Question

1. 1x –y +3z =5 ; 4x+2y-z =0 ; -x +3y + z = 5 by using Cramer’s rule and Matrix

inversion method.​

Answer 1

GIVEN :

1x –y +3z =5 ; 4x+2y-z =0 ; -x +3y + z = 5 by using Cramer’s rule and Matrix

inversion method.​

TO FIND :

The values of x,y and z by using Cramer’s rule and Matrix inversion method.​

SOLUTION :

Given equations are

1x –y +3z =5 ;

4x+2y-z =0 ;

-x +3y + z = 5

Now solving the given equations by Cramer's rule :

$D=\left|\begin{array}{ccc}1&-1&3\\4&2&-1\\-1&3&1\end{array}\right|$

= 1(2+3)+1(4-1)+3(12+2)

= 5+3+42

=50

D=50

The formula is $x=\frac{D_{x}}{D}$ , $y=\frac{D_{y}}{D}$ and $z=\frac{D_{z}}{D}$ provided $D\neq 0$

Now $x=\frac{D_{x}}{D}$

$x=\frac{\left|\begin{array}{ccc}5&-1&3\\0&2&-1\\5&3&1\end{array}\right|}{50}$

$=\frac{5(2+3)+1(0+5)+3(0-10)}{50}$

$=\frac{25+5-30}{50}$

$=\frac{0}{50}$

= 0

⇒ x=0

Now $y=\frac{D_{y}}{D}$

$y=\frac{\left|\begin{array}{ccc}1&-5&3\\4&20&-1\\-1&5&1\end{array}\right|}{50}$

$=\frac{1(0+5)-5(4-1)+3(20+0)}{50}$

$=\frac{5-15+60}{50}$

$=\frac{50}{50}$

= 1

⇒ y=1

Now $z=\frac{D_{z}}{D}$

$z=\frac{\left|\begin{array}{ccc}1&-1&5\\4&2&0\\-1&3&5\end{array}\right|}{50}$

$=\frac{1(10-0)+1(20+0)+5(12+2)}{50}$

$=\frac{10+20+70}{50}$

$=\frac{100}{50}$

= 2

⇒ z=2

∴ the values are x=0 , y=1 and z=2

Solving the given equations by Matrix inversion method :

1x –y +3z =5 ;

4x+2y-z =0 ;

-x +3y + z = 5

It is of the form AX=B where

$A=\left[\begin{array}{ccc}1&-1&3\\4&2&-1\\-1&3&1\end{array}\right]$ ,     $B=\left[\begin{array}{c}5\\0\\5\end{array}\right]$

AX=B

$X=A^{-1}B$

$A=\left|\begin{array}{ccc}1&-1&3\\4&2&-1\\-1&3&1\end{array}\right|$

= 1(2+3)+1(4-1)+3(12+2)

= 5+3+42

=50

|A|=50

The formula for $A^{-1}=\frac{adjA}{|A|}$

$adjA=\left[\begin{array}{ccc}5&10&-5\\-3&4&13\\14&2&6\end{array}\right]$

$A^{-1}=\frac{\left[\begin{array}{ccc}5&10&-5\\-3&4&13\\14&2&6\end{array}\right]}{50}$

Substituting in $X=A^{-1}B$ we get,

$\left[\begin{array}{c}x\\y\\z\end{array}\right]=\frac{\left[\begin{array}{ccc}5&10&-5\\-3&4&13\\14&2&6\end{array}\right]}{50}\times \left[\begin{array}{c}5\\0\\5\end{array}\right]$

$\left[\begin{array}{c}x\\y\\z\end{array}\right]=\frac{\left[\begin{array}{c}25+0-25\\-15+0+65\\70+0+30\end{array}\right]}{50}$

$=\frac{\left[\begin{array}{c}0\\50\\100\end{array}\right]}{50}$

$=\left[\begin{array}{c}0\\1\\2\end{array}\right]$

$\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}0\\1\\2\end{array}\right]$

∴ the values are x=0 , y=1 and z=2