In each of Problems 1–10, \(n+1\) data points are given. Find the \(n^{th}\) degree polynomial \(y=f\left ( x\right ) \) that fits these points.\[ \left ( x,y\right ) =\left \{ \left ( 0,3\right ) ,\left ( 1,1\right ) ,\left ( 2,-5\right ) \right \} \] Solution
Since \(n+1=3\), then \(n=2\). Therefore we need degree \(2\) polynomial\[ f\left ( x\right ) =A+Bx+Cx^{2}\] From the data given, we obtain the following three equations\begin {align*} 3 & =A\\ 1 & =A+B+C\\ -5 & =A+2B+4C \end {align*}
This gives the system\[\begin {bmatrix} 1 & 0 & 0\\ 1 & 1 & 1\\ 1 & 2 & 4 \end {bmatrix}\begin {bmatrix} A\\ B\\ C \end {bmatrix} =\begin {bmatrix} 3\\ 1\\ -5 \end {bmatrix} \] Augmented matrix is\[\begin {bmatrix} 1 & 0 & 0 & 3\\ 1 & 1 & 1 & 1\\ 1 & 2 & 4 & -5 \end {bmatrix} \] \(R_{2}\rightarrow -R_{1}+R_{2}\) gives\[\begin {bmatrix} 1 & 0 & 0 & 3\\ 0 & 1 & 1 & -2\\ 1 & 2 & 4 & -5 \end {bmatrix} \] \(R_{3}\rightarrow -R_{1}+R_{3}\) gives\[\begin {bmatrix} 1 & 0 & 0 & 3\\ 0 & 1 & 1 & -2\\ 0 & 2 & 4 & -8 \end {bmatrix} \] \(R_{3}\rightarrow -2R_{2}+R_{3}\) gives\[\begin {bmatrix} 1 & 0 & 0 & 3\\ 0 & 1 & 1 & -2\\ 0 & 0 & 2 & -4 \end {bmatrix} \] Hence we obtain the system in Echelon form as\[\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 2 \end {bmatrix}\begin {bmatrix} A\\ B\\ C \end {bmatrix} =\begin {bmatrix} 3\\ -2\\ -4 \end {bmatrix} \] Back substitution: Last row gives \(2C=-4\) or \(C=-2\). Second row gives \(B-C=-2\,\) or \(B=0\).
First row gives \(A=3\). Therefore the solution is \[\begin {bmatrix} A\\ B\\ C \end {bmatrix} =\begin {bmatrix} 3\\ 0\\ -2 \end {bmatrix} \] The polynomial is\begin {align*} f\left ( x\right ) & =A+Bx+Cx^{2}\\ & =3-2x^{2} \end {align*}
Here is plot of the solution fitted on the points
In Problems 1–4, find \(\left \vert \boldsymbol {a}-\boldsymbol {b}\right \vert ,2\boldsymbol {a}+\boldsymbol {b}\),\(3\boldsymbol {a}-4\boldsymbol {b}\)
\[ \boldsymbol {a}=\left ( 2,5,-4\right ) ,\boldsymbol {b}=\left ( 1,-2,-3\right ) \] Solution
\begin {align*} \left \vert \boldsymbol {a}-\boldsymbol {b}\right \vert & =\left \vert \left ( 2,5,-4\right ) -\left ( 1,-2,-3\right ) \right \vert \\ & =\left \vert \left ( 2-1,5+2,-4+3\right ) \right \vert \\ & =\left \vert \left ( 1,7,-1\right ) \right \vert \\ & =\sqrt {1+49+1}\\ & =\sqrt {51} \end {align*}
And\begin {align*} 2\boldsymbol {a}+\boldsymbol {b} & =2\left ( 2,5,-4\right ) +\left ( 1,-2,-3\right ) \\ & =\left ( 4,10,-8\right ) +\left ( 1,-2,-3\right ) \\ & =\left ( 4+1,10-2,-8-3\right ) \\ & =\left ( 5,8,-11\right ) \end {align*}
