Chapter 4: Determinants

Ans: The given system of equations is:

$x + 2y = 2$

$2x + 3y = 3$

The given system of equations is:

$\begin{array}{l} x+2y=2 \\ 2x+3y=3 \end{array}$

The given system of equations can be written in the form of $A X=B$, where $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right], X=\left[\begin{array}{l}2 \\ 3\end{array}\right]$ and $B=\left[\begin{array}{l}2 \\ 3\end{array}\right]$

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

$\therefore A$ is non-singular. Therefore, $A^{-1}$ exists.

Hence, the given system of equations is consistent.

Ans: The given system of equations is:

$\begin{array}{l} 2 x-y=5 \\ x+y=4 \end{array}$

The given system of equation can be written in the form of $A X=B$, where $A=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}5 \\ 4\end{array}\right]$ $|A|=2(1)-(-1)(1)=2+1=3 \neq 0$

$\therefore A$ is non-singular. Therefore, $A^{-1}$ exists.

Hence, the given system of equations is consistent.

Ans: The given system of equations is:

$\begin{array}{l} x+3 y=5 \\ 2 x+6 y=8 \end{array}$

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

where $A=\left[\begin{array}{ll}1 & 3 \\ 2 & 6\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}5 \\ 8\end{array}\right]$

Now, $|A|=1(6)-3(2)=6-6=0$

$\therefore A$ is a singular matrix. $(\operatorname{ad} j A)=\left[\begin{array}{cc}6 & -3 \\ -2 & 1\end{array}\right]$

$(a d j A) B=\left[\begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array}\right]\left[\begin{array}{l} 5 \\ 8 \end{array}\right]=\left[\begin{array}{l} 30-24 \\ -10+8 \end{array}\right]=\left[\begin{array}{c} 6 \\ -2 \end{array}\right] \neq 0$

Thus, the solution of the given system of equations does not exists. Hence, the given system of equations is inconsistent.

Ans: The given system of equations is:

$x + y + z = 1$

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

$ax + ay + 2az = 4$

The system of equation can be written in the form of $AX = B$, 

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

Now, $|A|=1(6 a-2 a)-1(4 a-2 a)+1(2 a-3 a)$

$=4 a-2 a-a=4 a-3 a=a \neq 0$

$\therefore A$ is a non-singular matrix. Therefore, $A^{-1}$ exists. Hence, the given system of equation is consistent.

Ans: The given system of equation is:

$3x - y - 2z = 2$

$2y - z =  - 1$

$3x - 5y = 3$

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

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

Now, $|A|=3(-5)-0+3(1+4)=-15+15=0$

$\therefore A$ is a singular matrix.

Now $(\operatorname{adj} A)=\left[\begin{array}{ccc}-5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6\end{array}\right]$

$\therefore(a d j A) B=\left[\begin{array}{ccc} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{array}\right]\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]$

$=\left[\begin{array}{c} -10-10+15 \\ -6-6+9 \\ -12-12+18 \end{array}\right]=\left[\begin{array}{l} -5 \\ -3 \\ -6 \end{array}\right] \neq 0$

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

Ans: The given system of equation is:

$5x - y + 4z = 5$

$2x + 3y + 5z = 2$

$5x - 2y + 6z =  - 1$

The system of equation can be written in the form of $AX = B$,

where $A=\left[\begin{array}{ccc}5 & -1 & 4 \\ 2 & 3 & 5 \\ 3 & -2 & 6\end{array}\right], X\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}5 \\ 2 \\ -1\end{array}\right]$

Now, $|A|=5(18+10)+1(12-25)+4(-4-15)$

$\begin{array}{l} =5(28)+1(-13)+4(-19) \\ =140-13-76 \\ =51 \neq 0 \end{array}$

$\therefore A$ is non-singular. Therefore, $A^{-1}$ exists. Hence, the given system of equations is consistent.

