Write the vector u = (1, −2, 5) ∈ R 3 as a linear combination of the vectors u1= (1, 1, 1), u2=
(1, 2, 3), u3= (2, −1, 1)
Let us determine the coefficients "c_1, c_2, c_3" such as "u=c_1u_1 + c_2u_2 + c_3u_3." We may rewrite this equality as a system of linear equations, where each equation concerns one coordinate.
"\\begin{cases}\n1 &= c_1\\cdot1 +c_2\\cdot1 + c_3\\cdot 2\\\\\n-2 &= c_1\\cdot1 +c_2\\cdot2 + c_3\\cdot (-1)\\\\\n5 &= c_1\\cdot1 +c_2\\cdot3 + c_3\\cdot1\\\\\n\\end{cases}"
We should solve this system. First we subtract the first equation from the second and the third and obtain
"\\begin{cases}\n1 &= c_1 +c_2 + 2c_3\\\\\n-3 &= c_2 -3c_3\\\\\n4 &= 2c_2 - c_3\\\\\n\\end{cases}"
Next, we subtract the second equation multiplied by two from the third equation
"\\begin{cases}\n1 &= c_1 +c_2 + 2c_3\\\\\n-3 &= c_2 -3c_3\\\\\n10 &= 5 c_3\\\\\n\\end{cases}"
Therefore, "c_3 = 2, \\;\\; c_2 = -3+3\\cdot2 = 3, \\;\\; c_1 = 1 - 3-2\\cdot2 = -6."
Therefore, "u = -6u_1 + 3u_2 + 2u_3."
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