5 Marks Questions

Question 15 Marks

Using determinants, show that the following system of linear equation is inconsistent:
$\begin{aligned}x-3 y+5 z & =4 \\2 x-6 y+10 z & =11 \\3 x-9 y+15 z & =12\end{aligned}$

Answer

For the given system of equations, we have
$D=\left|\begin{array}{ccc}1 & -3 & 5 \\ 2 & -6 & 10 \\ 3 & -9 & 15\end{array}\right|$
Taking $-3$ common from $C_{2'}$ we get
$D=\left|\begin{array}{rrr}1 & 1 & 5 \\ 2 & 2 & 10 \\ 3 & 3 & 15\end{array}\right|=0\quad[\because C_{1}$ and $C_{2}$ are identical$]$
$\text{Now, }D_x=\left|\begin{array}{ccc}4 & -3 & 5 \\ 11 & -6 & 10 \\ 12 & -9 & 15\end{array}\right|$
Taking common $(-3)$ and $5$ from $C_{2}$ and $C_{3'}$ we get
$D_x=-15\left|\begin{array}{ccc}4 & 1 & 1 \\ 11 & 2 & 2 \\ 12 & 3 & 3\end{array}\right|=0\quad[\because C_{2}$ and $C_{3}$ are identical$]$
$D_y=\left|\begin{array}{ccc}1 & 4 & 5 \\ 2 & 11 & 10 \\ 3 & 12 & 15\end{array}\right|$
Taking common (5) from $C_{3'}$ we get
$D_y=5\left|\begin{array}{ccc}1 & 4 & 1 \\ 2 & 11 & 2 \\ 3 & 12 & 3\end{array}\right|=0\quad[\because C_{1}$ and $C_{3}$ are identical$]$
$D_z=\left|\begin{array}{ccc}1 & -3 & 4 \\ 2 & -6 & 11 \\ 3 & -9 & 12\end{array}\right|$
Taking Common $-3$ from $C_{2'}$ we get
$D_z=(-3)\left|\begin{array}{ccc}1 & 1 & 4 \\ 2 & 2 & 11 \\ 3 & 3 & 12\end{array}\right|=0$
$\left[\because C_1 \quad\right.$ and $\quad C_2 \quad$ are identical]
So, given system of equations may or many not be consistent.
If we put $z=k$ in any two of three equations, we find that the two equations obtained are inconsistent as they represent a pair of parallel lines.
$\text{i.e.,}\quad x-3y=4-5k$
$\quad\quad2x-6y=11-10k\quad\text{[Taking first two equations]}$
$\text{or }x-3y=4-5k$
$x-3y= \frac{11-10k}{2},$
which represent pair of parallel lines.
Hence, the given system of equations is inconsistent..

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Question 25 Marks

Solve the following system of equations by Cramer's rule :
$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4, \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1 \text { and } \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

Answer

Let $\frac{1}{x}=p, \frac{1}{y}=q$ and $\frac{1}{z}=r$, then the above system of equations can be written as
$\begin{array}{r}2 p+3 q+10 r=4 \\4 p-6 q+5 r=1 \\6 p+9 q-20 r=2\end{array}$
$\begin{aligned}\text{Here,}\quad D & =\left|\begin{array}{ccc}2 & 3 & 10 \\4 & -6 & 5 \\6 & 9 & -20\end{array}\right| \\& =2(120-45)-3(-80-30)+10(36+36) \\& =150+330+720=1200 \\D_1 & =\left|\begin{array}{ccc}4 & 3 & 10 \\1 & -6 & 5 \\2 & 9 & -20\end{array}\right| \\& =4(120-45)-3(-20-10)+10(9+12) \\& =300+90+210=600 \\D_2 & =\left|\begin{array}{ccc}2 & 4 & 10 \\4 & 1 & 5 \\6 & 2 & -20\end{array}\right| \\& =2(-20-120)-4(-80-30)+10(8-6) \\& =-60+440+20=400\end{aligned}$
$\begin{aligned}\text{and}\quad D_3 & =\left|\begin{array}{ccc}2 & 3 & 4 \\ 4 & -6 & 1 \\ 6 & 9 & 2\end{array}\right| \\ & =2(-12-9)-3(8-6)+4(36+36) \\ & =-42-6+288=240\end{aligned}$
$\begin{aligned}\therefore \quad p & =\frac{D_1}{D}=\frac{600}{1200}=\frac{1}{2} \Rightarrow \frac{1}{x}=\frac{1}{2} \Rightarrow x=2, \\q & =\frac{D_2}{D}=\frac{400}{1200}=\frac{1}{3} \Rightarrow \frac{1}{y}=\frac{1}{3} \Rightarrow y=3, \\
r & =\frac{D_3}{D}=\frac{240}{1200}=\frac{1}{5} \Rightarrow \frac{1}{z}=\frac{1}{5} \Rightarrow z=5,\end{aligned}$
Hence, $x=2, y=3$ and $z=5.$

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Question 35 Marks

For what values of $a$ and $b$, the system of equations
$\begin{array}{r}2 x+a y+6 z=8 \\x+2 y+b z=5 \\x+y+3 z=4,\end{array}$
has (i) unique solution (ii) no solution?

