Exercise 1.3 Solutions – Applications of Matrices and Determinants (12th Maths)

IMPORTANT QUESTIONS

2 Marks : Nill

3 Marks : 2,3 & 5 (Rare)

5 Marks :

Q1(i) Solve by matrix inversion method

$\textbf{1(i). 2x+5y=-2 ; x+2y=-3}$

$\textbf{Step 1: Write in matrix form}$

$AX=B$

$A=\begin{pmatrix}2 & 5\\ 1 & 2\end{pmatrix}\;X=\begin{pmatrix}x\\ y\end{pmatrix}\;B=\begin{pmatrix}-2\\ -3\end{pmatrix}$

$\textbf{Step 2: Find } A^{-1}$

$|A|=4-5=-1\neq0$

$A^{-1}=\frac{1}{-1}\begin{pmatrix}2 & -5\\-1 & 2\end{pmatrix}=\begin{pmatrix}-2 & 5\\1 & -2\end{pmatrix}$

$\textbf{Step 3: Find } X=A^{-1}B$

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

$=\begin{pmatrix}4-15\\-2+6\end{pmatrix}=\begin{pmatrix}-11\\4\end{pmatrix}$

$\therefore\ x=-11,\; y=4$

Q1(ii) Solve by matrix inversion method

$\textbf{1(ii). 2x-y=8,\; 3x+2y=-2}$

$\textbf{Step 1: Write in matrix form}$

$A=\begin{pmatrix}2 & -1\\ 3 & 2\end{pmatrix},\;X=\begin{pmatrix}x\\ y\end{pmatrix},\;B=\begin{pmatrix}8\\ -2\end{pmatrix}$

$AX=B \Rightarrow X=A^{-1}B$

$\textbf{Step 2: Find } A^{-1}$

$|A|=4+3=7\neq0$

$A^{-1}=\frac{1}{7}\begin{pmatrix}2 & 1\\-3 & 2\end{pmatrix}$

$\textbf{Step 3: Find } X=A^{-1}B$

$\begin{pmatrix}x\\ y\end{pmatrix}=\frac{1}{7}\begin{pmatrix}2 & 1\\-3 & 2\end{pmatrix}\begin{pmatrix}8\\-2\end{pmatrix}$

$=\frac{1}{7}\begin{pmatrix}(2\times8)+(1\times-2)\\(-3\times8)+(2\times-2)\end{pmatrix}$

$=\frac{1}{7}\begin{pmatrix}14\\-28\end{pmatrix}=\begin{pmatrix}2\\-4\end{pmatrix}$

$\therefore\ x=2,\; y=-4$

Q1(iii) Solve by matrix inversion method

$\textbf{1(iii). 2x+3y-z=9,\; x+y+z=9,\; 3x-y-z=-1}$

$\textbf{Step 1: Write in matrix form}$

$A=\begin{pmatrix}2 & 3 & -1\\1 & 1 & 1\\3 & -1 & -1\end{pmatrix}\;X=\begin{pmatrix}x\\ y\\ z\end{pmatrix}\;B=\begin{pmatrix}9\\ 9\\ -1\end{pmatrix}$

$AX=B \Rightarrow X=A^{-1}B$

$\textbf{Step 2: Find } |A|$

$|A|=2(-1+1)-3(-1-3)-1(-1-3)$

$=2(0)-3(-4)-1(-4)$

$=0+12+4=16$

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

$A^{-1}=\frac{1}{16}\begin{pmatrix}+( -1+1 ) & -( -1-3 ) & +( -1-3 )\\-( -3-1 ) & +( -2+1 ) & -( -2-1 )\\+( -3+1 ) & -( 2+1 ) & +( 2-3 )\end{pmatrix}^{T}$

$=\frac{1}{16}\begin{pmatrix}0 & 4 & -4\\4 & 1 & 3\\-2 & -3 & -1\end{pmatrix}^{T}$

$=\frac{1}{16}\begin{pmatrix}0 & 4 & -2\\4 & 1 & -3\\-4 & 3 & -1\end{pmatrix}$

$\textbf{Step 4: Find } X=A^{-1}B$

$\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\frac{1}{16}\begin{pmatrix}0 & 4 & -2\\4 & 1 & -3\\-4 & 3 & -1\end{pmatrix}\begin{pmatrix}9\\9\\-1\end{pmatrix}$

$=\frac{1}{16}\begin{pmatrix}0+36+2\\36+9+3\\-36+27+1\end{pmatrix}$

$=\frac{1}{16}\begin{pmatrix}38\\48\\-8\end{pmatrix}=\begin{pmatrix}2\\3\\4\end{pmatrix}$

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

Q1(iv) Solve by matrix inversion method

$\textbf{1(iv). x+y+z-2=0,\; 6x-4y+5z-31=0,\; 5x+2y+2z=13}$

$\textbf{Step 1: Write in matrix form}$

$x+y+z=2,\quad 6x-4y+5z=31,\quad 5x+2y+2z=13$

$A=\begin{pmatrix}1 & 1 & 1\\6 & -4 & 5\\5 & 2 & 2\end{pmatrix}\;X=\begin{pmatrix}x\\ y\\ z\end{pmatrix}\;B=\begin{pmatrix}2\\ 31\\ 13\end{pmatrix}$

