🔬 Tutorial problems eta \(\eta\)#
Note
This problems are designed to help you practice the concepts covered in the lectures. Not all problems may be covered in the tutorial, those left out are for additional practice on your own.
\(\eta\).1#
Let \(I_{2}\) be the \((2 \times 2)\) identity matrix, and consider the following three matrices:
\[\begin{split}
A=\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right), B=\left(\begin{array}{cc}
4 & 3 \\
-7 & 0
\end{array}\right) \text {, and } C=\left(\begin{array}{ccc}
5 & -1 & -1 \\
12 & 0 & 2
\end{array}\right) \text {. }
\end{split}\]
(a) If possible, find \(A+B\).
(b) If possible, find \(A-B\).
(c) If possible, find \(A+4 B\).
(d) If possible, find \(A+I_{2}\).
(e) If possible, find \(A I_{2}\).
(f) If possible, find \(A+C\).
(g) If possible, find \(A+B^{T}\).
(h) If possible, find \(B C\).
(i) If possible, find \(C B\).
(j) If possible, find \(C B^{T}\).
(k) If possible, find \((A B)^{T}\).
(l) If possible, find \(C+5 I_{2}\).
(m) If possible, find \(C^{T} A\).
(n) If possible, find \((B C)^{T}\).
(o) If possible, find \(A C+B\).
[Bradley, 2013] Progress Exercises 9.2
(a)
We have
\[\begin{split}
\begin{array}{ll}
A+B & =\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right)+\left(\begin{array}{cc}
4 & 3 \\
-7 & 0
\end{array}\right) \\
& =\left(\begin{array}{cc}
1+4 & -4+3 \\
0+(-7) & 9+0
\end{array}\right) \\
& =\left(\begin{array}{cc}
5 & -1 \\
-7 & 9
\end{array}\right) .
\end{array}
\end{split}\]
(b)
We have
\[\begin{split}
\begin{array}{ll}
A-B & =\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right)-\left(\begin{array}{cc}
4 & 3 \\
-7 & 0
\end{array}\right) \\
& =\left(\begin{array}{cc}
1-4 & -4-3 \\
0-(-7) & 9-0
\end{array}\right) \\
& =\left(\begin{array}{cc}
1-4 & -4-3 \\
0+7 & 9-0
\end{array}\right) \\
& =\left(\begin{array}{cc}
-3 & -7 \\
7 & 9
\end{array}\right) .
\end{array}
\end{split}\]
(c)
We have
\[\begin{split}
\begin{array}{ll}
A+4 B & =\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right)+4\left(\begin{array}{cc}
4 & 3 \\
-7 & 0
\end{array}\right) \\
& =\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right)+\left(\begin{array}{cc}
(4)(4) & (4)(3) \\
(4)(-7) & (4)(0)
\end{array}\right) \\
& =\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right)+\left(\begin{array}{cc}
16 & 12 \\
-28 & 0
\end{array}\right) \\
& =\left(\begin{array}{cc}
1+16 & -4+12 \\
0+(-28) & 9+0
\end{array}\right) \\
& =\left(\begin{array}{cc}
17 & 8 \\
-28 & 9
\end{array}\right) .
\end{array}
\end{split}\]
(d)
We have
\[\begin{split}
\begin{array}{ll}
A+I_{2} & =\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right)+\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right) \\
& =\left(\begin{array}{cc}
1+1 & -4+0 \\
0+0 & 9+1
\end{array}\right) \\
& =\left(\begin{array}{cc}
2 & -4 \\
0 & 10
\end{array}\right) .
\end{array}
\end{split}\]
(e)
We have
\[\begin{split}
\begin{array}{ll}
A I_{2} & =\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right)\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right) \\
& =\left(\begin{array}{cc}
(1)(1)+(-4)(0) & (1)(0)+(-4)(1) \\
(0)(1)+(9)(0) & (0)(0)+(9)(1)
\end{array}\right) \\
& =\left(\begin{array}{cc}
1+0 & 0+(-4) \\
0+0 & 0+9
\end{array}\right) \\
& =\left(\begin{array}{cc}
1+0 & 0-4 \\
0+0 & 0+9
\end{array}\right) \\
& =\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right) \\
& =A .
