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Question: If \(2x + 3y - 5z = 7,x + y + z = 6\), \(3x - 4y + 2z = 1,\) then x =...

If 2x+3y−5z=7,x+y+z=62x + 3y - 5z = 7,x + y + z = 6, 3x−4y+2z=1,3x - 4y + 2z = 1, then x =

A

∣2−57116321∣÷∣73−56111−42∣\left| \begin{matrix} 2 & - 5 & 7 \\ 1 & 1 & 6 \\ 3 & 2 & 1 \end{matrix} \right| \div \left| \begin{matrix} 7 & 3 & - 5 \\ 6 & 1 & 1 \\ 1 & - 4 & 2 \end{matrix} \right|

B

$\left| \begin{matrix}

  • 7 & 3 & - 5 \
  • 6 & 1 & 1 \
  • 1 & - 4 & 2 \end{matrix} \right| \div \left| \begin{matrix} 2 & 3 & - 5 \ 1 & 1 & 1 \ 3 & - 4 & 2 \end{matrix} \right|$
C

∣73−56111−42∣÷∣23−51113−42∣\left| \begin{matrix} 7 & 3 & - 5 \\ 6 & 1 & 1 \\ 1 & - 4 & 2 \end{matrix} \right| \div \left| \begin{matrix} 2 & 3 & - 5 \\ 1 & 1 & 1 \\ 3 & - 4 & 2 \end{matrix} \right|

D

None of these

Answer

∣73−56111−42∣÷∣23−51113−42∣\left| \begin{matrix} 7 & 3 & - 5 \\ 6 & 1 & 1 \\ 1 & - 4 & 2 \end{matrix} \right| \div \left| \begin{matrix} 2 & 3 & - 5 \\ 1 & 1 & 1 \\ 3 & - 4 & 2 \end{matrix} \right|

Explanation

Solution

For the given set of equation, by Cramer’s Rule

x=DxD=∣73−56111−42∣÷∣23−51113−42∣x = \frac{D_{x}}{D} = \left| \begin{matrix} 7 & 3 & - 5 \\ 6 & 1 & 1 \\ 1 & - 4 & 2 \end{matrix} \right| \div \left| \begin{matrix} 2 & 3 & - 5 \\ 1 & 1 & 1 \\ 3 & - 4 & 2 \end{matrix} \right|.