$AX=B \Rightarrow X=A^{-1}B$

$\textbf{Step 2: Find } |A|$

$|A|=( -8-10 )-1(12-25)+1(12+20)$

$=-18-(-13)+32$

$=-18+13+32=27$

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

$A^{-1}=\frac{1}{27}\begin{pmatrix}+( -8-10 ) & -(12-25) & +(12+20)\\-(2-2) & +(2-5) & -(2-5)\\+(4+5) & -(6-5) & +(4-2)\end{pmatrix}^{T}$

$=\frac{1}{27}\begin{pmatrix}-18 & 13 & 32\\0 & -3 & 3\\9 & -1 & 2\end{pmatrix}^{T}$

$=\frac{1}{27}\begin{pmatrix}-18 & 0 & 9\\13 & -3 & -1\\32 & 3 & 2\end{pmatrix}$

$\textbf{Step 4: Find } X=A^{-1}B$

$\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\frac{1}{27}\begin{pmatrix}-18 & 0 & 9\\13 & -3 & -1\\32 & 3 & 2\end{pmatrix}\begin{pmatrix}2\\31\\13\end{pmatrix}$

$=\frac{1}{27}\begin{pmatrix}-36+0+117\\26-93-13\\64+93+26\end{pmatrix}$

$=\frac{1}{27}\begin{pmatrix}81\\-80\\183\end{pmatrix}=\begin{pmatrix}3\\-2\\1\end{pmatrix}$

$\therefore\ x=3,\; y=-2,\; z=1$

Q2

$\textbf{2. If }A=\begin{pmatrix}-5 & 1 & 3\\ 7 & 1 & -5\\ 1 & -1 & 2\end{pmatrix},\;B=\begin{pmatrix}1 & 1 & 2\\ 3 & 2 & 1\\ 2 & 1 & 3\end{pmatrix},\text{ find } AB \text{ and } BA \text{ and hence solve}$

$x+y+2z=1,\quad 3x+2y+z=7,\quad 2x+y+3z=2$

[/latex]

$\textbf{Step 1: Find } AB$

$AB=\begin{pmatrix}-5 & 1 & 3\\ 7 & 1 & -5\\ 1 & -1 & 2\end{pmatrix}\begin{pmatrix}1 & 1 & 2\\ 3 & 2 & 1\\ 2 & 1 & 3\end{pmatrix}$

$=\begin{pmatrix}-5+3+6 & -5+2+3 & -10+1+9\\7+3-10 & 7+2-5 & 14+1-15\\1-3+4 & 1-2+2 & 2-1+6\end{pmatrix}$

$=\begin{pmatrix}4 & 0 & 0\\0 & 4 & 0\\0 & 0 & 4\end{pmatrix}=4I_3$

—

$\textbf{Step 2: Find } BA$

$BA=\begin{pmatrix}1 & 1 & 2\\ 3 & 2 & 1\\ 2 & 1 & 3\end{pmatrix}\begin{pmatrix}-5 & 1 & 3\\ 7 & 1 & -5\\ 1 & -1 & 2\end{pmatrix}$

$=\begin{pmatrix}-5+7+2 & 1+1-2 & 3-5+4\\-15+14+1 & 3+2-1 & 9-10+2\\-10+7+3 & 2+1-3 & 6-5+6\end{pmatrix}$

$=\begin{pmatrix}4 & 0 & 0\\0 & 4 & 0\\0 & 0 & 4\end{pmatrix}=4I_3$

—

$\textbf{Step 3: Hence } AB=BA=4I_3\Rightarrow B^{-1}=\frac{1}{4}A$

—

$\textbf{Step 4: Solve the system}$

$\begin{pmatrix}1 & 1 & 2\\3 & 2 & 1\\2 & 1 & 3\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}1\\ 7\\ 2\end{pmatrix}$

$BX=C \Rightarrow X=B^{-1}C$

$\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\frac{1}{4}\begin{pmatrix}-5 & 1 & 3\\ 7 & 1 & -5\\ 1 & -1 & 2\end{pmatrix}\begin{pmatrix}1\\ 7\\ 2\end{pmatrix}$

$=\frac{1}{4}\begin{pmatrix}-5+7+6\\7+7-10\\1-7+4\end{pmatrix}=\frac{1}{4}\begin{pmatrix}8\\4\\-2\end{pmatrix}$

$=\begin{pmatrix}2\\1\\-1\end{pmatrix}$

$\therefore\ x=2,\; y=1,\; z=-1$

Q3 : A man is appointed in a job with a monthly salary of certain amount and a fixed amount of annual increment. If his salary was ₹19,800 per month at the end of the first month after 3 years of service and ₹23,400 per month at the end of the first month after 9 years of service, find his starting salary and his annual increment. (Use matrix inversion method)