\end{array}
\end{split}\]
(f)
Note that \(A\) is a \((2 \times 2)\) matrix and \(C\) is a \((2 \times 3)\) matrix. Since \(A\) and \(C\) do not have the same dimensions, their sum \((A+C)\) is undefined.
(g)
We have
\[\begin{split}
\begin{array}{ll}
A+B^{T} &=\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right)+\left(\begin{array}{cc}
4 & 3 \\
-7 & 0
\end{array}\right)^{T} \\
& =\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right)+\left(\begin{array}{cc}
4 & -7 \\
3 & 0
\end{array}\right) \\
& =\left(\begin{array}{cc}
1+4 & -4+(-7) \\
0+3 & 9+0
\end{array}\right) \\
& =\left(\begin{array}{cc}
1+4 & -4-7 \\
0+3 & 9+0
\end{array}\right) \\
& =\left(\begin{array}{cc}
5 & -11 \\
3 & 9
\end{array}\right) \text {. }
\end{array}
\end{split}\]
(h)
We have
\[\begin{split}
\begin{array}{ll}
BC &=\left(\begin{array}{cc}
4 & 3 \\
-7 & 0
\end{array}\right)\left(\begin{array}{ccc}
5 & -1 & -1 \\
12 & 0 & 2
\end{array}\right) \\
& =\left(\begin{array}{ccc}
(4)(5)+(3)(12) & (4)(-1)+(3)(0) & (4)(-1)+(3)(2) \\
(-7)(5)+(0)(12) & (-7)(-1)+(0)(0) & (-7)(-1)+(0)(2)
\end{array}\right) \\
& =\left(\begin{array}{ccc}
20+36 & -4+0 & -4+6 \\
-35+0 & 7+0 & 7+0
\end{array}\right) \\
& =\left(\begin{array}{ccc}
56 & -4 & 2 \\
-35 & 7 & 7
\end{array}\right) .
\end{array}
\end{split}\]
(i)
Note that \(C\) is a \((2 \times 3)\) matrix and \(B\) is a \((2 \times 2)\) matrix. Since the number of columns in \(C\) (three) does not equal the number of rows in \(B\) (two), their dot product \((C B)\) is undefined.
(j)
Note that \(C\) is a \((2 \times 3)\) matrix and \(B\) is a \((2 \times 2)\) matrix. Since \(B\) is a \((2 \times 2)\) matrix, we know that \(B^{T}\) is a \((2 \times 2)\) matrix. The reason for this is that the number of rows in \(B^{T}\) is equal to the number of columns in \(B\) and the number of columns in \(B^{T}\) is equal to the number of rows in \(B\). Since the number of columns in \(C\) (three) does not equal the number of rows in \(B^{T}\) (two), their dot product \((C B\) ) is undefined.
(k)
We have
\[\begin{split}
\begin{array}{ll}
A B & =\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right)\left(\begin{array}{cc}
4 & 3 \\
-7 & 0
\end{array}\right) \\
& =\left(\begin{array}{cc}
(1)(4)+(-4)(-7) & (1)(3)+(-4)(0) \\
(0)(4)+(9)(-7) & (0)(3)+(9)(0)
\end{array}\right) \\
& =\left(\begin{array}{cc}
4+28 & 3+0 \\
0-63 & 0+0
\end{array}\right) \\
& =\left(\begin{array}{cc}
32 & 3 \\
-63 & 0
\end{array}\right) .
\end{array}
\end{split}\]
This means that
\[\begin{split}
\begin{array}{ll}
(A B)^{T} & =\left(\begin{array}{cc}
32 & 3 \\
-63 & 0
\end{array}\right)^{T} \\
& =\left(\begin{array}{cc}
32 & -63 \\
3 & 0
\end{array}\right) .
\end{array}
\end{split}\]
(l)
Note that \(C\) is a \((2 \times 3)\) matrix, \(I_{2}\) is a \((2 \times 2)\) matrix and \(I_{3}\) is a \((3 \times 3)\) matrix. In fact, all identity matrices are square matrices. As such, \(5 I\) will also be a square matrix, regardless of its dimension. Since \(C\) is not a square matrix and \(5 I\) is a square matrix, we know that \(C\) and \(5 I\) do not have the same dimensions. As such, their sum \((C+5 I)\) is not defined.
