Question

Using cramers rule find the values of x , y & z x+y -z =0 ; 2x+y+3z =9 , x-y+z = 2​

Answer 1

Given Question :-

Given Question :- Solve using Cramer's Rule :

x + y - z = 0

2x + y + 3z = 9

x - y + z = 2

$─━─━─━─━─━─━─━─━─━─━─━$

$\large\underline\purple{\bold{Solution :- }}$

$\begin{gathered}\sf A=\left[\begin{array}{ccc}1&1& - 1\\2&1&3\\1&-1&1\end{array}\right]\end{gathered}$

$\begin{gathered}\sf B=\left[\begin{array}{c}0\\9\\2\end{array}\right]\end{gathered}$

$\begin{gathered}\sf X=\left[\begin{array}{c}x\\y\\z\end{array}\right]\end{gathered}$

$\large\underline{\bold{❥︎Step :- 1 }}$

$\tt \: \to \: |A| = \begin{array}{|ccc|}</p><p>1 & 1 & - 1 \\</p><p>2 & 1 & 3 \\</p><p>1 &-1 & 1 \\</p><p>\end{array}$

$\bf\implies \: |A| = 1(1 + 3) - 1(2 - 3) - 1( - 2 - 1)$

$\bf\implies \: |A| = 4 + 1+ 3 = 8$

$\bf\implies \:system \: of \: equations \: have \: unique \: solution$

$\large\underline{\bold{❥︎Step :- 2 }}$

$\tt \: \to \: |D_1| = \begin{array}{|ccc|}</p><p>0 & 1 & - 1 \\</p><p>9 & 1 &3 \\</p><p>2 &-1 & 31 \\</p><p>\end{array}$

$\bf\implies \: |D_1| = - 1(9 - 6) - 1( - 9 - 2) = 8$

$\large\underline{\bold{❥︎Step :- 3 }}$

$\tt \: \to \: |D_2| = \begin{array}{|ccc|}</p><p>1 & 0 & - 1 \\</p><p>2 & 9 & 3 \\</p><p>1 &2 & 1 \\</p><p>\end{array}$

$\bf\implies \: |D_2| = 1(9 - 6) - 1(4 - 9) = 8$

$\large\underline{\bold{❥︎Step :- 4 }}$

$\tt \: \to \: |D_3| = \begin{array}{|ccc|}</p><p>1 & 1 & 0 \\</p><p>2 & 1 & 9 \\</p><p>1 &-1 & 2 \\</p><p>\end{array}$

$\bf\implies \: |D_3| = 1(2 + 9) - 1( 4 - 9) = 16$

$\large\underline{\bold{❥︎Step :- 5 }}$

$\tt \: \to \:x = \dfrac{ |D_1| }{ |A| } = \dfrac{8}{8} = 1$

$\tt \: \to \:y = \dfrac{ |D_2| }{ |A| } = \dfrac{8}{8} = 1$

$\tt \: \to \:z = \dfrac{ |D_3| }{ |A| } = \dfrac{16}{8} = 2$