Question

Solve the equation by matrix method (by finding inverse) 3x+4y+2z=8 , 2y-3z=3 , x-2y+6z=-2


Answer it ​

Answer 1

Answer:

Given as 3x + 4y + 2z = 8 2y – 3z = 3 x – 2y + 6z = -2 The given equation can be written in matrix form = 2(– 6 + 1) – 1(– 2 + 3) + 1(1 – 9) = – 10 – 1 – 8 = – 19 Therefore, the above system has a unique solution, given by X = A – 1B Co-factors of A are C11 = (– 1)1 + 1 – 6 + 1 = – 5 C21 = (– 1)2 + 1(24 + 4) = – 28 C31 = (– 1)3 + 1 – 1 – 3 = – 4 C12 = (– 1)1 + 2 – 2 + 3 = – 1 C22 = (– 1)2 + 1 – 4 – 3 = – 7 C32 = (– 1)3 + 1 – 2 – 1 = 3 C13 = (– 1)1 + 21 – 9 = – 8 C23 = (– 1)2 + 12 – 3 = – 1 C33 = (– 1)3 + 1 6 – 1 = 5 So, X = 1, Y = 1 and Z = 1Read more on Sarthaks.com - https://www.sarthaks.com/620966/solve-the-system-of-equations-by-matrix-method-3x-4y-2z-8-2y-3z-3-x-2y-6z-2?show=620985#a620985

Answer 2

Answer:

$\omega$ be a complex number such that $2 \omega + 1 = z$ , where $z = \sqrt{-3}. If \ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ [tex] \left | \begin{array}{ccdc} 1 & 1 & 1 \\ 1 & - {\omega}^{2} -1 & {\omega}^{2} \\ 1 & {\omega}^{2} & {\omega}^{7} \end{array} \right |$