And\begin {align*} 3\boldsymbol {a}-4\boldsymbol {b} & =3\left ( 2,5,-4\right ) -4\left ( 1,-2,-3\right ) \\ & =\left ( 6,15,-12\right ) -\left ( 4,-8,-12\right ) \\ & =\left ( 6-4,15+8,-12+12\right ) \\ & =\left ( 2,23,0\right ) \end {align*}
In Problems 19–24, use the method of Example 3 to determine whether the given vectors \(\boldsymbol {u},\boldsymbol {v}\), and \(\boldsymbol {w}\) are linearly independent or dependent. If they are linearly dependent, find scalars \(a,b\), and \(c\) not all zero such that \(a\boldsymbol {u}+b\boldsymbol {v}+c\boldsymbol {w}=\boldsymbol {0}\)\begin {align*} \boldsymbol {u} & =\left ( 2,0,1\right ) \\ \boldsymbol {v} & =\left ( -3,1,-1\right ) \\ \boldsymbol {w} & =\left ( 0,-2,-1\right ) \end {align*}
Solution
We set up \(Ax=0\) and solve for \(x\) where \(x\) here is \(\left ( a,b,c\right ) \) vector. If \(x\) is the trivial solution, then the vectors are linearly independent. If we find non-trivial solution, then the vectors are linearly dependent.\begin {align*} a\boldsymbol {u}+b\boldsymbol {v}+c\boldsymbol {w} & =\boldsymbol {0}\\ a\begin {bmatrix} 2\\ 0\\ 1 \end {bmatrix} +b\begin {bmatrix} -3\\ 1\\ -1 \end {bmatrix} +c\begin {bmatrix} 0\\ -2\\ -1 \end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \\\begin {bmatrix} 2 & -3 & 0\\ 0 & 1 & -2\\ 1 & -1 & -1 \end {bmatrix}\begin {bmatrix} a\\ b\\ c \end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \end {align*}
Augmented matrix\[\begin {bmatrix} 2 & -3 & 0\\ 0 & 1 & -2\\ 1 & -1 & -1 \end {bmatrix} \] \(R_{3}\rightarrow -\frac {1}{2}R_{1}+R_{3}\) gives\[\begin {bmatrix} 2 & -3 & 0\\ 0 & 1 & -2\\ 0 & \frac {1}{2} & -1 \end {bmatrix} \] \(R_{3}\rightarrow \frac {-1}{2}R_{2}+R_{3}\) gives\[\begin {bmatrix} 2 & -3 & 0\\ 0 & 1 & -2\\ 0 & 0 & 0 \end {bmatrix} \] Hence the system becomes\[\begin {bmatrix} 2 & -3 & 0\\ 0 & 1 & -2\\ 0 & 0 & 0 \end {bmatrix}\begin {bmatrix} a\\ b\\ c \end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \] \(a,b\) are leading variables and \(c\) is free variable. Let \(c=t\) which can be any value. Then \(b=2t\) and \(2a-3b=0\) or \(a=3t\). Hence solution is\[\begin {bmatrix} a\\ b\\ c \end {bmatrix} =t\begin {bmatrix} 3\\ 2\\ 1 \end {bmatrix} \] There are infinite solutions. We need only one non-zero solution to show that the vectors are linearly dependent. Let \(t=1\)\[\begin {bmatrix} a\\ b\\ c \end {bmatrix} =\begin {bmatrix} 3\\ 2\\ 1 \end {bmatrix} \] Hence vectors are linearly dependent\[ 3\boldsymbol {u}+2\boldsymbol {v}+\boldsymbol {w}=\boldsymbol {0}\]
