Question #59826

(Q3) Use Gaussian reduction to solve the following system of equations and verify your
results by using Mat Lab

X + Y + Z = 1
X + 2Y + 3Z = 2
2X + Y + 4Z = 5

Expert's answer

Answer on Question #59826 – Math – Linear Algebra

Question

(Q3) Use Gaussian reduction to solve the following system of equations and verify your results by using Mat Lab


X+Y+Z=1X+2Y+3Z=22X+Y+4Z=5\begin{array}{l} X + Y + Z = 1 \\ X + 2Y + 3Z = 2 \\ 2X + Y + 4Z = 5 \\ \end{array}

Solution

\left[ \begin{array}{cccc} x & +y & +z & = 1 \\ x & +2y & +3z & = 2 \\ 2x & +y & +4z & = 5 \end{array} \right] \begin{array}{cccc} R_2 \to R_2 - R_1 & x & +y & +z = 1 \\ y & +2z & = 1 \\ 2x & +y & +4z & = 5 \end{array} \right] \begin{array}{c} R_3 \to R_3 - 2R_1 \\ R_3 \to R_3 + R_2 \\ R_3 \to R_3 / 4 \end{array}[x+y+z=1y+2z=1y+2z=3]x+y+z=1y+2z=1x+4z=4\left[ \begin{array}{cccc} x & +y & +z & = 1 \\ y & +2z & = 1 \\ -y & +2z & = 3 \end{array} \right] \begin{array}{cccc} x & +y & +z & = 1 \\ y & +2z & = 1 \\ x & +4z & = 4 \end{array}[x+y+z=1y+2z=1z=1]R2R22R3R2R22R3\left[ \begin{array}{cccc} x & +y & +z & = 1 \\ y & +2z & = 1 \\ z & = 1 \end{array} \right] \begin{array}{c} R_2 \to R_2 - 2R_3 \\ R_2 \to R_2 - 2R_3 \end{array}[x+y+z=1y=1z=1]R1R1R2x+z=2y=1z=1R1R1R2x=1y=1z=1\left[ \begin{array}{cccccc} x & +y & +z & = & 1 \\ y & & & = & -1 \\ z & & & = & 1 \end{array} \right] \begin{array}{cccc} R_1 \to R_1 - R_2 & x & +z & = & 2 \\ y & & & = & -1 \\ z & = & 1 \end{array} \begin{array}{cccc} R_1 \to R_1 - R_2 & x & & = & 1 \\ y & & = & -1 \\ z & = & 1 \end{array}


where R1,R2,R3R_1, R_2, R_3 are row 1, row 2, row 3 respectively, R2R2R1R_2 \to R_2 - R_1 means that the first row is subtracted from the second row and the result is placed in the second row.

Verify the result using Mat Lab:



Answer: (X,Y,Z)=(1,1,1)(X, Y, Z) = (1, -1, 1).

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