3.7 Applications of Matrices in Real World Problems

3.7 Applications of Matrices in Real World Problems

Example 12

Three forces act on a particle which must be in equilibrium i.e. $F_1 + F_2 + F_3 = 0$, where $F_1 = \begin{bmatrix} 5 \\ 2 \end{bmatrix}, F_2 = \begin{bmatrix} -3 \\ x \end{bmatrix}, F_3 = \begin{bmatrix} -2 \\ -4 \end{bmatrix}$. Find the value of $x$.

Solution Since $F_1 + F_2 + F_3 = 0$

$\begin{bmatrix} 5 \\ 2 \end{bmatrix} + \begin{bmatrix} -3 \\ x \end{bmatrix} + \begin{bmatrix} -2 \\ -4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

$\begin{bmatrix} 5 - 3 - 2 \\ 2 + x - 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \Rightarrow \begin{bmatrix} 0 \\ x - 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

$x - 2 = 0$

$x = 2$

Example 13

Eight years ago Huria's was $\dfrac{3}{4}$ of Jannat's age. After four years Huria's age will be $\dfrac{6}{7}$ of Jannat's age. Find their present ages by using matrices.

Solution Suppose Huria's present age is $x$ years and Jannat's present age is $y$ years. Eight years ago their ages were $x - 8$ and $y - 8$. According to first condition:

$x - 8 = \dfrac{3}{4}(y - 8)$

or $4x - 32 = 3y - 24$

or $4x - 3y = -24 + 32$

$4x - 3y = 8$ ...(i)

After $4$ years their ages will be $x + 4$ and $y + 4$.

Applying the second condition:

$x + 4 = \dfrac{6}{7}(y + 4)$

or $7x + 28 = 6y + 24$

$7x - 6y = -4$ ...(ii)

Matrix form of (i) and (ii) is

$\begin{bmatrix} 4 & -3 \\ 7 & -6 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ -4 \end{bmatrix}$

Here, $A = \begin{bmatrix} 4 & -3 \\ 7 & -6 \end{bmatrix}, A_x = \begin{bmatrix} 8 & -3 \\ -4 & -6 \end{bmatrix}, A_y = \begin{bmatrix} 4 & 8 \\ 7 & -4 \end{bmatrix}$

$|A| = \begin{vmatrix} 4 & -3 \\ 7 & -6 \end{vmatrix}$

$= (4)(-6) - (7)(-3)$

$= -24 + 21$

$= -3 \neq 0$

Applying Cramer's rule

$x = \dfrac{|A_x|}{|A|} = \dfrac{\begin{vmatrix} 8 & -3 \\ -4 & -6 \end{vmatrix}}{-3}$ $y = \dfrac{|A_y|}{|A|} = \dfrac{\begin{vmatrix} 4 & 8 \\ 7 & -4 \end{vmatrix}}{-3}$

$= \dfrac{-48 - 12}{-3} = \dfrac{-60}{-3} = 20$ $= \dfrac{-16 - 56}{-3} = \dfrac{-72}{-3} = 24$

$\therefore$ Present ages of Huria and Jannat are $20$ years and $24$ years respectively.