MATH SOLVE

4 months ago

Q:
# Solve the system of linear equations using the Gauss-Jordan elimination method.(x,y,z)=__________________2x + 2y − 3z = 162x − 3y + 2z = −44x − y + 3z = −4

Accepted Solution

A:

Answer:(x,y,z)=(2,0,-4)Step-by-step explanation:First we create the extended matrix from the equations[tex]\left[\begin{array}{ccc|c}2&2&-3&16\\2&-3&2&-4\\4&-1&3&-4\end{array}\right][/tex]Using the elementary operationsSubstract to the 2nd line the first one, and the 3rd one twice the first:[tex]\left[\begin{array}{ccc|c}2&2&-3&16\\0&-5&5&-20\\0&-5&9&-36\end{array}\right][/tex]Divide the first line by 2, the 2nd one by -5 and substract to the 3rd the 2nd:[tex]\left[\begin{array}{ccc|c}1&1&-3/2&8\\0&1&-1&4\\0&0&4&-16\\\end{array}\right][/tex]Divide the 3rd by 4:[tex]\left[\begin{array}{ccc|c}1&1&-3/2&8\\0&1&-1&4\\0&0&1&-4\\\end{array}\right][/tex]Add the 3rd to the 2nd:[tex]\left[\begin{array}{ccc|c}1&1&-3/2&8\\0&1&0&0\\0&0&1&-4\\\end{array}\right][/tex]Substract the 2nd to the 1st[tex]\left[\begin{array}{ccc|c}1&0&-3/2&8\\0&1&0&0\\0&0&1&-4\\\end{array}\right][/tex]Add the 3rd multiplied by 3/2:[tex]\left[\begin{array}{ccc|c}1&0&0&2\\0&1&0&0\\0&0&1&-4\\\end{array}\right][/tex]The answer is determined:x=2y=0z=-4You can check they are correct, by entering in the original formulas.