Chapter 4: Determinants

Ans: The given system can be written as $A X=B$, where

$A=\left[\begin{array}{ccc} 2 & 1 & 1 \\ 2 & -4 & -2 \\ 0 & 3 & -5 \end{array}\right], X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{l} 1 \\ 3 \\ 9 \end{array}\right]$

$\begin{array}{l}\left|\begin{array}{lll}2 & 1 & 1 \\ 2 & -4 & -2 \\ 0 & 3 & -5\end{array}\right| \\ = & 2(20+6)-1(-10-0)+1(6-0) \\ = & 52+10+6=68 \neq 0\end{array}$

Thus, $\mathrm{A}$ is non-singular, Therefore, its inverse exists.

Therefore, the given system is consistent and has a unique solution given by $X=$ $A^{-1} B$

Cofactors of $A$ are 

$\begin{array}{l} A_{11}=20+6=26 \\ A_{12}=-(-10+0)=10 \\ A_{13}=6+0=6 \\ A_{21}=-(-5-3)=8 \\ A_{22}=-10-0=-10 \\ A_{23}=-(6-0)=-6 \\ A_{31}=(-2+4)=2 \\ A_{32}=-(-4-2)=6 \\ A_{33}=-8-2=-10 \end{array}$

$\operatorname{adj}(A)=\left[\begin{array}{ccc}26 & 10 & 6 \\ 8 & -10 & -6 \\ 2 & 6 & -10\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}26 & 8 & 2 \\ 10 & -10 & 6 \\ 6 & -6 & -10\end{array}\right]$

$\therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{68}\left[\begin{array}{ccc}26 & 8 & 2 \\ 10 & -10 & 6 \\ 6 & -6 & -10\end{array}\right]$

Now, $X=A^{-1} B \Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$

$=\frac{1}{68}\left[\begin{array}{ccc}26 & 8 & 2 \\ 10 & -10 & 6 \\ 6 & -6 & -10\end{array}\right]\left[\begin{array}{l}1 \\ 3 \\ 9\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{68}\left[\begin{array}{c}26+24+18 \\ 10-30+54 \\ 6-18-90\end{array}\right]$

$=\frac{1}{68}\left[\begin{array}{c}68 \\ 34 \\ -102\end{array}\right]=\left[\begin{array}{c}1 \\ \frac{1}{2} \\ \frac{-3}{2}\end{array}\right]$

Hence, $x=1, y=\frac{1}{2}$ and $z=\frac{-3}{2}$

Ans: The given system of equations can be written in the form of $AX = B$,

where $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$

Now, $|A|=1(1+3)+1(2+3)+1(2-1)=4+5+1=10 \neq 0$

Thus $A$ is non-singular. Therefore, its inverse exists. Now, $A_{11}=4, A_{12}=-5, A_{13}=1$

$\begin{array}{l} A_{21}=2, A_{22}=0, A_{23}=-2 \\ A_{31}=2, A_{32}=5, A_{33}=3 \\ \therefore A^{-1}=\dfrac{1}{|A|}(a d j A)=\dfrac{1}{10}\left[\begin{array}{ccc} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array}\right] \end{array}$

$\begin{array}{l} \therefore X=A^{-1} B=\dfrac{1}{10}\left[\begin{array}{ccc} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array}\right]\left[\begin{array}{l} 4 \\ 0 \\ 2 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]-\dfrac{1}{10}\left[\begin{array}{c} 16+0+4 \\ -20+0+10 \\ 4+0+6 \end{array}\right] \\ =\dfrac{1}{10}\left[\begin{array}{c} 20 \\ -10 \\ 10 \end{array}\right] \end{array}$

$=\left[\begin{array}{c} 2 \\ -1 \\ 1 \end{array}\right]$

Hence, $x=2,\; y=-1,\;\text{&}\; z=1$

Ans: The given system of equation can be written in the form of $A X=B$ where

$\begin{array}{c} |A|=2(4+1)-3(2-3)+3(-1+6) \\ \quad=2(5)-3(-5)+3(5) \\ =10+15+15=40 \neq 0 \end{array}$

Thus, $A$ is non-singular. Therefore, its inverse exists. Now.

$\begin{array}{l} A_{11}=5, A_{2}=5, A_{13}=5 \\ A_{21}=3, A_{22}=-13, A_{23}-11 \\ A_{34}=9, A_{12}=1, A_{35}=-7 \\ \therefore A^{-1}=\dfrac{1}{|A|}(a d j A)=\dfrac{1}{40}\left[\begin{array}{ccc} 5 & 3 & 9 \\5 & -13 & 1 \\ 5 & 11 & -7 \end{array}\right] \end{array}$

$\begin{array}{l} \therefore X=A^{-1} B=\dfrac{1}{40}\left[\begin{array}{ccc} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{array}\right]\left[\begin{array}{c} 5 \\ -4 \\ 3 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} y \\ z \end{array}\right]=\dfrac{1}{40}\left[\begin{array}{c} 25-12+27 \\ 25+52+3 \\ 25-44-21 \end{array}\right] \end{array}$