In Problems 19–24, use the method of Example 3 to determine whether the given vectors \(\boldsymbol {u},\boldsymbol {v}\), and \(\boldsymbol {w}\) are linearly independent or dependent. If they are linearly dependent, find scalars \(a,b\), and \(c\) not all zero such that \(a\boldsymbol {u}+b\boldsymbol {v}+c\boldsymbol {w}=\boldsymbol {0}\)\begin {align*} \boldsymbol {u} & =\left ( 2,0,3\right ) \\ \boldsymbol {v} & =\left ( 5,4,-2\right ) \\ \boldsymbol {w} & =\left ( 2,-1,1\right ) \end {align*}
Solution
We set up \(Ax=0\) and solve for \(x\) where \(x\) here is \(\left ( a,b,c\right ) \) vector. If \(x\) is the trivial solution, then the vectors are linearly independent. If we find non-trivial solution, then the vectors are linearly dependent.\begin {align*} a\boldsymbol {u}+b\boldsymbol {v}+c\boldsymbol {w} & =\boldsymbol {0}\\ a\begin {bmatrix} 2\\ 0\\ 3 \end {bmatrix} +b\begin {bmatrix} 5\\ 4\\ -2 \end {bmatrix} +c\begin {bmatrix} 2\\ -1\\ 1 \end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \\\begin {bmatrix} 2 & 5 & 2\\ 0 & 4 & -1\\ 3 & -2 & 1 \end {bmatrix}\begin {bmatrix} a\\ b\\ c \end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \end {align*}
Augmented matrix\[\begin {bmatrix} 2 & 5 & 2\\ 0 & 4 & -1\\ 3 & -2 & 1 \end {bmatrix} \] \(R_{3}\rightarrow -3R_{1}+2R_{3}\) gives\[\begin {bmatrix} 2 & 5 & 2\\ 0 & 4 & -1\\ 0 & -19 & -4 \end {bmatrix} \] \(R_{3}\rightarrow -19R_{2}+4R_{3}\) gives\[\begin {bmatrix} 2 & 5 & 2\\ 0 & 4 & -1\\ 0 & 0 & 3 \end {bmatrix} \] Hence the system in Echelon form becomes\[\begin {bmatrix} 2 & 5 & 2\\ 0 & 4 & -1\\ 0 & 0 & 3 \end {bmatrix}\begin {bmatrix} a\\ b\\ c \end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \] Last row gives \(c=0\). Second row gives \(4b=0\) or \(b=0\). First row gives \(2a=0\) or \(a=0\).
Therefore the vectors are linearly independent because only the trivial solution exist.
In Problems 25–28, express the vector \(\boldsymbol {t}\) as a linear combination of the vectors \(\boldsymbol {u},\boldsymbol {v}\), and \(\boldsymbol {w}\).\[ \boldsymbol {t}=\left ( 0,0,19\right ) ,\boldsymbol {u}=\left ( 1,4,3\right ) ,\boldsymbol {v}=\left ( -1,-2,2\right ) ,\boldsymbol {w}=\left ( 4,4,1\right ) \] Solution
In system form we are looking for\begin {align*} a\boldsymbol {u}+b\boldsymbol {v}+c\boldsymbol {w} & =\boldsymbol {t}\\ a\begin {bmatrix} 1\\ 4\\ 3 \end {bmatrix} +b\begin {bmatrix} -1\\ -2\\ 2 \end {bmatrix} +c\begin {bmatrix} 4\\ 4\\ 1 \end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 19 \end {bmatrix} \\\begin {bmatrix} 1 & -1 & 4\\ 4 & -2 & 4\\ 3 & 2 & 1 \end {bmatrix}\begin {bmatrix} a\\ b\\ c \end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 19 \end {bmatrix} \end {align*}