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

where $A=\left[\begin{array}{ll}5 & 2 \\ 7 & 3\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 5\end{array}\right]$

Now $|A|=15-14-1 \neq 0$

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

Now,

$\begin{array}{l} A^{-1}=\dfrac{1}{|\mathrm{~A}|}(\operatorname{adj} A) \\ \therefore A^{-1}=\left[\begin{array}{cc} 3 & -2 \\ -7 & 5 \end{array}\right] \\ \therefore X=A^{-1} B=\left[\begin{array}{cc} 3 & -2 \\ -7 & 5 \end{array}\right]\left[\begin{array}{l} 4 \\ 5 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 12-10 \\ -28+25 \end{array}\right]=\left[\begin{array}{c} 2 \\ -3 \end{array}\right] \end{array}$

Hence, $x=2$ and $y=-3$

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

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

Now, $|A|=8+3=11 \neq 0$

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

$\begin{array}{l} A^{-1}=|A|^{1}(\operatorname{adj} A)=\dfrac{1}{11}\left[\begin{array}{cc} 4 & 1 \\ -3 & 2 \end{array}\right] \\ \therefore X=A^{-1} B=\dfrac{1}{11}\left[\begin{array}{cc} 4 & 1 \\ -3 & 2 \end{array}\right]\left[\begin{array}{c} -2 \\ 3 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\dfrac{1}{11}\left[\begin{array}{c} -8+3 \\ 6+6 \end{array}\right]=\dfrac{1}{11}\left[\begin{array}{l} -5 \\ 12 \end{array}\right]=\left[\begin{array}{c} -\dfrac{5}{11} \\ \dfrac{12}{11} \end{array}\right] \end{array}$

Hence, $x=\dfrac{-5}{11}$ and $y=\dfrac{12}{11}$.

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

where $A=\left[\begin{array}{ll}4 & -3 \\ 3 & -5\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 7\end{array}\right]$

Now, $|A|=-20+9=-11 \neq 0$

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

Now, $A^{-1}=\dfrac{1}{\mid A}(a d i-A)=-\dfrac{1}{11}\left[\begin{array}{ll}-5 & 3 \\ -3 & 4\end{array}\right]=\dfrac{1}{11}\left[\begin{array}{cc}5 & -3 \\ 3 & -4\end{array}\right]$

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

$\left[\begin{array}{l} x \\ y  \end{array}\right]=\dfrac{1}{11}\left[\begin{array}{ll} 5 & -3 \\ 3 & -4 \end{array}\right]\left[\begin{array}{l} 3 \\ 7 \end{array}\right]$ 

$\begin{aligned} =& \dfrac{1}{11}\left[\begin{array}{c} 15-21 \\ 9-28 \end{array}\right] \\ &=\dfrac{1}{11}\left[\begin{array}{c} -6 \\ -19 \end{array}\right] \\ =\left[\begin{array}{r} -\dfrac{6}{11} \\ -\dfrac{19}{11} \end{array}\right] \end{aligned}$ 

Hence, $x=\dfrac{-6}{11}$ and $y=\dfrac{-19}{11}$

Ans: The system of equation is

$\begin{array}{l} 5 x+2 y=3 \\ 3 x+2 y=5 \end{array}$

Writing the above equation as $\mathrm{AX}=\mathrm{B}$

$\left[\begin{array}{ll} 5 & 2 \\ 3 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3 \\ 5 \end{array}\right]$

Hence $A=\left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right] \&\; B=\left[\begin{array}{l}3 \\ 5\end{array}\right]$

Calculating |A|

$\begin{aligned} |\mathrm{A}| &=\left|\begin{array}{ll} 5 & 2 \\ 3 & 2 \end{array}\right| \\ &=5(2)-3(2)=10-6=4 \end{aligned}$

Since $|\mathrm{A}| \neq 0$

The System of equation is consistent and has a unique solution

Now,

$\begin{array}{l} A X=B \\ X=A^{-1} B \end{array}$

Calculating $\mathrm{A}^{-1}$ 

$A^{-1}=\dfrac{1}{|A|} \operatorname{adj}(A)$

Interchange sign

$\operatorname{adj} A=\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]$

Now,

$\begin{array}{l} \mathrm{A}^{-1}=\dfrac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A} \\ \mathrm{A}^{-1}=\dfrac{1}{4}\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right] \end{array}$

Thus,

$X=A^{-1} B$

$\left[\begin{array}{l} x \\ y \end{array}\right]=\dfrac{1}{4}\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right]\left[\begin{array}{l} 3 \\ 5 \end{array}\right]=\dfrac{1}{4}\left[\begin{array}{c} 2(3)+(-2) 5 \\ -3(3)+5(5) \end{array}\right]$

$\left[\begin{array}{l} x \\ y \end{array}\right]=\dfrac{1}{4}\left[\begin{array}{c} 6-10 \\ -9+25 \end{array}\right]=\dfrac{1}{4}\left[\begin{array}{c} -4 \\ 16 \end{array}\right]$

$\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} -1 \\ 4 \end{array}\right]$ 

Hence, $x=-1 \;\&\; y=4$