(m)
We have
\[\begin{split}
\begin{array}{ll}
C^{T} A & =\left(\begin{array}{ccc}
5 & -1 & -1 \\
12 & 0 & 2
\end{array}\right)^{T}\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right) \\
& =\left(\begin{array}{cc}
5 & 12 \\
-1 & 0 \\
-1 & 2
\end{array}\right)\left(\begin{array}{cc}
1 & -4 \\
0 & 9
\end{array}\right) \\
& =\left(\begin{array}{cc}
(5)(1)+(12)(0) & (5)(-4)+(12)(9) \\
(-1)(1)+(0)(0) & (-1)(-4)+(0)(9) \\
(-1)(1)+(2)(0) & (-1)(-4)+(2)(9)
\end{array}\right) \\
& =\left(\begin{array}{cc}
5+0 & -20+108 \\
-1+0 & 4+0 \\
-1+0 & 4+18
\end{array}\right) \\
& =\left(\begin{array}{cc}
5 & 88 \\
-1 & 4 \\
-1 & 22
\end{array}\right) .
\end{array}
\end{split}\]
(n)
Recall from Part (h) that
\[\begin{split}
B C=\left(\begin{array}{ccc}
56 & -4 & 2 \\
-35 & 7 & 7
\end{array}\right)
\end{split}\]
Thus we have
\[\begin{split}
\begin{array}{ll}
(B C)^{T} & =\left(\begin{array}{ccc}
56 & -4 & 2 \\
-35 & 7 & 7
\end{array}\right)^{T} \\
& =\left(\begin{array}{cc}
56 & -35 \\
-4 & 7 \\
2 & 7
\end{array}\right) .
\end{array}
\end{split}\]
(o)
Note that \(A\) is a \((2 \times 2)\) matrix, \(B\) is a \((2 \times 2)\) matrix and \(C\) is a \((2 \times 3)\) matrix. Since \(A\) is a \((2 \times 2)\) matrix and \(C\) is a \((2 \times 3)\) matrix, we know that the matrix dot product \(A C\) is defined and will be a \((2 \times 3)\) matrix. It is defined because the number of columns in \(A\) (two) is the same as the number of rows in \(B\) (two). The resulting matrix \((A C)\) will have the same number of rows as \(A\) (two) and the same number of columns as \(B\) (two). It is for this reason that it is a \((2 \times 3)\) matrix. Since \(A C\) is a \((2 \times 3)\) matrix and \(B\) is a \((2 \times 2)\) matrix, we know that they do not have the same dimensions. As such, their sum \((A C+B)\) is not defined.
\(\eta\).2#
The Real Estate Institute wants to develop a model which explains the relationship between the price of land and the distance from the central business district. The price per square metre of the last five blocks of land sold are shown in the following vector:
\[\begin{split}
y=\left(\begin{array}{l}
6 \\
4 \\
7 \\
5 \\
9
\end{array}\right)
\end{split}\]
The distance of these blocks from the central business district are shown in the second column of the following matrix:
\[\begin{split}
X=\left(\begin{array}{cc}
1 & 15 \\
1 & 20 \\
1 & 5 \\
1 & 16 \\
1 & 1
\end{array}\right)
\end{split}\]
It can be shown that
\[\begin{split}
\left(X^{T} X\right)^{-1}=\left(\begin{array}{cc}
\frac{4,535}{6,430} & \frac{-57}{1,286} \\
\frac{-57}{1,286} & \frac{5}{1,286}
\end{array}\right)
\end{split}\]
(a) Find \(X^{T} y\).
(b) Find \(X^{T} X\).
(c) Find \(\left(X^{T} X\right)^{-1} X^{T} y\). (Note that this is the formula for the ordinary least squares (and maximum likelihood) estimator of the coefficient parameter vector in the classical linear regression model.)
(d) Find the hat matrix, \(P=X\left(X^{T} X\right)^{-1} X^{T}\).
(e) Calculate \(P^{T}\). Is the hat matrix symmetric?
(f) Calculate \(P P\). Is the hat matrix idempotent?
(g) Find the residual-making matrix, \(M=I-P\).
