} = \frac{\text{adj. } A}{|A|} = \frac{1}{5} \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}$

Given system of equations is $AX = B$, where

$A = \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}, X = \begin{bmatrix} x \\ y \end{bmatrix}, B = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$

$\begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$

$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} -1 \\ 2 \end{bmatrix}$

$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{5} \begin{bmatrix} -1+4 \\ 2+2 \end{bmatrix} = \begin{bmatrix} 3/5 \\ 4/5 \end{bmatrix}$

$x = \frac{3}{5}$ and $y = \frac{4}{5}$