Answer to Question #293865 in Linear Algebra for Himanshu

Question #293865

Express v = t2+4t-3 in P(t) as a linear combination of the









polynomials p =t2−2t+5,p =2t2 −3tand p =t+1.

1
Expert's answer
2022-02-06T15:20:19-0500

"v=t^2+4t-3\\\\\np_1=t^2-2t+5\\\\\np_2=2t^2-3t\\\\\np_3=t+1\\\\\nv=Ap_1+Bp_2+Cp_3"

"t^2+4t-3=\\\\\n=A(t^2-2t+5)+B(2t^2-3t)+C(t+1)\\\\"

"t^2: A+2B=1\\\\\nt:-2A-3B+C=4\\\\\nt^0: 5A+C=-3"

"\\begin{pmatrix}\n 1 & 2&0&|&1 \\\\\n -2&-3&1&|&4\\\\\n5&0&1&|&-3\n\\end{pmatrix}\\mapsto"

2r+1r(2)

3r+1r(-5)

"\\begin{pmatrix}\n 1 & 2&0&|&1 \\\\\n 0&1&1&|&6\\\\\n0&-10&1&|&-8\n\\end{pmatrix}\\mapsto"

3r+2r(10)

"\\begin{pmatrix}\n 1 & 2&0&|&1 \\\\\n 0&1&1&|&6\\\\\n0&0&11&|&52\n\\end{pmatrix}\\mapsto"

3r("\\frac{1}{11}" )

"\\begin{pmatrix}\n 1 & 2&0&|&1 \\\\\n 0&1&1&|&6\\\\\n0&0&1&|&\\frac{52}{11}\n\\end{pmatrix}\\mapsto"

2r+3r(-1)

"\\begin{pmatrix}\n 1 & 2&0&|&1 \\\\\n 0&1&0&|&\\frac{14}{11}\\\\\n0&0&1&|&\\frac{52}{11}\n\\end{pmatrix}\\mapsto"

1r+2r(-2)

"\\begin{pmatrix}\n 1 & 0&0&|&-\\frac{17}{11} \\\\\n 0&1&0&|&\\frac{14}{11}\\\\\n0&0&1&|&\\frac{52}{11}\n\\end{pmatrix}"

"A=-\\frac{17}{11}, B=\\frac{14}{11}, C=\\frac{52}{11}\\\\\nv=t^2+4t-3=-\\frac{17}{11}(t^2-2t+5)+\\\\\n+\\frac{14}{11}(2t^2-3t)+\\frac{52}{11}(t+1)"



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