(h) Calculate \(M^{T}\). Is the residual-making matrix symmetric?
(i) Calculate \(M M\). Is the residual-making matrix idempotent?
[Shannon, 1995] p. 228, Question 12. Some additional parts have been added to that question here.*
Review the classic linear regression model (CLRM) in the lecture notes.
\[
b=\left(X^{T} X\right)^{-1} X^{T} Y
\]
(a)
Note that
\[\begin{split}
\begin{array}{ll}
X^{T} y &=\left(\begin{array}{cc}
1 & 15 \\
1 & 20 \\
1 & 5 \\
1 & 16 \\
1 & 1
\end{array}\right)^{T}\left(\begin{array}{l}
6 \\
4 \\
7 \\
5 \\
9
\end{array}\right) \\
& =\left(\begin{array}{ccccc}
1 & 1 & 1 & 1 & 1 \\
15 & 20 & 5 & 16 & 1
\end{array}\right)\left(\begin{array}{l}
6 \\
4 \\
7 \\
5 \\
9
\end{array}\right) \\
& =\left(\begin{array}{c}
(1)(6)+(1)(4)+(1)(7)+(1)(5)+(1)(9) \\
(15)(6)+(20)(4)+(5)(7)+(16)(5)+(1)(9)
\end{array}\right) \\
& =\left(\begin{array}{c}
6+4+7+5+9 \\
90+80+35+80+9
\end{array}\right) \\
& =\left(\begin{array}{c}
31 \\
294
\end{array}\right) \text {. }
\end{array}
\end{split}\]
(b)
Note that
\[\begin{split}
\begin{array}{ll}
X^{T} X & =\left(\begin{array}{cc}
1 & 15 \\
1 & 20 \\
1 & 5 \\
1 & 16 \\
1 & 1
\end{array}\right)^{T}\left(\begin{array}{cc}
1 & 15 \\
1 & 20 \\
1 & 5 \\
1 & 16 \\
1 & 1
\end{array}\right) \\
& =\left(\begin{array}{ccccc}
1 & 1 & 1 & 1 & 1 \\
15 & 20 & 5 & 16 & 1
\end{array}\right)\left(\begin{array}{cc}
1 & 15 \\
1 & 20 \\
1 & 5 \\
1 & 16 \\
1 & 1
\end{array}\right) \\
& =\left(\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right) .
\end{array}
\end{split}\]
We have
\[\begin{split}
\begin{array}{ll}
a_{11} & =(1)(1)+(1)(1)+(1)(1)+(1)(1)+(1)(1) \\
& =1+1+1+1+1 \\
& =5, \\ \\
a_{12} & =(1)(15)+(1)(20)+(1)(5)+(1)(16)+(1)(1) \\
& =15+20+5+16+1 \\
& =57, \\ \\
a_{21} & =(15)(1)+(20)(1)+(5)(1)+(16)(1)+(1)(1) \\
& =15+20+5+16+1 \\
& =57, \\ \\
a_{22} & =(15)(15)+(20)(20)+(5)(5)+(16)(16)+(1)(1) \\
& =225+400+25+256+1 \\
& =907
\end{array}
\end{split}\]
This means that
\[\begin{split}
X^{T} X=\left(\begin{array}{cc}
5 & 57 \\
57 & 907
\end{array}\right)
\end{split}\]
(c)
Note that
\[\begin{split}
\begin{array}{ll}
\left(X^{T} X\right)^{-1} X^{T} y & =\left(\begin{array}{cc}
\frac{4,535}{6,430} & \frac{-57}{1,286} \\
\frac{-57}{1,286} & \frac{5}{1,286}
\end{array}\right)\left(\begin{array}{c}
31 \\
294
\end{array}\right) \\
& =\left(\begin{array}{l}
\left(\frac{4,435}{6,430}\right)(31)+\left(\frac{-57}{1,286}\right)(294) \\
\left(\frac{-57}{1,286}\right)(31)+\left(\frac{5}{1,286}\right)(294)
\end{array}\right) \\
& =\left(\left(\begin{array}{c}
\left.\frac{907}{1,286}\right)(31)+\left(\frac{-57}{1,286}\right)(294) \\
\left(\frac{-57}{1,286}\right)(31)+\left(\frac{5}{1,286}\right)(294)
\end{array}\right)\right. \\
& =\left(\begin{array}{c}
\frac{28,117}{1,286}-\frac{16,758}{1,286} \\
\frac{-1,767}{1,286}+\frac{1,470}{1,286}
\end{array}\right) \\
& =\left(\begin{array}{c}
\frac{11,359}{1,286} \\
\frac{-297}{1,286}
\end{array}\right) .
