Answer to Question #110819 in Linear Algebra for Hetisani Sewela

Question #110819
Consider the linear system
x + 2y + 3z = a
x + 3y + 8z = b
x + 2y + 2z = c



where a, b and c are arbitrary constants. Find all solutions of this system.
1
Expert's answer
2020-04-21T13:47:14-0400

x + 2y + 3z = a

x + 3y + 8z = b

x + 2y + 2z = c

factorizing

"\\begin{bmatrix}\n 1 & 2&3 \\\\\n 1 & 3&8\\\\\n 1 &2&2\n\\end{bmatrix}" "\\begin{bmatrix}\n x \\\\\n y \\\\\nz\n\\end{bmatrix}" ="\\begin{bmatrix}\n a \\\\\n b\\\\\nc\n\\end{bmatrix}"

writing the argument matrix

"\\begin{bmatrix}\n 1 & 2&3&a \\\\\n 1 & 3&8&b\\\\\n1&2&2&c\n\\end{bmatrix}"

reducing to row echelon

R1 to R1, R2-R1, R3-R1 we obtain

"\\begin{bmatrix}\n 1 & 2&3&a \\\\\n 0 & 1&5&b-a\\\\\n0&0&-1&c-a\n\\end{bmatrix}"

replacing the variables to form system of equations

x+2y+3Z=a

y+5z=b-a

-z=c-a

taking back substitution

  • z=a-c

y+5(a-c)=b-a

y=b-a-5a+5c

  • y=b-6a+5c

x+2(b-6a+5c)+3(a-c)=a

x+2b-12a+10c+3a-3c=a

x+2b-9a+7c=a

  • x=10a-2b-7c

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