Exercise 1.5 Solutions – Applications of Matrices and Determinants (12th Maths)

IMPORTANT QUESTIONS

2 MARKS : Nill
3 Marks : Nill
5 Marks : 2,3,4(Rare)

Q1 (i). Solve the following systems of linear equations by
Gaussian elimination method:

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

$\text{Matrix form}$

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

$AX=B$

$\text{Augmented matrix } (A|B)$

$\left[\begin{array}{ccc|c}2 & -2 & 3 & 2\\1 & 2 & -1 & 3\\3 & -1 & 2 & 1\end{array}\right]$

$R_1 \leftrightarrow R_2$

$\left[\begin{array}{ccc|c}1 & 2 & -1 & 3\\2 & -2 & 3 & 2\\3 & -1 & 2 & 1\end{array}\right]$

$R_2 \to R_2-2R_1$

$R_3 \to R_3-3R_1$

$\left[\begin{array}{ccc|c}1 & 2 & -1 & 3\\0 & -6 & 5 & -4\\0 & -7 & 5 & -8\end{array}\right]$

$R_3 \to 6R_3-7R_2$

$\left[\begin{array}{ccc|c}1 & 2 & -1 & 3\\0 & -6 & 5 & -4\\0 & 0 & -5 & -20\end{array}\right]$

$\text{Equivalent equations}$

$x+2y-z=3 \;\rightarrow\; (1)$

$-6y+5z=-4 \;\rightarrow\; (2)$

$-5z=-20 \;\rightarrow\; (3)$

$\text{From (3), } z=4$

$\text{Substitute } z=4 \text{ in (2)}$

$-6y+5(4)=-4$

$-6y+20=-4$

$-6y=-24 \Rightarrow y=4$

$\text{Substitute } y \text{ and } z \text{ in (1)}$

$x+2(4)-4=3$

$x+8-4=3$

$x+4=3 \Rightarrow x=-1$

$\therefore\ x=-1,\; y=4,\; z=4$

Q1 (i). Solve the following systems of linear equations by
Gaussian elimination method:

$\textbf{1(ii). 2x+4y+6z=22,\; 3x+8y+5z=27,\; -x+y+2z=2}$

$\text{Divide (1) by 2}$

$x+2y+3z=11,\quad 3x+8y+5z=27,\quad -x+y+2z=2$

$\text{Matrix form}$

$\begin{pmatrix}1 & 2 & 3\\3 & 8 & 5\\-1 & 1 & 2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}11\\27\\2\end{pmatrix}$

$AX=B$

$\text{Augmented matrix } (A|B)$

$\left[\begin{array}{ccc|c}1 & 2 & 3 & 11\\3 & 8 & 5 & 27\\-1 & 1 & 2 & 2\end{array}\right]$

$R_2 \rightarrow R_2-3R_1$

$R_3 \rightarrow R_3+R_1$

$\left[\begin{array}{ccc|c}1 & 2 & 3 & 11\\0 & 2 & -4 & -6\\0 & 3 & 5 & 13\end{array}\right]$

$R_3 \rightarrow 2R_3-3R_2$

$\left[\begin{array}{ccc|c}1 & 2 & 3 & 11\\0 & 2 & -4 & -6\\0 & 0 & 22 & 44\end{array}\right]$

$\text{Equivalent equations}$

$x+2y+3z=11 \;\rightarrow\; (1)$

$2y-4z=-6 \;\rightarrow\; (2)$

$22z=44 \;\rightarrow\; (3)$

$\text{From (3), } z=2$

$\text{Substitute } z=2 \text{ in (2)}$

$2y-8=-6$

$2y=2 \Rightarrow y=1$

$\text{Substitute } y \text{ and } z \text{ in (1)}$

$x+2(1)+3(2)=11$

$x+2+6=11$

$x=11-8=3$

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

Q2.

