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
y+5(a-c)=b-a
y=b-a-5a+5c
x+2(b-6a+5c)+3(a-c)=a
x+2b-12a+10c+3a-3c=a
x+2b-9a+7c=a
Comments
Leave a comment