Augmented matrix\[\begin {bmatrix} 1 & -1 & 4 & 0\\ 4 & -2 & 4 & 0\\ 3 & 2 & 1 & 19 \end {bmatrix} \] \(R_{2}\rightarrow -4R_{1}+R_{2}\) gives\[\begin {bmatrix} 1 & -1 & 4 & 0\\ 0 & 2 & -12 & 0\\ 3 & 2 & 1 & 19 \end {bmatrix} \] \(R_{3}\rightarrow -3R_{1}+R_{3}\) gives \[\begin {bmatrix} 1 & -1 & 4 & 0\\ 0 & 2 & -12 & 0\\ 0 & 5 & -11 & 19 \end {bmatrix} \] \(R_{3}\rightarrow -\frac {5}{2}R_{2}+R_{3}\) gives\[\begin {bmatrix} 1 & -1 & 4 & 0\\ 0 & 2 & -12 & 0\\ 0 & 0 & 19 & 19 \end {bmatrix} \] The above is Echelon form. Hence the system is\[\begin {bmatrix} 1 & -1 & 4\\ 0 & 2 & -12\\ 0 & 0 & 19 \end {bmatrix}\begin {bmatrix} a\\ b\\ c \end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 19 \end {bmatrix} \] Last row gives \(19c=19\) or \(c=1\). Second row gives \(2b-12c=0\) or \(b=6\). First row gives \(a-b+4c=0\) or \(a=b-4c\) or \(a=6-4=2\). Hence\begin {align*} a\boldsymbol {u}+b\boldsymbol {v}+c\boldsymbol {w} & =\boldsymbol {t}\\ 2\boldsymbol {u}+6\boldsymbol {v}+\boldsymbol {w} & =\boldsymbol {t} \end {align*}
Apply Theorem 1 to determine whether or not \(W\) is a subspace of \(\mathbb {R} ^{n}\).
\(W\) is the set of all vectors in \(\mathbb {R} ^{3}\) such that \(x_{1}=5x_{2}\)
Solution
Theorem 1 at page 225 gives conditions for subspace:
The non empty subset \(W\) of the vector space \(V\) is a subspace of \(V\) if and only if it satisfies the following two conditions:
Let \(\boldsymbol {u}=\left ( x_{1},x_{2},x_{3}\right ) \) and \(\boldsymbol {v}=\left ( y_{1},y_{2},y_{3}\right ) \) where \(x_{1}=5x_{2}\) and \(y_{1}=5y_{2}\). then (1) becomes \begin {align*} \boldsymbol {u}+\boldsymbol {v} & =\left ( x_{1},x_{2},x_{3}\right ) +\left ( y_{1},y_{2},y_{3}\right ) \\ & =\left ( x_{1}+y_{1},x_{2}+y_{2},x_{3}+y_{3}\right ) \end {align*}
Then\begin {align*} x_{1}+y_{1} & =5x_{2}+5y_{2}\\ & =5\left ( x_{2}+y_{2}\right ) \end {align*}
Hence closed under addition. Condition (2) says\begin {align*} c\boldsymbol {u} & =c\left ( x_{1},x_{2},x_{3}\right ) \\ & =\left ( cx_{1},cx_{2},cx_{3}\right ) \end {align*}
Hence \(cx_{1}=c\left ( 5x_{2}\right ) =5\left ( cx_{2}\right ) \). Therefore closed under scalar multiplication as well. Therefore this is a subspace.
Apply Theorem 1 to determine whether or not \(W\) is a subspace of \(\mathbb {R} ^{n}\).
