Answer to Question #265802 in Math for Terro

Question #265802

Consider the following matrix equation.

( "\\begin{matrix}\n 2 & -1 & 0 \\\\\n 1 & 2 & -1 \\\\\n 0& 1&3\n\\end{matrix}" ) ( "\\begin{matrix}\n x1 \\\\\n x2 \\\\\nx3\n\\end{matrix}" ) = ( "\\begin{matrix}\n 2 \\\\\n 1 \\\\\n3\n\\end{matrix}" )

Solve the equation by using Thomas algorithm.

Expert's answer

Solution. From the first equation of the system

"2x_1-x_2=2"

Divide through by 2 get

"x_1 -0.5x_2=1."

We write the system of equations as

"\\begin{pmatrix}\n 1 & -0.5 & 0 \\\\\n 1 & 2 & -1 \\\\\n 0 & 1 & 3 \n\\end{pmatrix} \n \\begin{pmatrix}\n x_1 \\\\\n x_2 \\\\\n x_3 \n\\end{pmatrix} = \\begin{pmatrix}\n 1 \\\\\n 1 \\\\\n 3 \n\\end{pmatrix} ."

From the second equation of the system

"x_1 +2x_2-x_3=1"

Replacing x_1 -0.5*x_2=1 get

"1 +2.5x_2-x_3=1 \\to 2.5x_2-x_3=0"

Divide through by 2.5 get

"x_2-\\frac{2}{5}x_3=0."

We write the system of equations as

"\\begin{pmatrix}\n 1 & -0.5 & 0 \\\\\n 0 & 1 & -\\frac{2}{5} \\\\\n 0 & 1 & 3 \n\\end{pmatrix} \n \\begin{pmatrix}\n x_1 \\\\\n x_2 \\\\\n x_3 \n\\end{pmatrix} = \\begin{pmatrix}\n 1 \\\\\n 0 \\\\\n 3 \n\\end{pmatrix} ."

From the third equation of the system

"x_2+3x_3=3"

Replacing "x_2-\\frac{2}{5}x_3=0" get

"\\frac {17}{5} x_3=3"

Divide through by "\\frac {17}{5}" get

"x_3= \\frac{15}{17}"

We write the system of equations as

"\\begin{pmatrix}\n 1 & -0.5 & 0 \\\\\n 0 & 1 & -\\frac{2}{5} \\\\\n 0 & 0 & 1 \n\\end{pmatrix} \n \\begin{pmatrix}\n x_1 \\\\\n x_2 \\\\\n x_3 \n\\end{pmatrix} = \\begin{pmatrix}\n 1 \\\\\n 0 \\\\\n \\frac {15}{17} \n\\end{pmatrix} ."

Let us find the roots of the equations of the system

"x_3= \\frac{15}{17}"

"x_2=0+\\frac{2}{5}x_3 = \\frac{2}{5}\\times \\frac{15}{17}= \\frac{6}{17}"

"x_1 = \\frac{1}{2}x_2+1=\\frac{1}{2} \\times \\frac{6}{17} +1=\\frac{3}{17} + 1 = \\frac{20}{17}"

Answer. "x_1 = \\frac{20}{17}"; "x_2= \\frac{6}{17}"; "x_3= \\frac{15}{17}".