\end{array}
\end{split}\]
(d)
Note that
\[\begin{split}
\begin{array}{ll}
P & =X\left(X^{T} X\right)^{-1} X^{T} \\
& =X\left[\left(X^{T} X\right)^{-1} X^{T}\right] .
\end{array}
\end{split}\]
We have
\[\begin{split}
\begin{array}{ll}
& \left(X^{T} X\right)^{-1} X^{T}=\left(\begin{array}{cc}
\frac{4,535}{6,430} & \frac{-57}{1,286} \\
\frac{-57}{1,286} & \frac{5}{1,286}
\end{array}\right)\left(\begin{array}{cc}
1 & 15 \\
1 & 20 \\
1 & 5 \\
1 & 16 \\
1 & 1
\end{array}\right)^{T} \\
& =\left(\begin{array}{ll}
\frac{4,535}{6,430} & \frac{-57}{1,286} \\
\frac{-57}{1,286} & \frac{5}{1,286}
\end{array}\right)\left(\begin{array}{ccccc}
1 & 1 & 1 & 1 & 1 \\
15 & 20 & 5 & 16 & 1
\end{array}\right) \\
& =\left(\begin{array}{lllll}
b_{11} & b_{12} & b_{13} & b_{14} & b_{15} \\
b_{21} & b_{22} & b_{23} & b_{24} & b_{25}
\end{array}\right) .
\end{array}
\end{split}\]
The elements of this matrix are
\[\begin{split}
\begin{array}{ll}
b_{11} & =\left(\frac{4,535}{6,430}\right)(1)+\left(\frac{-57}{1,286}\right) (15)\\
& =\frac{907}{1,286}-\frac{855}{1,286} \\
& =\frac{52}{1,286}, \\ \\
b_{12} & =\left(\frac{4,535}{6,430}\right)(1)+\left(\frac{-57}{1,286}\right) (20)\\
& =\frac{907}{1,286}-\frac{1,140}{1,286} \\
& =\frac{-233}{1,286}, \\ \\
b_{13} & =\left(\frac{4,535}{6,430}\right)(1)+\left(\frac{-57}{1,286}\right) (5)\\
& =\frac{907}{1,286}-\frac{285}{1,286} \\
& =\frac{622}{1,286},\\ \\
b_{14}&=\left(\frac{4,535}{6,430}\right)(1)+\left(\frac{-57}{1,286}\right)(16) \\
& =\frac{907}{1,286}-\frac{912}{1,286} \\
& =\frac{-5}{1,286}, \\ \\
b_{15}&=\left(\frac{4,535}{6,430}\right)(1)+\left(\frac{-57}{1,286}\right)(1) \\
& =\frac{907}{1,286}-\frac{57}{1,286} \\
& =\frac{850}{1,286} \\ \\
b_{21}&=\left(\frac{-57}{1,286}\right)(1)+\left(\frac{5}{1,286}\right)(15) \\
& =\frac{-57}{1,286}+\frac{75}{1,286} \\
& =\frac{18}{1,286}, \\ \\
b_{22}&=\left(\frac{-57}{1,286}\right)(1)+\left(\frac{5}{1,286}\right)(20) \\
& =\frac{-57}{1,286}+\frac{100}{1,286} \\
& =\frac{43}{1,286}, \\ \\
b_{23}&=\left(\frac{-57}{1,286}\right)(1)+\left(\frac{5}{1,286}\right)(5) \\
& =\frac{-57}{1,286}+\frac{25}{1,286} \\
& =\frac{-32}{1,286}, \\ \\
b_{24}&=\left(\frac{-57}{1,286}\right)(1)+\left(\frac{5}{1,286}\right)(16) \\
& =\frac{-57}{1,286}+\frac{80}{1,286} \\
& =\frac{23}{1,286}, \\ \\
b_{25} & =\left(\frac{-57}{1,286}\right)(1)+\left(\frac{5}{1,286}\right)(1)\\
& =\frac{-57}{1,286}+\frac{5}{1,286} \\
& =\frac{-52}{1,286} .