\(W\) is the set of all vectors in \(\mathbb {R} ^{3}\) such that \(x_{1}+x_{2}+x_{3}=1\)
Solution
Theorem 1 at page 225 gives conditions for subspace:
The non empty subset \(W\) of the vector space \(V\) is a subspace of \(V\) if and only if it satisfies the following two conditions:
Let \(\boldsymbol {u}=\left ( x_{1},x_{2},x_{3}\right ) \) and \(\boldsymbol {v}=\left ( y_{1},y_{2},y_{3}\right ) \) where \(x_{1}+x_{2}+x_{3}=1\) and \(y_{1}+y_{2}+y_{3}=1\). then (1) becomes \begin {align*} \boldsymbol {u}+\boldsymbol {v} & =\left ( x_{1},x_{2},x_{3}\right ) +\left ( y_{1},y_{2},y_{3}\right ) \\ & =\left ( x_{1}+y_{1},x_{2}+y_{2},x_{3}+y_{3}\right ) \end {align*}
Then\begin {align*} x_{1}+y_{1}+x_{2}+y_{2}+x_{3}+y_{3} & =\left ( x_{1}+x_{2}+x_{3}\right ) +\left ( y_{1}+y_{2}+y_{3}\right ) \\ & =1+1\\ & =2 \end {align*}
Therefore this is not closed under addition since \(\boldsymbol {u}+\boldsymbol {v}\) does not satisfy (1). Hence not a subspace.
In Problems 15–18, apply the method of Example 5 to find two solution vectors \(\boldsymbol {u}\) and \(\boldsymbol {v}\) such that the solution space is the set
of all linear combinations of the form \(s\boldsymbol {u}+t\boldsymbol {v}\)\begin {align*} x_{1}+3x_{2}+8x_{3}-x_{4} & =0\\ x_{1}-3x_{2}-10x_{3}+5x_{4} & =0\\ x_{1}+4x_{2}+11x_{3}-2x_{4} & =0 \end {align*}
(notice: typo in book. Last term in second equation is \(5x_{5}\) in book, but it should be \(5x_{4}\)).
Solution
System is\[\begin {bmatrix} 1 & 3 & 8 & -1\\ 1 & -3 & -10 & 5\\ 1 & 4 & 11 & -2 \end {bmatrix}\begin {bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \] Augmented matrix\[\begin {bmatrix} 1 & 3 & 8 & -1\\ 1 & -3 & -10 & 5\\ 1 & 4 & 11 & -2 \end {bmatrix} \] \(R_{2}\rightarrow -R_{1}+R_{2}\) gives \[\begin {bmatrix} 1 & 3 & 8 & -1\\ 0 & -6 & -18 & 6\\ 1 & 4 & 11 & -2 \end {bmatrix} \] \(R_{3}\rightarrow -R_{1}+R_{3}\) gives\[\begin {bmatrix} 1 & 3 & 8 & -1\\ 0 & -6 & -18 & 6\\ 0 & 1 & 3 & -1 \end {bmatrix} \] \(R_{3}\rightarrow \frac {1}{6}R_{2}+R_{3}\) gives\[\begin {bmatrix} 1 & 3 & 8 & -1\\ 0 & -6 & -18 & 6\\ 0 & 0 & 0 & 0 \end {bmatrix} \] Leading variables are \(x_{1},x_{2}\). Free variables are \(x_{3},x_{4}\). Let \(x_{4}=t,x_{3}=s\). The system becomes\[\begin {bmatrix} 1 & 3 & 8 & -1\\ 0 & -6 & -18 & 6\\ 0 & 0 & 0 & 0 \end {bmatrix}\begin {bmatrix} x_{1}\\ x_{2}\\ s\\ t \end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \] From second row, \(-6x_{2}-18s+6t=0\) or \(x_{2}=-\frac {18s-6t}{6}=-3s+t\).
