Real numbers $a,\\;b,\\;c$ and $d$ are such that $a+b+c+d=1$ and for the given vectors $\\textbf{v}_1,\\;\\textbf{v}_2,\\;\\textbf{v}_3,\\;\\textbf{v}_4$ $a\\textbf{v}_1+b\\textbf{v}_2+c\\textbf{v}_3+d\\textbf{v}_4=\\textbf{0}$. Find $a,\\;b,\\;c,\\;d$.
", "licence": "Creative Commons Attribution 4.0 International"}, "preamble": {"css": "", "js": ""}, "parts": [{"scripts": {}, "showCorrectAnswer": true, "marks": 0, "variableReplacementStrategy": "originalfirst", "type": "gapfill", "unitTests": [], "customName": "", "prompt": "$a=$ [[0]]
\n$b=$ [[1]]
\n$c=$ [[2]]
\n$d=$ [[3]]
\nInput all values as exact decimals.
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except 0)"}, "c1": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "c1", "definition": "if(t=1,1/(1-del),-al/(1-del))"}, "v4": {"group": "Ungrouped variables", "templateType": "anything", "description": "", "name": "v4", "definition": "switch(t=4,w,z)"}}, "name": "Simon's copy of Combine linearly dependent vectors to get zero", "variable_groups": [], "ungrouped_variables": ["a", "x", "c", "b", "ga", "al", "y", "v1", "v2", "v3", "v4", "t", "w", "c2", "c3", "del", "c1", "z", "c4"], "statement": "You are given the following four vectors in $\\mathbb{R}^4$: \\[\\begin{align} \\textbf{v}_1&=\\var{v1}\\\\ \\textbf{v}_2&=\\var{v2}\\\\ \\textbf{v}_3&=\\var{v3}\\\\ \\textbf{v}_4&=\\var{v4}\\end{align}\\]
\nShow that the vectors $\\textbf{v}_1,\\;\\textbf{v}_2,\\;\\textbf{v}_3,\\;\\textbf{v}_4$ are linearly dependent by finding real numbers $a,\\;b,\\;c$ and $d$ such that \\[a\\textbf{v}_1+b\\textbf{v}_2+c\\textbf{v}_3+d\\textbf{v}_4=\\textbf{0},\\;\\;\\; a+b+c+d=1\\]where $\\textbf{0}=\\var{rowvector(0,0,0,0)}$ is the zero vector.
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On putting $a=1-b-c-d$ we have
\n\\[(1-b-c-d)\\textbf{v}_1+b\\textbf{v}_2+c\\textbf{v}_3+d\\textbf{v}_4=0\\]
\nHence on combining the vectors and equating the four components each to $0$ we have the four equations:
\n\\[\\begin{align} \\simplify[std]{{v1[0][0]} * (1 -b -c -d) + {v2[0][0]} * b + {v3[0][0]} * c + {v4[0][0]} * d }&= 0\\\\ \\simplify[std]{{v1[0][1]} * (1 -b -c -d) + {v2[0][1]} * b + {v3[0][1]} * c + {v4[0][1]} * d }&= 0\\\\ \\simplify[std]{{v1[0][2]} * (1 -b -c -d) + {v2[0][2]} * b + {v3[0][2]} * c + {v4[0][2]} * d }&= 0\\\\ \\simplify[std]{{v1[0][3]} * (1 -b -c -d) + {v2[0][3]} * b + {v3[0][3]} * c + {v4[0][3]} * d }&= 0\\end{align}\\]
\nOn rearranging these equations in the unknowns $b,\\;c,\\;d$ we get:
\n\\[\\begin{align} \\simplify[std]{{v2[0][0] -v1[0][0]} * b + {v3[0][0] -v1[0][0]} * c + {v4[0][0] -v1[0][0]} * d} &= \\var{-v1[0][0]}\\\\ \\simplify[std]{{v2[0][1] -v1[0][1]} * b + {v3[0][1] -v1[0][1]} * c + {v4[0][1] -v1[0][1]} * d }&= \\var{-v1[0][1]}\\\\ \\simplify[std]{{v2[0][2] -v1[0][2]} * b + {v3[0][2] -v1[0][2]} * c + {v4[0][2] -v1[0][2]} * d }&= \\var{-v1[0][2]}\\\\ \\simplify[std]{{v2[0][3] -v1[0][3]} * b + {v3[0][3] -v1[0][3]} * c + {v4[0][3] -v1[0][3]} * d}&= \\var{-v1[0][3]}\\end{align}\\]
\n\nWe have 4 equations in 3 unknowns. However, we can solve the first 3 equations simultaneously. One method is by writing the equations in matrix form and inverting the $3 \\times 3$ matrix obtained (an alternative method is Gaussian elimination):
\n\n\\[\\begin{pmatrix} \\var{v2[0][0] -v1[0][0]}& \\var{v3[0][0] -v1[0][0]} & \\var{v4[0][0] -v1[0][0]}\\\\ \\var{v2[0][1] -v1[0][1]}&\\var{v3[0][1] -v1[0][1]} &\\var{v4[0][1] -v1[0][1]} \\\\ \\var{v2[0][2] -v1[0][2]}& \\var{v3[0][2] -v1[0][2]}&\\var{v4[0][2] -v1[0][2]} \\end{pmatrix} \\begin{pmatrix} b \\\\ c \\\\ d \\end{pmatrix} = \\begin{pmatrix} \\var{-v1[0][0]}\\\\ \\var{-v1[0][1]} \\\\ \\var{-v1[0][2]} \\end{pmatrix} \\]
\n\n\\[\\begin{pmatrix} b \\\\ c \\\\ d \\end{pmatrix} = \\begin{pmatrix} \\var{v2[0][0] -v1[0][0]}& \\var{v3[0][0] -v1[0][0]} & \\var{v4[0][0] -v1[0][0]}\\\\ \\var{v2[0][1] -v1[0][1]}&\\var{v3[0][1] -v1[0][1]} &\\var{v4[0][1] -v1[0][1]} \\\\ \\var{v2[0][2] -v1[0][2]}& \\var{v3[0][2] -v1[0][2]}&\\var{v4[0][2] -v1[0][2]} \\end{pmatrix}^{-1} \\begin{pmatrix} \\var{-v1[0][0]}\\\\ \\var{-v1[0][1]} \\\\ \\var{-v1[0][2]} \\end{pmatrix}=\\begin{pmatrix} \\var{c2} \\\\ \\var{c3} \\\\ \\var{c4} \\end{pmatrix} \\]
\n\nHence we obtain
\n$b=\\var{c2}$, $c=\\var{c3}$, $d=\\var{c4}$ and hence $a=1- b-c-d= \\var{c1}$.
\n\nSubstituting these values also correctly satisfy the 4th equation, so the equations are consistent and they have the unique solution given.
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