$\textbf{2. If } ax^2+bx+c \text{ is divided by } x+3,\; x-5,\; x-1,$

$\text{the remainders are } 21,\; 61 \text{ and } 9 \text{ respectively.}$

$\text{Find } a,b,c.$

$\text{Given } P(x)=ax^2+bx+c$

$P(-3)=21$

$a(-3)^2+b(-3)+c=21$

$9a-3b+c=21 \;\rightarrow\; (1)$

$P(5)=61$

$a(5)^2+b(5)+c=61$

$25a+5b+c=61 \;\rightarrow\; (2)$

$P(1)=9$

$a(1)^2+b(1)+c=9$

$a+b+c=9 \;\rightarrow\; (3)$

—

$\text{Matrix form}$

$\begin{pmatrix}9 & -3 & 1\\25 & 5 & 1\\1 & 1 & 1\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}21\\61\\9\end{pmatrix}$

$\text{Augmented matrix } (A|B)$

$\left[\begin{array}{ccc|c}9 & -3 & 1 & 21\\25 & 5 & 1 & 61\\1 & 1 & 1 & 9\end{array}\right]$

$R_1 \leftrightarrow R_3$

$\left[\begin{array}{ccc|c}1 & 1 & 1 & 9\\25 & 5 & 1 & 61\\9 & -3 & 1 & 21\end{array}\right]$

$R_2 \rightarrow R_2-25R_1$

$R_3 \rightarrow R_3-9R_1$

$\left[\begin{array}{ccc|c}1 & 1 & 1 & 9\\0 & -20 & -24 & -164\\0 & -12 & -8 & -60\end{array}\right]$

$R_3 \rightarrow -R_3$

$\left[\begin{array}{ccc|c}1 & 1 & 1 & 9\\0 & -20 & -24 & -164\\0 & 12 & 8 & 60\end{array}\right]$

$R_3 \rightarrow 5R_3+3R_2$

$\left[\begin{array}{ccc|c}1 & 1 & 1 & 9\\0 & -20 & -24 & -164\\0 & 0 & -32 & -192\end{array}\right]$

—

$\text{Equivalent equations}$

$a+b+c=9 \;\rightarrow\; (1)$

$-20b-24c=-164 \;\rightarrow\; (2)$

$-32c=-192 \;\rightarrow\; (3)$

$\text{From (3), } c=6$

$\text{Substitute } c=6 \text{ in (2)}$

$-20b-24(6)=-164$

$-20b-144=-164$

$-20b=-20 \Rightarrow b=1$

$\text{Substitute } b \text{ and } c \text{ in (1)}$

$a+1+6=9$

$a+7=9 \Rightarrow a=2$

$\therefore\ a=2,\; b=1,\; c=6$

Q3.An amount of `65,000 is invested in three bonds at the rates of 6%, 8% and 9% per annum respectively. The total annual income is` 4,800. The income from the third bond is 600 more than that from the second bond. Determine the price of each bond.

$\text{Let the amounts invested be } x,\; y,\; z$

$x+y+z=65000 \;\rightarrow\; (1)$

$\frac{6}{100}x+\frac{8}{100}y+\frac{9}{100}z=4800$

$6x+8y+9z=480000 \;\rightarrow\; (2)$

$\frac{9}{100}z-\frac{8}{100}y=600$

$-8y+9z=60000 \;\rightarrow\; (3)$

—

$\text{Matrix form}$

$\begin{pmatrix}1 & 1 & 1\\6 & 8 & 9\\0 & -8 & 9\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}65000\\480000\\60000\end{pmatrix}$

$\text{Augmented matrix } (A|B)$

$\left[\begin{array}{ccc|c}1 & 1 & 1 & 65000\\6 & 8 & 9 & 480000\\0 & -8 & 9 & 60000\end{array}\right]$

$R_2 \rightarrow R_2-6R_1$

$\left[\begin{array}{ccc|c}1 & 1 & 1 & 65000\\0 & 2 & 3 & 90000\\0 & -8 & 9 & 60000\end{array}\right]$

$R_3 \rightarrow R_3+4R_2$

$\left[\begin{array}{ccc|c}1 & 1 & 1 & 65000\\0 & 2 & 3 & 90000\\0 & 0 & 21 & 420000\end{array}\right]$