Answer

For the given system of equations, we have
$\begin{aligned} D & =\left|\begin{array}{lll}2 & a & 6 \\ 1 & 2 & b \\ 1 & 1 & 3\end{array}\right|=2(6-b)-a(3-b)+6(1-2) \\ & =12-2 b-3 a+a b-6 \\ & =6-3 a-2 b+a b \\ & =(b-3)(a-2)\end{aligned}$
$\begin{aligned} D_x & =\left|\begin{array}{lll}8 & a & 6 \\ 5 & 2 & b \\ 4 & 1 & 3\end{array}\right|=8(6-b)-a(15-4 b)+6(5-8) \\ & =48-8 b-15 a+4 a b-18 \\ & =30-15 a-8 b+4 a b \\ & =(a-2)(4 b-15)\end{aligned}$
$\begin{aligned} D_y & =\left|\begin{array}{lll}2 & 8 & 6 \\ 1 & 5 & b \\ 1 & 4 & 3\end{array}\right|=2(15-4 b)-8(3-b)+6(4-5) \\ & =30-8 b-24+8 b-6 \\ & =0\end{aligned}$
$\begin{aligned} D_z & =\left|\begin{array}{lll}2 & a & 8 \\ 1 & 2 & 5 \\ 1 & 1 & 4\end{array}\right|=2(8-5)-a(4-5)+8(1-2) \\ & =6+a-8=a-2\end{aligned}$
(i) For unique solution, we must have
$D \neq 0 \Rightarrow(a-2)(b-3) \neq 0 \Rightarrow a \neq 2 \text { or } b \neq 3$
(ii) For no solution, we must have $D=0$ and at least one of $D_{x^{\prime}} D_y$ and $D_z$ is non-zero.
Clearly, $b=3$, we have
$D=0 \text { and } D_{z} \neq 0$
Hence, the system has no solution for $b=3.$

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Question 45 Marks

In $A=\left(\begin{array}{ccc}2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{array}\right)$ find $A^{-1}$. Using $A^{-1}$ solve the system of equations $\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=2 ; \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=5$; $\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=-4$

Answer

Here, $|A|=1200$
Co-factors are
$\left.\begin{array}{lll}C_{11}=75, & C_{21}=150, & C_{31}=75 \\C_{12}=110, & C_{22}=-100, & C_{32}=30 \\C_{13}=72, & C_{23}=0, & C_{33}=-24\end{array}\right\}$
$A^{-1}=\frac{1}{1200}\left[\begin{array}{ccc}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{array}\right]$
Given equation in matrix from is :
$\left[\begin{array}{ccc}2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{array}\right]\left[\begin{array}{l}\frac{1}{x} \\ \frac{1}{y} \\ \frac{1}{z}\end{array}\right]=\left[\begin{array}{c}2 \\ 5 \\ -4\end{array}\right]$
$\text{or }AX=B$
$\text{or }X=A^{-1}B$
$\text{or}\left[\begin{array}{c}\frac{1}{x} \\ \frac{1}{y} \\ \frac{1}{z}\end{array}\right]=\left[\begin{array}{c}\frac{1}{2} \\ \frac{-1}{3} \\ \frac{1}{5}\end{array}\right]$
$\text{or }x=2,y=-3,z=5$

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Question 55 Marks

Use product $\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$ to solve the system of equations $x+3 z=9,-x+2 y-2 z=4$, $2 x-3 y+4 z=-3$

Answer

$\begin{array}{c}AB={\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]} \\ A B=I\end{array}$
$\text{or }A^{-1}=B=\left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$
Given equations in matrix form are:
$\begin{aligned} {\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right] } & =\left[\begin{array}{c}9 \\ 4 \\ -3\end{array}\right] \\ A X & =C\end{aligned}$
$\text{or}\quad X=(A)^{-1}C=(A^{-1})C$
$\text{or }\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{ccc}-2 & 9 & 6 \\ 0 & 2 & 1 \\ 1 & -3 & -2\end{array}\right]\left[\begin{array}{c}9 \\ 4 \\ -3\end{array}\right]=\left[\begin{array}{l}0 \\ 5 \\ 3\end{array}\right]$
$\text{or}\quad x=0,y=5,z=3$

