Exercise 1.4 Solutions – Applications of Matrices and Determinants (12th Maths)

IMPORTANT QUESTIONS

2 MARKS

Q1 (i). Solve by Cramer’s rule

$\textbf{1(i). Solve by Cramer's rule}\quad 5x-2y+16=0,\; x+3y-7=0$

$5x-2y=-16,\quad x+3y=7$

$\begin{pmatrix}5 & -2\\1 & 3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-16\\7\end{pmatrix}$

$AX=B$

$\Delta=\begin{vmatrix}5 & -2\\1 & 3\end{vmatrix}=15+2=17\neq0$

$\Delta x=\begin{vmatrix}-16 & -2\\7 & 3\end{vmatrix}=-48+14=-34$

$\Delta y=\begin{vmatrix}5 & -16\\1 & 7\end{vmatrix}=35+16=51$

$x=\frac{\Delta x}{\Delta}=\frac{-34}{17}=-2$

$y=\frac{\Delta y}{\Delta}=\frac{51}{17}=3$

$\therefore\ x=-2,\; y=3$

Q1 (ii).Solve by Cramer’s rule

$\textbf{1(ii). Solve by Cramer's rule}\quad \frac{3}{x}+2y=12,\; \frac{2}{x}+3y=13$

$\text{Let } \frac{1}{x}=a$

$3a+2y=12,\quad 2a+3y=13$

$\begin{pmatrix}3 & 2\\2 & 3\end{pmatrix}\begin{pmatrix}a\\y\end{pmatrix}=\begin{pmatrix}12\\13\end{pmatrix}$

$AX=B$

$\Delta=\begin{vmatrix}3 & 2\\2 & 3\end{vmatrix}=9-4=5\neq0$

$\Delta a=\begin{vmatrix}12 & 2\\13 & 3\end{vmatrix}=36-26=10$

$\Delta y=\begin{vmatrix}3 & 12\\2 & 13\end{vmatrix}=39-24=15$

$a=\frac{\Delta a}{\Delta}=\frac{10}{5}=2$

$y=\frac{\Delta y}{\Delta}=\frac{15}{5}=3$

$x=\frac{1}{a}=\frac{1}{2}$

$\therefore\ x=\frac{1}{2},\; y=3$

Q1 (iii)

$\textbf{1(iii). Solve by Cramer's rule}\quad 3x+3y-z=11,\; 2x-y+2z=9,\; 4x+3y+2z=25$

$\text{Matrix form}$

$\begin{pmatrix}3 & 3 & -1\\2 & -1 & 2\\4 & 3 & 2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}11\\9\\25\end{pmatrix}$

$AX=B$

$\Delta=\begin{vmatrix}3 & 3 & -1\\2 & -1 & 2\\4 & 3 & 2\end{vmatrix}$

$=3(-2-6)-3(4-8)-1(6+4)$

$=3(-8)-3(-4)-10$

$=-24+12-10=-22$

—

$\Delta x=\begin{vmatrix}11 & 3 & -1\\9 & -1 & 2\\25 & 3 & 2\end{vmatrix}$

$=11(-2-6)-3(18-50)-1(27+25)$

$=11(-8)-3(-32)-52$

$=-88+96-52=-44$

—

$\Delta y=\begin{vmatrix}3 & 11 & -1\\2 & 9 & 2\\4 & 25 & 2\end{vmatrix}$

$=3(18-50)-11(4-8)-1(50-36)$

$=3(-32)-11(-4)-14$

$=-96+44-14=-66$

—

$\Delta z=\begin{vmatrix}3 & 3 & 11\\2 & -1 & 9\\4 & 3 & 25\end{vmatrix}$

$=3(-25-27)-3(50-36)+11(6+4)$

$=3(-52)-3(14)+110$

$=-156-42+110=-88$

—

$x=\frac{\Delta x}{\Delta}=\frac{-44}{-22}=2$

$y=\frac{\Delta y}{\Delta}=\frac{-66}{-22}=3$

$z=\frac{\Delta z}{\Delta}=\frac{-88}{-22}=4$

$\therefore\ x=2,\; y=3,\; z=4$

Q1 (iv)