—

$\text{Equivalent equations}$

$x+y+z=65000 \;\rightarrow\; (1)$

$2y+3z=90000 \;\rightarrow\; (2)$

$21z=420000 \;\rightarrow\; (3)$

$\text{From (3), } z=\frac{420000}{21}=20000$

$\text{Substitute } z=20000 \text{ in (2)}$

$2y+3(20000)=90000$

$2y+60000=90000$

$2y=30000 \Rightarrow y=15000$

$\text{Substitute } y \text{ and } z \text{ in (1)}$

$x+15000+20000=65000$

$x+35000=65000$

$x=30000$

—

$\therefore\ x=30000,\; y=15000,\; z=20000$

Q4

$\textbf{4. A boy is walking along the path } y=ax^2+bx+c$

$\text{through the points } (-6,8),\; (-2,-12),\; (3,8).$

$\text{He wants to meet his friend at } P(7,60). \text{ Will he meet his friend?}$

—

$\text{Substitute } (-6,8)$

$a(-6)^2+b(-6)+c=8$

$36a-6b+c=8 \;\rightarrow\; (1)$

$\text{Substitute } (-2,-12)$

$a(-2)^2+b(-2)+c=-12$

$4a-2b+c=-12 \;\rightarrow\; (2)$

$\text{Substitute } (3,8)$

$a(3)^2+b(3)+c=8$

$9a+3b+c=8 \;\rightarrow\; (3)$

—

$\text{Matrix form}$

$\begin{pmatrix}36 & -6 & 1\\4 & -2 & 1\\9 & 3 & 1\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}8\\-12\\8\end{pmatrix}$

$\text{Augmented matrix } (A|B)$

$\left[\begin{array}{ccc|c}36 & -6 & 1 & 8\\4 & -2 & 1 & -12\\9 & 3 & 1 & 8\end{array}\right]$

$R_2 \rightarrow 9R_2-R_1$

$R_3 \rightarrow 4R_3-R_1$

$\left[\begin{array}{ccc|c}36 & -6 & 1 & 8\\0 & -12 & 8 & -116\\0 & 18 & 3 & 24\end{array}\right]$

$R_3 \rightarrow 2R_3+3R_2$

$\left[\begin{array}{ccc|c}36 & -6 & 1 & 8\\0 & -12 & 8 & -116\\0 & 0 & 30 & -300\end{array}\right]$

—

$\text{Equivalent equations}$

$36a-6b+c=8 \;\rightarrow\; (1)$

$-12b+8c=-116 \;\rightarrow\; (2)$

$30c=-300 \;\rightarrow\; (3)$

$\text{From (3), } c=-10$

$\text{Substitute } c=-10 \text{ in (2)}$

$-12b+8(-10)=-116$

$-12b-80=-116$

$-12b=-36 \Rightarrow b=3$

$\text{Substitute } b \text{ and } c \text{ in (1)}$

$36a-6(3)-10=8$

$36a-18-10=8$

$36a=36 \Rightarrow a=1$

—

$\text{Equation of path: } y=x^2+3x-10$

$\text{At } x=7$

$y=7^2+3(7)-10$

$=49+21-10=60$

$\therefore\ (7,60) \text{ lies on the path, hence he will meet his friend.}$

Exercise 1.5 Solutions – Applications of Matrices and Determinants (12th Maths)

IMPORTANT QUESTIONS

Q1 (i). Solve the following systems of linear equations byGaussian elimination method:

Q1 (i). Solve the following systems of linear equations byGaussian elimination method:

Q2.

Q3.An amount of 65,000 is invested in three bonds at the rates of 6%, 8% and 9% per annum respectively. The total annual income is 4,800. The income from the third bond is 600 more than that from the second bond. Determine the price of each bond.

Q4

Q1 (i). Solve the following systems of linear equations by
Gaussian elimination method:

Q1 (i). Solve the following systems of linear equations by
Gaussian elimination method:

Q3.An amount of `65,000 is invested in three bonds at the rates of 6%, 8% and 9% per annum respectively. The total annual income is` 4,800. The income from the third bond is 600 more than that from the second bond. Determine the price of each bond.