\end{array}
\end{split}\]
This means that
\[\begin{split}
\left(X^{T} X\right)^{-1} X^{T}=\left(\begin{array}{lllll}
\frac{52}{1,286} & \frac{-233}{1,286} & \frac{622}{1,286} & \frac{-5}{1,286} & \frac{850}{1,286} \\
\frac{18}{1,286} & \frac{43}{1,286} & \frac{-32}{1,286} & \frac{23}{1,286} & \frac{-52}{1,286}
\end{array}\right)
\end{split}\]
Thus we have
\[\begin{split}
\begin{array}{ll}
P & =X\left(X^{T} X\right)^{-1} X^{T} \\
& =X\left[\left(X^{T} X\right)^{-1} X^{T}\right] \\
& =\left(\begin{array}{cc}
1 & 15 \\
1 & 20 \\
1 & 5 \\
1 & 16 \\
1 & 1
\end{array}\right)\left(\begin{array}{ccccc}
\frac{52}{1,286} & \frac{-233}{1,286} & \frac{622}{1,286} & \frac{-5}{1,286} & \frac{850}{1,286} \\
\frac{18}{1,286} & \frac{43}{1,286} & \frac{-32}{1,286} & \frac{23}{1,286} & \frac{-52}{1,286}
\end{array}\right) \\
& =\left(\begin{array}{ccccc}
c_{11} & c_{12} & c_{13} & c_{14} & c_{15} \\
c_{21} & c_{22} & c_{23} & c_{24} & c_{25} \\
c_{31} & c_{32} & c_{33} & c_{34} & c_{35} \\
c_{41} & c_{42} & c_{43} & c_{44} & c_{45} \\
c_{51} & c_{52} & c_{53} & c_{54} & c_{55}
\end{array}\right) .
\end{array}
\end{split}\]
The elements of this matrix are
\[\begin{split}
\begin{array}{ll}
c_{11} & =(1)\left(\frac{52}{1,286}\right)+(15)\left(\frac{18}{1,286}\right) \\
& =\frac{52}{1,286}+\frac{270}{1,286} \\
& =\frac{322}{1,286}, \\
c_{12} & =(1)\left(\frac{-233}{1,286}\right)+(15)\left(\frac{43}{1,286}\right) \\
& =\frac{-233}{1,286}+\frac{645}{1,286} \\
& =\frac{412}{1,286}, \\ \\
c_{13} &=(1)\left(\frac{622}{1,286}\right)+(15)\left(\frac{-32}{1,286}\right) \\
& =\frac{622}{1,286}-\frac{480}{1,286} \\
& =\frac{142}{1,286}, \\ \\
c_{14}&=(1)\left(\frac{-5}{1,286}\right)+(15)\left(\frac{23}{1,286}\right) \\
& =\frac{-5}{1,286}+\frac{345}{1,286} \\
& =\frac{340}{1,286}, \\ \\
c_{15}&=(1)\left(\frac{850}{1,286}\right)+(15)\left(\frac{-52}{1,286}\right) \\
& =\frac{850}{1,286}-\frac{780}{1,286} \\
& =\frac{70}{1,286}, \\ \\
c_{21}&=(1)\left(\frac{52}{1,286}\right)+(20)\left(\frac{18}{1,286}\right) \\
& =\frac{52}{1,286}+\frac{360}{1,286} \\
& =\frac{412}{1,286}, \\ \\
c_{22}&=(1)\left(\frac{-233}{1,286}\right)+(20)\left(\frac{43}{1,286}\right) \\
& =\frac{-233}{1,286}+\frac{860}{1,286} \\
& =\frac{627}{1,286}, \\ \\
c_{23}&=(1)\left(\frac{622}{1,286}\right)+(20)\left(\frac{-32}{1,286}\right) \\
& =\frac{622}{1,286}-\frac{640}{1,286} \\
& =\frac{-18}{1,286}, \\ \\
c_{24}&=(1)\left(\frac{-5}{1,286}\right)+(20)\left(\frac{23}{1,286}\right) \\