From first row, \(x_{1}+3x_{2}+8s-t=0\). Hence \(x_{1}=-3x_{2}-8s+t\) or \(x_{1}=-3\left ( -3s+t\right ) -8s+t\allowbreak \) or \(x_{1}=s-2t\). Therefore the solution is\begin {align*} \begin {bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end {bmatrix} & =\begin {bmatrix} s-2t\\ -3s+t\\ s\\ t \end {bmatrix} \\ & =s\begin {bmatrix} 1\\ -3\\ 1\\ 0 \end {bmatrix} +t\begin {bmatrix} -2\\ 1\\ 0\\ 1 \end {bmatrix} \\ & =s\boldsymbol {u}+t\boldsymbol {v} \end {align*}
Therefore he solution space is the set of all linear combinations of the form \(s\boldsymbol {u}+t\boldsymbol {v}\)
In Problems 19–22, reduce the given system to echelon form to find a single solution vector \(\boldsymbol {u}\) such that the solution space is
the set of all scalar multiples of \(\boldsymbol {u}\).\begin {align*} x_{1}+7x_{2}+2x_{3}-3x_{4} & =0\\ 2x_{1}+7x_{2}+x_{3}-4x_{4} & =0\\ 3x_{1}+5x_{2}-x_{3}-5x_{4} & =0 \end {align*}
Solution
System is\[\begin {bmatrix} 1 & 7 & 2 & -3\\ 2 & 7 & 1 & -4\\ 3 & 5 & -1 & -5 \end {bmatrix}\begin {bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \] Augmented matrix\[\begin {bmatrix} 1 & 7 & 2 & -3\\ 2 & 7 & 1 & -4\\ 3 & 5 & -1 & -5 \end {bmatrix} \] \(R_{2}\rightarrow -2R_{1}+R_{2}\) gives\[\begin {bmatrix} 1 & 7 & 2 & -3\\ 0 & -7 & -3 & 2\\ 3 & 5 & -1 & -5 \end {bmatrix} \] \(R_{3}\rightarrow -3R_{1}+R_{3}\) gives\[\begin {bmatrix} 1 & 7 & 2 & -3\\ 0 & -7 & -3 & 2\\ 0 & -16 & -7 & 4 \end {bmatrix} \] \(R_{3}\rightarrow \frac {-16}{7}R_{2}+R_{3}\) gives\[\begin {bmatrix} 1 & 7 & 2 & -3\\ 0 & -7 & -3 & 2\\ 0 & 0 & -\frac {1}{7} & -\frac {4}{7}\end {bmatrix} \] Hence the system in Echelon form is\[\begin {bmatrix} 1 & 7 & 2 & -3\\ 0 & -7 & -3 & 2\\ 0 & 0 & -\frac {1}{7} & -\frac {4}{7}\end {bmatrix}\begin {bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \] Leading variables are \(x_{1},x_{2},x_{3}\). Free variable is \(x_{4}=t\). Last row gives \(-\frac {1}{7}x_{3}-\frac {4}{7}t=0\). Hence \(x_{3}=-4t\). Second row gives \(-7x_{2}-3x_{3}+2x_{4}=0\) or \(-7x_{2}=3x_{3}-2x_{4}\) or \(-7x_{2}=3\left ( -4t\right ) -2\left ( t\right ) \). Hence \(-7x_{2}=-14t\) or \(x_{2}=2t\).
First row gives \(x_{1}+7x_{2}+2x_{3}-3x_{4}=0\) or \(x_{1}=-7\left ( 2t\right ) -2\left ( -4t\right ) +3\left ( t\right ) \). Hence \(x_{1}=-3t\). Therefore the solution is\begin {align*} \begin {bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end {bmatrix} & =\begin {bmatrix} -3t\\ 2t\\ -4t\\ t \end {bmatrix} \\ & =t\begin {bmatrix} -3\\ 2\\ -4\\ 1 \end {bmatrix} \\ & =t\boldsymbol {u} \end {align*}
The solution space is the set of all scalar multiples of \(\boldsymbol {u}\).
My fictional company Linear Algebra Inc had a stock price of $10 on day 1, $15 on day 2, and $10 on day 3. Interpolate this data with a quadratic polynomial \(f(t)=a+bt+ct^{2}\),where \(t\) is the day and \(f(t)\) is the price on day \(t\). Is it a good idea to use \(f(t)\) to predict the stock price of Linear Algebra Inc on day \(4\)?