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Question 65 Marks

Given $A=\left[\begin{array}{lll}5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1\end{array}\right], B^{-1}=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right],$ compute $(A B)^{-1}.$

Answer

$|A|=5(-1)+4(1)=-1$
$\begin{array}{lll}C_{11}=-1 & C_{21}=8 & C_{31}=-12 \\ C_{12}=0 & C_{22}=1 & C_{32}=-2 \\ C_{13}=1 & C_{23}=-10 & C_{33}=15\end{array}$
$\begin{aligned} A^{-1} & =\left[\begin{array}{ccc}1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15\end{array}\right] \\ (A B)^{-1} & =B^{-1} A^{-1}\end{aligned}$
$\begin{array}{l}=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]\left[\begin{array}{ccc}1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15\end{array}\right] \\ =\left[\begin{array}{lll}-2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42\end{array}\right]\end{array}$

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Question 75 Marks

If $A=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1\end{array}\right]$, find $A^{-1}$. Hence, solve the system of equations $x+y+z=6, x+2 z=7,3 x+y+z=12.$

Answer

$|A| =4\neq0 \Rightarrow A^{-1}$ exists.
$\text{Adj }A=\left[\begin{array}{crc}-2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1\end{array}\right]$
$\therefore A^{-1}=\frac{1}{|A|} \cdot$$\text{adj }A=\frac{1}{4}\left[\begin{array}{crc}-2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1\end{array}\right]$
Given system of equations can be written as $Ax=B$
where, $X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right], B=\left[\begin{array}{l}6 \\ 7 \\ 12\end{array}\right]$
$\therefore X=A^{-1}\cdot B$
$=\frac{1}{4}\left[\begin{array}{crc}-2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1\end{array}\right]\left[\begin{array}{l}6 \\ 7 \\ 12\end{array}\right]=\left[\begin{array}{l}3 \\ 1 \\ 2\end{array}\right]$
$\therefore x=3,y=1,z=2$

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Question 85 Marks

If $A=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right],$ Find $A^{-1}.$ Use it to solve the system of equations.
Hence, solve the system of equations:
$2 x-3 y+5 z=11$
$\begin{array}{r}3 x+2 y-4 z=-5 \\ x+y-2 z=-3\end{array}$

Answer

$|A|=-1 \neq 0 \therefore A^{-1} \text { exists }$
Co-factors of A are:
$\begin{array}{lll}A_{11}=0 ; & A_{12}=2 ; & A_{13}=1 \\A_{21}=-1 ; & A_{22}=-9 ; & A_{23}=-5 \\A_{31}=2 ; & A_{32}=23 ; & A_{33}=13\end{array}$
$\operatorname{adj}(A)=\left[\begin{array}{ccc}0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{array}\right]$
$\Rightarrow A^{-1}=\frac {1}{|A|}.\text{adj}(A)$
$=\left[\begin{array}{ccc}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right]$
For, $X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}11 \\ -5 \\ -3\end{array}\right],$ the system of equation is $A\cdot X=B$
$\therefore X=A^{-1} \cdot B=\left[\begin{array}{ccc}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right]\left[\begin{array}{l}11 \\ -5 \\ -3\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
$\therefore x=1,y=2,z=3$

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Question 95 Marks

If $A=\left[\begin{array}{ccc}3 & 1 & 2 \\ 3 & 2 & -3 \\ 2 & 0 & -1\end{array}\right],$ Find $A^{-1}.$ Hence, solve the system of equations:
$\begin{array}{r}3 x+3 y+2 z=1 \\x+2 y=4 \\2 x-3 y-z=5\end{array}$

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Question 105 Marks

Using properties of determinants, prove that
$\left|\begin{array}{ccc}\frac{(a+b)^2}{c} & c & c \\a & \frac{(b+c)^2}{a} & a \\b & b & \frac{(c+a)^2}{b}\end{array}\right|=2(a+b+c)^3 .$

