Letv1 = (0,3,6,0),v2 = (0,2,4,6), and v3 = (1,−1,−2,1). Express (4,−1,−2,−11) as a
linear combination of v1,v2, and v3
To find a linear combination we should solve a system
"\\begin{bmatrix}\n 0 & 0 & 1 \\\\\n 3 & 2 & -1 \\\\\n 6 & 4 & -2 \\\\\n 0 & 6 & 1\n\\end{bmatrix} \\cdot \n\\begin{bmatrix}\n x1 \\\\\n x2 \\\\\n x3\n\\end{bmatrix} =\n\\begin{bmatrix}\n 4 \\\\\n -1 \\\\\n -2 \\\\\n -11\n\\end{bmatrix}"
That is equivalent to the system of four equations:
"\\begin{cases}\n x3 = 4 \\\\\n 3x1+2x2-x3 = -1 \\\\\n6x1 +4x2 -2x3 = -2 \\\\\n6x2+x3 = -11\n\\end{cases}"
From the first equation get x3 = 4. Substitute this value into the fourth equation obtain 6 x2 = -11 -4, or x2 = -5/2. And from the second equation 3 x1 = -1 +5 +4, so x1 = 8/3. Substituting these values into the third equation, we make sure that it is also fulfilled.
So (4,−1,−2,−11) = 8*v1 -5/2*v + 4*v3
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