Solution
Data is \(\left ( 1,10\right ) ,\left ( 2,15\right ) ,\left ( 3,10\right ) \). Therefore we obtain 3 equations using \(f(t)=a+bt+ct^{2}\) as\begin {align*} 10 & =a+b+c\\ 15 & =a+2b+4c\\ 10 & =a+3b+9c \end {align*}
Which gives the system\[\begin {bmatrix} 1 & 1 & 1\\ 1 & 2 & 4\\ 1 & 3 & 9 \end {bmatrix}\begin {bmatrix} a\\ b\\ c \end {bmatrix} =\begin {bmatrix} 10\\ 15\\ 10 \end {bmatrix} \] The augmented matrix is \[\begin {bmatrix} 1 & 1 & 1 & 10\\ 1 & 2 & 4 & 15\\ 1 & 3 & 9 & 10 \end {bmatrix} \] \(R_{2}\rightarrow -R_{1}+R_{2}\) gives\[\begin {bmatrix} 1 & 1 & 1 & 10\\ 0 & 1 & 3 & 5\\ 1 & 3 & 9 & 10 \end {bmatrix} \] \(R_{3}\rightarrow -R_{1}+R_{3}\) gives\[\begin {bmatrix} 1 & 1 & 1 & 10\\ 0 & 1 & 3 & 5\\ 0 & 2 & 8 & 0 \end {bmatrix} \] \(R_{3}\rightarrow -2R_{2}+R_{3}\) gives\[\begin {bmatrix} 1 & 1 & 1 & 10\\ 0 & 1 & 3 & 5\\ 0 & 0 & 2 & -10 \end {bmatrix} \] Hence the system in Echelon form is\[\begin {bmatrix} 1 & 1 & 1\\ 0 & 1 & 3\\ 0 & 0 & 2 \end {bmatrix}\begin {bmatrix} a\\ b\\ c \end {bmatrix} =\begin {bmatrix} 10\\ 5\\ -10 \end {bmatrix} \] Leading variables are \(a,b,c\). There are no free variables. From last row \(2c=-10\) hence \(c=-5\). From second row \(b+3c=5\) or \(b=5-3c\) or \(b=5-3\left ( -5\right ) \) or \(b=20\). From first row \(a+b+c=10\). Hence \(a=10-b-c\) or \(a=10-20+5\) or \(a=-5\). The solution is\[\begin {bmatrix} a\\ b\\ c \end {bmatrix} =\begin {bmatrix} -5\\ 20\\ -5 \end {bmatrix} \] Therefore, the interpolation polynomial is\[ f\left ( t\right ) =a+bt+ct^{2}\] Or\[ \fbox {$f\left ( t\right ) =-5+20t-5t^2$}\] It is not good idea to use \(f\left ( t\right ) \) to predict the price outside the range of interpolation, which is \(t=1\cdots 3\). Doing so is extrapolation and can produce wrong prediction. For example, using \(t=4\) gives \(f\left ( 4\right ) =-5\) dollars as stock price, which is not possible. The lowest value a stock can have is zero dollars, which is when the company go bankrupt.
Here is plot of the solution fitted on the points
Geometrically, what do subspaces of \(\mathbb {R} ^{2}\) look like?
Solution
A Subspace of \(\mathbb {R} ^{2}\) is all straight lines that pass through the origin. So each straight lines that pass through the origin is a subspace. This shows there are infinite number of subspaces.
Another subspace of \(\mathbb {R} ^{2}\) is just the origin \(\boldsymbol {0}\). And \(\mathbb {R} ^{2}\) itself is subspace of itself.
Let \(A\) be an \(n\times n\) matrix and consider the linear system \(A\boldsymbol {x}=\boldsymbol {b}\). If I know that the solution set to this linear system is a subspace of \(\mathbb {R} ^{n}\), what can you say about \(\boldsymbol {b}\)?
Solution
The vector \(\boldsymbol {b}\) is the zero vector. This is by theorem 2, page 226 in the textbook.