I was given the following problem:
Let \(A=\{a_1,a_2,a_3\}\) and \(B=\{b_1,b_2,b_3\}\) be bases for a vector space \(V\) and suppose \(a_1=3b_1-b_2,a_2=-b_1+5b_2+b_3,a_3=b_2-6b_3\). Find the change of coordinates from \(A\) to \(B\). Find \([x]_B\) for \(x=4a_1+5a_2+a_3\).
It seems to me the matrix that changes the vector from \(A\) to \(B\) is \(\left(\begin{matrix}3&-1&0\\-1&5&1\\0&1&-6\\\end{matrix}\right)\). I can take the inverse of that matrix and multiply it by \(\left(\begin{matrix}4&5&1\\\end{matrix}\right)\). The only problem is that doing this gives me \(\left(\begin{matrix}\frac{155}{87}&\frac{39}{29}&\frac{5}{87}\\\end{matrix}\right)\)- a different answer than given in the answer key. I will attach a picture of the answer in the answer key.
Let \(A=\{a_1,a_2,a_3\}\) and \(B=\{b_1,b_2,b_3\}\) be bases for a vector space \(V\) and suppose \(a_1=3b_1-b_2,a_2=-b_1+5b_2+b_3,a_3=b_2-6b_3\). Find the change of coordinates from \(A\) to \(B\). Find \([x]_B\) for \(x=4a_1+5a_2+a_3\).
It seems to me the matrix that changes the vector from \(A\) to \(B\) is \(\left(\begin{matrix}3&-1&0\\-1&5&1\\0&1&-6\\\end{matrix}\right)\). I can take the inverse of that matrix and multiply it by \(\left(\begin{matrix}4&5&1\\\end{matrix}\right)\). The only problem is that doing this gives me \(\left(\begin{matrix}\frac{155}{87}&\frac{39}{29}&\frac{5}{87}\\\end{matrix}\right)\)- a different answer than given in the answer key. I will attach a picture of the answer in the answer key.
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