$=\dfrac{1}{40}\left[\begin{array}{c} 40 \\ 80 \\ -40 \end{array}\right]$

$=\left[\begin{array}{c} 1 \\ 2 \\ -1 \end{array}\right]$

Hence, $x=1, y=2$ and $z=-1$ 

Ans: The given system of equations can be written in the form of $AX = B$,

where $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}7 \\ -5 \\ 12\end{array}\right]$

Now,

$|A|=1(12-5)+1(9+10)+2(-3-8)=7+19-22=4 \neq 0$

Thus, $A$ is non-singular. Therefore, its inverse exists. Now, $A_{11}=7, A_{12}=-19, A_{3}=11$

$\begin{array}{l} A_{21}=1, A_{22}=-1, A_{23}=-1 \\ A_{31}=-3, A_{12}=11, A_{35}=7 \end{array}$

$\therefore A^{-1}=\left.\left.\right|_{A}\right|^{1}(\operatorname{adj} A)=\dfrac{1}{4}\left[\begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array}\right]$

$\therefore X=A^{-1} B=\dfrac{1}{4}\left[\begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array}\right]\left[\begin{array}{c} 7 \\ -5 \\ 12 \end{array}\right]$ 

$\Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\dfrac{1}{4}\left[\begin{array}{c} 49-5-36 \\ -133+5+132 \\ -77+5+84 \end{array}\right]$

$=\dfrac{1}{4}\left[\begin{array}{c} 8 \\ 4 \\ 12 \end{array}\right]=\left[\begin{array}{l} 2 \\ 1 \\ 3 \end{array}\right]$

Hence, $x=2, y=1$ and $z=3$.

Ans: $A=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]$

$\therefore A \mid=2(-4+4)+3(-6+4)+5(3-2)=0-6+5=-1 \neq 0$

Now, $A_{11}=0, A_{2}=2, A_{3}=1$

$\begin{array}{l} A_{31}=-1, A_{22}=-9, A_{23}=-5 \\ A_{31}=2, A_{32}=23, A_{33}=13 \\ \therefore A^{-1}=\dfrac{1}{|A|}(\operatorname{adj} A)=-\left[\begin{array}{lll} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right] \end{array}$

Now, the given system of equations can be written in the form of $A X=B$,

where $A=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}11 \\ -5 \\ -3\end{array}\right]$

The solution of the system of equations is given by $X=A^{-1} B\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right]\left[\begin{array}{l}11 \\ -5 \\ -3\end{array}\right]$ Using (1)

$=\left[\begin{array}{c} 0-5+6 \\ -22-45+69 \\ -11-25+39 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]$

Hence $ x=1,\; y=2$, and $z=3$

Ans: Let the cost of onions, wheat and rice per ${\text{kg}}$ be Rs. X and Rs. Z respectively.

Then, the given situation can be represented by a system of equations as:

$4x + 3y + 2z = 60$

$2x + 4y + 6z = 90$

$6x + 2y + 3z - 70$

This system of equations can be written in the form of $AX = B$,

where $A=\left[\begin{array}{lll}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{array}\right], X\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}60 \\ 90 \\ 70\end{array}\right]$

$|A|=4(12-12)-3(6-36)+2(4-24)=0+90-40=50 \neq 0$

Now,

$\begin{array}{l} A_{11}=0, A_{2}=30, A_{13}=-20 \\ A_{21}=-5, A_{22}=0, A_{23}=10 \\ A_{31}=10, A_{32}=-20, A_{33}=10 \\ \therefore \operatorname{adj} A=\left[\begin{array}{ccc} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{array}\right] \\ \therefore A^{-1}=\left.A\right|^{1} \operatorname{adj} A=\dfrac{1}{50}\left[\begin{array}{ccc} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{array}\right] \end{array}$

Now,

$\begin{array}{l} X=A^{-1} B \\ \Rightarrow X=\dfrac{1}{50}\left[\begin{array}{ccc} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{array}\right]\left[\begin{array}{l} 60 \\ 90 \\ 70 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\dfrac{1}{50}\left[\begin{array}{c} 0-450+700 \\ 1800+0-1400 \\ -1200+900+700 \end{array}\right] \\ =\dfrac{1}{50}\left[\begin{array}{l} 250 \\ 400 \\ 400 \end{array}\right] \end{array}$ 

$=\left[\begin{array}{l}5 \\ 8 \\ 8\end{array}\right]$

$\therefore x=5, y=8$, and $z=8$

Hence, the cost of onions is $5 R s$ per $\mathrm{kg}$, the cost of wheat is $8 \mathrm{Rs}$ per $\mathrm{kg}$, and the cost of rice is $8 \mathrm{Rs}$ per $\mathrm{kg}$.