Answer

$\text{LHS}=\frac{1}{a b c}\left|\begin{array}{ccc}(a+b)^2 & c^2 & c^2 \\ a^2 & (b+c)^2 & a^2 \\ b^2 & b^2 & (c+a)^2\end{array}\right|$
$\text{C}_{1} \rightarrow \text{C}_{1}-\text{C}_{3'}\text{C}_{2} \rightarrow \text{C}_{2}-\text{C}_{3}$
$=\frac{1}{abc}$
$\left|\begin{array}{ccc}(a+b+c)(a+b-c) & 0 & c^2 \\ 0 & (b+c+a)(b+c-a) & a^2 \\ (b+c+a)(b-c-a) & (b+c+a)(b-c-a) & (c+a)^2\end{array}\right|$
$=\frac{(a+b+c)^2}{a b c}\left|\begin{array}{ccc}(a+b-c) & 0 & c^2 \\ 0 & (b+c-a) & a^2 \\ (b-c-a) & (b-c-a) & (c+a)^2\end{array}\right|$
$\text{R}_{3} \rightarrow \text{R}_{3}-\text{R}_{1}-\text{R}_{2}$
$=\frac{(a+b+c)^2}{a b c}\left|\begin{array}{ccc}(a+b-c) & 0 & c^2 \\ 0 & (b+c-a) & a^2 \\ (-2 a) & (-2 c) & 2 c a\end{array}\right|$
Taking common $2ac$ from $\text{R}_{3}$
$=\frac{(2 a c)(a+b+c)^2}{b a^2 c^2}\left|\begin{array}{ccc}a c+b c-c^2 & 0 & c^2 \\ 0 & a b+c a-a^2 & a^2 \\ -1 & -1 & 1\end{array}\right|$
$(\text{C}_1 \rightarrow \text{C}_1+\text{C}_{3'} \text{C}_2 \rightarrow \text{C}_2+\text{C}_3)$
$=\frac{2 a c(a+b+c)^2}{b a^2 c^2}\left|\begin{array}{ccc}(a c+b c) & c^2 & c^2 \\ a^2 & (b a+c a) & a^2 \\ 0 & 0 & 2 c a\end{array}\right|$
Taking $c$ from $R_1$ & $a$ from $R_2$
$=\frac{2 a c(a+b+c)^2}{(a c) b}\left|\begin{array}{ccc}a+b & c & c \\ a & b+c & a \\ 0 & 0 & 1\end{array}\right|$
Expanding along $\text{R}_{3}$
$=\frac{2 a c(a+b+c)^2}{a b c}\left[a b+a c+b^2+b c-a c\right]$
$=\frac{2abc(a+b+c)^{3}}{abc}$
$=2(a+b+c)^{3}$

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Question 115 Marks

Using properties of determinants, prove that:
$\left|\begin{array}{ccc}(y+z)^2 & x y & z x \\x y & (x+z)^2 & y z \\x z & y z & (x+y)^2\end{array}\right|=2 x y z(x+y+z)^3 .$

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Question 125 Marks

Prove that $\left|\begin{array}{lll}y z-x^2 & z x-y^2 & x y-z^2 \\ z x-y^2 & x y-z^2 & y z-x^2 \\ x y-z^2 & y z-x^2 & z x-y^2\end{array}\right|$ is divisible by $(x+y+z)$, and hence find the quotient.

Answer

$\text { L.H.S. }=\left|\begin{array}{lll}y z-x^2 & z x-y^2 & x y-z^2 \\z x-y^2 & x y-z^2 & y z-x^2 \\x y-z^2 & y z-x^2 & z x-y^2\end{array}\right|$
Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$
$=\left|\begin{array}{ccc}y z-x^2 & z x-y^2 & x y-z^2 \$x-y)(x+y+z) & (y-z)(x+y+z) & (z-x)(x+y+z) \$x-z)(x+y+z) & (y-x)(x+y+z) & (z-y)(x+y+z)\end{array}\right|$
Taking $(x+y+z)$ common from $R_2$ and $R_3$
$\begin{array}{l}=(x+y+z)^2\left|\begin{array}{ccc}y z-x^2 & z x-y^2 & x y-z^2 \\ x-y & y-z & z-x \\ x-z & y-x & z-y\end{array}\right| \\ \text { Applying } C_1 \rightarrow C_1+C_2+C_3 \\ =(x+y+z)^2\end{array}$
$\left|\begin{array}{ccc}(x y+y z+z x)-\left(x^2+y^2+z^2\right) & z x-y^2 & x y-z^2 \\ 0 & y-z & z-x \\ 0 & y-x & z-y\end{array}\right|$
Expanding along $C_1$
$=(x+y+z)^{2}[xy+yz+zx-(x^{2}+y^{2}+z^{2}]$ $[(y-z)(z-y)-(z-x)(y-x)]$
$=(x+y+z)^2\left[x y+y z+z x-\left(x^2+y^2+z^2\right)\right]$ $\left[y z-y^2-z^2+y z-y z+x z+x y-x^2\right]$
$=(x+y+z)^{2}[xy+yz+zx-(x^{2}+y^{2}+z^{2})]^{2}$
Hence, it is divisible by $(x+y+z)$ and the quotient is $(x+y+z)[xy+yz+zx-(x^{2}+y^{2}+z^{2})]^{2}.$

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Applied Maths 5 Marks Questions

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