$\textbf{1(iv). Solve by Cramer's rule}\quad \frac{3}{x}-\frac{4}{y}-\frac{2}{z}=1,\; \frac{1}{x}+\frac{2}{y}+\frac{1}{z}=2,\; \frac{2}{x}-\frac{5}{y}-\frac{4}{z}=-1$

$\text{Let } \frac{1}{x}=a,\; \frac{1}{y}=b,\; \frac{1}{z}=c$

$3a-4b-2c=1,\quad a+2b+c=2,\quad 2a-5b-4c=-1$

$\Delta=\begin{vmatrix}3 & -4 & -2\\1 & 2 & 1\\2 & -5 & -4\end{vmatrix}$

$=3(-8+5)+4(-4-2)-2(-5-4)$

$=3(-3)+4(-6)-2(-9)$

$=-9-24+18=-15$

—

$\Delta a=\begin{vmatrix}1 & -4 & -2\\2 & 2 & 1\\-1 & -5 & -4\end{vmatrix}$

$=(-8+5)+4(-8-1)-2(-10+2)$

$=-3-36+16=-15$

—

$\Delta b=\begin{vmatrix}3 & 1 & -2\\1 & 2 & 1\\2 & -1 & -4\end{vmatrix}$

$=3(-8+1)-1(-4-2)-2(-1-4)$

$=-21+6+10=-5$

—

$\Delta c=\begin{vmatrix}3 & -4 & 1\\1 & 2 & 2\\2 & -5 & -1\end{vmatrix}$

$=3(-2+10)+4(-1-2)+1(-5-4)$

$=24-12-9=-5$

—

$a=\frac{\Delta a}{\Delta}=\frac{-15}{-15}=1$

$b=\frac{\Delta b}{\Delta}=\frac{-5}{-15}=\frac{1}{3}$

$c=\frac{\Delta c}{\Delta}=\frac{-5}{-15}=\frac{1}{3}$

—

$x=\frac{1}{a}=1,\quad y=\frac{1}{b}=3,\quad z=\frac{1}{c}=3$

$\therefore\ x=1,\; y=3,\; z=3$

2.In a competitive examination, one mark is awarded for every correct answer while 1/4 mark is deducted for every wrong answer. A student answered 100 questions and got 80 marks. How many questions did he answer correctly ?

$\text{Let number of correct answers } = x$

$\text{Let number of wrong answers } = y$

$x+y=100 \;\rightarrow\; (1)$

$x-\frac{1}{4}y=80$

$4x-y=320 \;\rightarrow\; (2)$

$\text{Matrix form}$

$\begin{pmatrix}1 & 1\\4 & -1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}100\\320\end{pmatrix}$

$AX=B$

$\Delta=\begin{vmatrix}1 & 1\\4 & -1\end{vmatrix}=-1-4=-5$

$\Delta x=\begin{vmatrix}100 & 1\\320 & -1\end{vmatrix}=-100-320=-420$

$\Delta y=\begin{vmatrix}1 & 100\\4 & 320\end{vmatrix}=320-400=-80$

$x=\frac{\Delta x}{\Delta}=\frac{-420}{-5}=84$

$y=\frac{\Delta y}{\Delta}=\frac{-80}{-5}=16$

$\therefore\ \text{Number of correct answers } = 84$

$\text{Number of wrong answers } = 16$

3.A chemist has one solution which is 50% acid and another solution which is 25% acid. How much each should be mixed to make 10 litres of a 40% acid solution ?

$\text{Let quantity of 50\% acid solution } = x \text{ litres}$

$\text{Let quantity of 25\% acid solution } = y \text{ litres}$

$x+y=10 \;\rightarrow\; (1)$

$\frac{50}{100}x+\frac{25}{100}y=\frac{40}{100}\times10$

$\frac{x}{2}+\frac{y}{4}=4$

$2x+y=16 \;\rightarrow\; (2)$

$\text{Matrix form}$

$\begin{pmatrix}1 & 1\\2 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}10\\16\end{pmatrix}$

$AX=B$

$\Delta=\begin{vmatrix}1 & 1\\2 & 1\end{vmatrix}=1-2=-1$

$\Delta x=\begin{vmatrix}10 & 1\\16 & 1\end{vmatrix}=10-16=-6$

$\Delta y=\begin{vmatrix}1 & 10\\2 & 16\end{vmatrix}=16-20=-4$

$x=\frac{\Delta x}{\Delta}=\frac{-6}{-1}=6$

$y=\frac{\Delta y}{\Delta}=\frac{-4}{-1}=4$

$\therefore\ \text{Mix } 6 \text{ litres of 50\% solution and } 4 \text{ litres of 25\% solution}$

4.A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself ?