& =\frac{-5}{1,286}+\frac{460}{1,286} \\
& =\frac{455}{1,286}, \\ \\
c_{25}&=(1)\left(\frac{850}{1,286}\right)+(20)\left(\frac{-52}{1,286}\right) \\
& =\frac{850}{1,286}-\frac{1,040}{1,286} \\
& =\frac{-190}{1,286}, \\ \\
c_{31}&=(1)\left(\frac{52}{1,286}\right)+(5)\left(\frac{18}{1,286}\right) \\
& =\frac{52}{1,286}+\frac{90}{1,286} \\
& =\frac{142}{1,286}, \\ \\
c_{32}&=(1)\left(\frac{-233}{1,286}\right)+(5)\left(\frac{43}{1,286}\right) \\
& =\frac{-233}{1,286}+\frac{215}{1,286} \\
& =\frac{-18}{1,286} \\ \\
c_{33}&=(1)\left(\frac{622}{1,286}\right)+(5)\left(\frac{-32}{1,286}\right) \\
& =\frac{622}{1,286}-\frac{160}{1,286} \\
& =\frac{462}{1,286}, \\ \\
c_{34}&=(1)\left(\frac{-5}{1,286}\right)+(5)\left(\frac{23}{1,286}\right) \\
& =\frac{-5}{1,286}+\frac{115}{1,286} \\
& =\frac{110}{1,286}, \\ \\
c_{35}&=(1)\left(\frac{850}{1,286}\right)+(5)\left(\frac{-52}{1,286}\right) \\
&=\frac{850}{1,286}-\frac{260}{1,286} \\
&=\frac{590}{1,286}, \\ \\
c_{41} &=(1)\left(\frac{52}{1,286}\right)+(16)\left(\frac{18}{1,286}\right) \\
&=\frac{52}{1,286}+\frac{288}{1,286} \\
&=\frac{340}{1,286}, \\ \\
c_{42} &=(1)\left(\frac{-233}{1,286}\right) + (16)\left(\frac{43}{1,286}\right) \\
&=\frac{-233}{1,286}+\frac{688}{1,286} \\
&=\frac{455}{1,286}, \\ \\
c_{43} &= (1)\left(\frac{622}{1,286} \right) + (16)\left(\frac{-32}{1,286}\right) \\
&= \frac{622}{1,286}-\frac{512}{1,286} \\
&= \frac{110}{1,286}, \\ \\
c_{44} &= (1) \left(\frac{-5}{1,286}\right) + (16) \left(\frac{25}{1,286}\right) \\
&= \frac{-5}{1,286} + \frac{368}{1,286} \\
&=\frac{363}{1,286}, \\ \\
c_{45} &=(1)\left(\frac{850}{1,286}\right)+(16)\left(\frac{-52}{1,286}\right) \\
&= \frac{850}{1,286} - \frac{832}{1,286} \\
&= \frac{18}{1,286}, \\ \\
c_{51} & =(1)\left(\frac{52}{1,286}\right)+(1)\left(\frac{18}{1,286}\right) \\
& =\frac{52}{1,286}+\frac{18}{1,286} \\
& =\frac{70}{1,286}, \\ \\
c_{52} & =(1)\left(\frac{-233}{1,286}\right)+(1)\left(\frac{43}{1,286}\right) \\
& =\frac{-233}{1,286}+\frac{43}{1,286} \\
& =\frac{-190}{1,286}, \\ \\
c_{53} & =(1)\left(\frac{622}{1,286}\right)+(1)\left(\frac{-32}{1,286}\right) \\
& =\frac{622}{1,286}-\frac{32}{1,286} \\
& =\frac{590}{1,286}, \\ \\
c_{54} & =(1)\left(\frac{-5}{1,286}\right)+(1)\left(\frac{23}{1,286}\right) \\
& =\frac{-5}{1,286}+\frac{23}{1,286} \\
& =\frac{18}{1,286}, \\ \\
c_{55} & =(1)\left(\frac{850}{1,286}\right)+(1)\left(\frac{-52}{1,286}\right) \\
& =\frac{850}{1,286}-\frac{52}{1,286} \\
& =\frac{798}{1,286} .