$\text{Let the monthly salary be } x \text{ and annual increment be } y$

$\text{Writing the equation in matrix form}$

$\begin{pmatrix}1 & 3\\1 & 9\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}19800\\23400\end{pmatrix}$

$AX=B \Rightarrow X=A^{-1}B$

$\text{To find } A^{-1}$

$|A|=9-3=6$

$A^{-1}=\frac{1}{|A|}\text{adj }A=\frac{1}{6}\begin{pmatrix}9 & -3\\-1 & 1\end{pmatrix}$

$\begin{pmatrix}x\\y\end{pmatrix}=\frac{1}{6}\begin{pmatrix}9 & -3\\-1 & 1\end{pmatrix}\begin{pmatrix}19800\\23400\end{pmatrix}$

$=\frac{1}{6}\begin{pmatrix}178200-70200\\-19800+23400\end{pmatrix}$

$=\frac{1}{6}\begin{pmatrix}108000\\3600\end{pmatrix}=\begin{pmatrix}18000\\600\end{pmatrix}$

$\therefore\ \text{Monthly salary } = 18000,\quad \text{Annual increment } = 600$

Q4.Four men and 4 women can finish a piece of work jointly in 3 days while 2 men and 5 women can finish the same work jointly in 4 days. Find the time taken by one man alone and that of one woman alone to finish the work by using matrix inversion method.

$\text{Let the work done by one man in 1 day } = x$

$\text{Let the work done by one woman in 1 day } = y$

$4x+4y=\frac{1}{3}\;\Rightarrow\;12x+12y=1$

$2x+5y=\frac{1}{4}\;\Rightarrow\;8x+20y=1$

$\text{Matrix form}$

$\begin{pmatrix}12 & 12\\8 & 20\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}$

$AX=B \Rightarrow X=A^{-1}B$

$\text{To find } A^{-1}$

$|A|=(240-96)=144$

$A^{-1}=\frac{1}{144}\begin{pmatrix}20 & -12\\-8 & 12\end{pmatrix}$

$\begin{pmatrix}x\\y\end{pmatrix}=\frac{1}{144}\begin{pmatrix}20 & -12\\-8 & 12\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}$

$=\frac{1}{144}\begin{pmatrix}20-12\\-8+12\end{pmatrix}=\frac{1}{144}\begin{pmatrix}8\\4\end{pmatrix}$

$=\begin{pmatrix}\frac{1}{18}\\\frac{1}{36}\end{pmatrix}$

$x=\frac{1}{18},\quad y=\frac{1}{36}$

$\text{Time taken by one man alone } = 18 \text{ days}$

$\text{Time taken by one woman alone } = 36 \text{ days}$

Q5 : The prices of three commodities A B, and C are `x y, and z per units respectively. A person P purchases 4 units of B and sells two units of A and 5 units of C . Person Q purchases 2 units of C and sells 3 units of A and one unit of B . Person R purchases one unit of A and sells 3 unit of B and one unit of C . In the process, P Q, and R earn` 15,000, `1,000 and` 4,000 respectively. Find the prices per unit of A B, and C . (Use matrix inversion method to
solve the problem.)

$\text{Price of } A = x,\quad \text{Price of } B = y,\quad \text{Price of } C = z$

$\text{Given}$

$2x-y+5z=15000$

$3x+y-2z=1000$

$-x+3y+z=4000$

$\text{Matrix form}$

$\begin{pmatrix}2 & -1 & 5\\3 & 1 & -2\\-1 & 3 & 1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}15000\\1000\\4000\end{pmatrix}$

$AX=B \Rightarrow X=A^{-1}B$

$\text{To find } A^{-1}$

$|A|=2(1+6)-(-1)(3-2)+5(9+1)$

$=2(7)+1(1)+5(10)$

$=14+1+50=65$

$\text{adj }A=\begin{pmatrix}7 & 19 & 3\\-1 & 7 & 19\\10 & -2 & 14\end{pmatrix}$

$A^{-1}=\frac{1}{65}\begin{pmatrix}7 & 19 & 3\\-1 & 7 & 19\\10 & -2 & 14\end{pmatrix}$

$\begin{pmatrix}x\\y\\z\end{pmatrix}=\frac{1}{65}\begin{pmatrix}7 & 19 & 3\\-1 & 7 & 19\\10 & -2 & 14\end{pmatrix}\begin{pmatrix}15000\\1000\\4000\end{pmatrix}$

$=\frac{1}{65}\begin{pmatrix}105000+19000+12000\\-15000+7000+76000\\150000-2000+56000\end{pmatrix}$

$=\frac{1}{65}\begin{pmatrix}136000\\68000\\204000\end{pmatrix}$

$=\begin{pmatrix}2000\\1000\\3000\end{pmatrix}$

$\therefore\ x=2000,\; y=1000,\; z=3000$