$\text{Let time taken by pump A } = x \text{ minutes}$

$\text{Let time taken by pump B } = y \text{ minutes}$

$\frac{1}{x}+\frac{1}{y}=\frac{1}{10}$

$\frac{1}{x}-\frac{1}{y}=\frac{1}{30}$

$\text{Let } \frac{1}{x}=a,\; \frac{1}{y}=b$

$a+b=\frac{1}{10} \;\rightarrow\; (1)$

$a-b=\frac{1}{30} \;\rightarrow\; (2)$

$100a+100b=10 \;\Rightarrow\; 10a+10b=1$

$30a-30b=1$

$\text{Matrix form}$

$\begin{pmatrix}10 & 10\\30 & -30\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}$

$AX=B$

$\Delta=\begin{vmatrix}10 & 10\\30 & -30\end{vmatrix}=-300-300=-600$

$\Delta a=\begin{vmatrix}1 & 10\\1 & -30\end{vmatrix}=-30-10=-40$

$\Delta b=\begin{vmatrix}10 & 1\\30 & 1\end{vmatrix}=10-30=-20$

$a=\frac{\Delta a}{\Delta}=\frac{-40}{-600}=\frac{1}{15}$

$b=\frac{\Delta b}{\Delta}=\frac{-20}{-600}=\frac{1}{30}$

$x=\frac{1}{a}=15,\quad y=\frac{1}{b}=30$

$\therefore\ \text{Pump A takes 15 minutes and Pump B takes 30 minutes}$

5.A family of 3 people went out for dinner in a restaurant. The cost of two dosai, three idlies and two vadais is `150. The cost of the two dosai, two idlies and four vadais is` 200. The cost of five dosai, four idlies and two vadais is `250. The family has` 350 in hand and they ate 3 dosai and six idlies and six vadais. Will they be able to manage to pay the bill within the amount they had ?

$\text{Let cost of 1 dosai } = x,\quad \text{1 idli } = y,\quad \text{1 vadai } = z$

$2x+3y+2z=150$

$2x+2y+4z=200$

$5x+4y+2z=250$

$\text{Matrix form}$

$\begin{pmatrix}2 & 3 & 2\\2 & 2 & 4\\5 & 4 & 2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}150\\200\\250\end{pmatrix}$

$AX=B$

$\Delta=2(4-16)-3(4-20)+2(8-10)$

$=2(-12)-3(-16)+2(-2)$

$=-24+48-4=20$

—

$\Delta x=\begin{vmatrix}150 & 3 & 2\\200 & 2 & 4\\250 & 4 & 2\end{vmatrix}$

$=150(4-16)-3(400-1000)+2(800-500)$

$=150(-12)-3(-600)+2(300)$

$=-1800+1800+600=600$

—

$\Delta y=\begin{vmatrix}2 & 150 & 2\\2 & 200 & 4\\5 & 250 & 2\end{vmatrix}$

$=2(400-1000)-150(4-20)+2(500-1000)$

$=2(-600)-150(-16)+2(-500)$

$=-1200+2400-1000=200$

—

$\Delta z=\begin{vmatrix}2 & 3 & 150\\2 & 2 & 200\\5 & 4 & 250\end{vmatrix}$

$=2(500-800)-3(500-1000)+150(8-10)$

$=2(-300)-3(-500)+150(-2)$

$=-600+1500-300=600$

—

$x=\frac{\Delta x}{\Delta}=\frac{600}{20}=30$

$y=\frac{\Delta y}{\Delta}=\frac{200}{20}=10$

$z=\frac{\Delta z}{\Delta}=\frac{600}{20}=30$

—

$\text{Cost of 1 dosai } = 30,\quad \text{1 idli } = 10,\quad \text{1 vadai } = 30$

$\text{Total cost for 3 dosai, 6 idlies and 6 vadais}$

$=3(30)+6(10)+6(30)$

$=90+60+180=330$

$\text{They have ₹350, hence they can pay the bill.}$

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