For the system $x-y+z=4,2 x+y-3 z=0$, $x+y+z=2$, the values of $x, y, z$ respectively are given by

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Aicharya · Accepted Answer

Given equation in matrix form $\left[\begin{array}{ccc} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 \end{array}ight]\left[\begin{array}{l} x \ y \ z \end{array}ight]=\left[\begin{array}{l} 4 \ 0 \ 2 \end{array}ight]$ It is of the form $\mathrm{AX}=\mathrm{B}$ Now, $|A|=\left|\begin{array}{ccc}1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1\end{array}ight|=10 
eq 0$ $\begin{aligned} 	herefore \quad & A^{-1} 	ext { exist } \ & \operatorname{adj} A=\left[\begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3\end{array}ight] \ A^{-1} & =\frac{1}{|A|} 	ext { adj } A \ & =\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3\end{array}ight]\end{aligned}$ $\begin{array}{ll} & 	ext {Now, } A X=B \ 	herefore \quad & A^{-1}(A X)=A^{-1} B \ 	herefore \quad & X=A^{-1} B\end{array}$ $\begin{aligned} 	herefore \quad X & =\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3\end{array}ight]\left[\begin{array}{l}4 \ 0 \ 2\end{array}ight] \ & =\frac{1}{10}\left[\begin{array}{c}16+0+4 \ -20+0+10 \ 4+0+6\end{array}ight] \ & 	herefore=\frac{1}{10}\left[\begin{array}{c}20 \ -10 \ 10\end{array}ight]=\left[\begin{array}{c}2 \ -1 \ 1\end{array}ight] \ 	herefore \quad & x=2, y=-1, z=1\end{aligned}$ Alternative Solution: Sure, let's solve this system of equations step by step. We have three equations: 1) $x - y + z = 4$ 2) $2x + y - 3z = 0$ 3) $x + y + z = 2$ First, let's add the first and second equations: $(x - y + z) + (2x + y - 3z) = 4 + 0$ We simplify that to: $3x - 2z = 4$ Next, let's subtract the third equation from the first: $(x - y + z) - (x + y + z) = 4 - 2$ We simplify that to: $-2y = 2$ Solving this equation we find that $y = -1$. Substitute $y = -1$ into the third equation, we get: $x - 1 + z = 2$ Solving this we get: $x + z = 3$ Now we have a system of two equations: $3x - 2z = 4$ and $x + z = 3$ Solving this system we find that $x = 2$ and $z = 1$. So, the solution to the system is $(x, y, z) = (2, -1, 1)$, which corresponds to option B.