\end{array}
\end{split}\]
As such, the hat matrix is given by
\[\begin{split}
P=\left(\begin{array}{ccccc}
\frac{322}{1,286} & \frac{412}{1,286} & \frac{142}{1,286} & \frac{340}{1,286} & \frac{70}{1,286} \\
\frac{412}{1,286} & \frac{627}{1,286} & \frac{-18}{1,286} & \frac{455}{1,286} & \frac{-190}{1,286} \\
\frac{142}{1,286} & \frac{-18}{1,286} & \frac{462}{1,286} & \frac{110}{1,286} & \frac{590}{1,286} \\
\frac{340}{1,286} & \frac{455}{1,286} & \frac{110}{1,286} & \frac{363}{1,286} & \frac{18}{1,286} \\
\frac{70}{1,286} & \frac{-190}{1,286} & \frac{590}{1,286} & \frac{18}{1,286} & \frac{798}{1,286}
\end{array}\right) .
\end{split}\]
(e)
Note that
\[\begin{split}
\begin{aligned}
P^{T} & =\left(\begin{array}{ccccc}
\frac{322}{1,286} & \frac{412}{1,286} & \frac{142}{1,286} & \frac{340}{1,286} & \frac{70}{1,286} \\
\frac{412}{1,286} & \frac{627}{1,286} & \frac{-18}{1,286} & \frac{455}{1,286} & \frac{-190}{1,286} \\
\frac{142}{1,286} & \frac{-18}{1,286} & \frac{462}{1,286} & \frac{110}{1,286} & \frac{590}{1,286} \\
\frac{340}{1,286} & \frac{455}{1,286} & \frac{110}{1,286} & \frac{363}{1,286} & \frac{18}{1,286} \\
\frac{70}{1,286} & \frac{-190}{1,286} & \frac{590}{1,286} & \frac{18}{1,286} & \frac{798}{1,286}
\end{array}\right)^{T} \\
& =\left(\begin{array}{ccccc}
\frac{322}{1,286} & \frac{412}{1,286} & \frac{142}{1,286} & \frac{340}{1,286} & \frac{70}{1,286} \\
\frac{412}{1,286} & \frac{627}{1,286} & \frac{-18}{1,286} & \frac{455}{1,286} & \frac{-190}{1,286} \\
\frac{142}{1,286} & \frac{-18}{1,286} & \frac{462}{1,286} & \frac{110}{1,286} & \frac{590}{1,286} \\
\frac{340}{1,286} & \frac{455}{1,286} & \frac{110}{1,286} & \frac{363}{1,286} & \frac{18}{1,286} \\
\frac{70}{1,286} & \frac{-190}{1,286} & \frac{590}{1,286} & \frac{18}{1,286} & \frac{798}{1,286}
\end{array}\right) \\
& =P .
\end{aligned}
\end{split}\]
Thus we can conclude that the hat matrix is symmetric.
(f)
Note that
\[\begin{split}
\begin{array}{ll}
PP &= \left(\begin{array}{ccccc}
\frac{322}{1,286} & \frac{412}{1,286} & \frac{142}{1,286} & \frac{340}{1,286} & \frac{70}{1,286} \\
\frac{412}{1,286} & \frac{627}{1,286} & \frac{-18}{1,286} & \frac{455}{1,286} & \frac{-190}{1,286} \\
\frac{142}{1,286} & \frac{-18}{1,286} & \frac{462}{1,286} & \frac{110}{1,286} & \frac{590}{1,286} \\
\frac{340}{1,286} & \frac{455}{1,286} & \frac{110}{1,286} & \frac{363}{1,286} & \frac{18}{1,286} \\
\frac{70}{1,286} & \frac{-190}{1,286} & \frac{590}{1,286} & \frac{18}{1,286} & \frac{798}{1,286}
\end{array}\right)
\left(\begin{array}{ccccc}
\frac{322}{1,286} & \frac{412}{1,286} & \frac{142}{1,286} & \frac{340}{1,286} & \frac{70}{1,286} \\
\frac{412}{1,286} & \frac{627}{1,286} & \frac{-18}{1,286} & \frac{455}{1,286} & \frac{-190}{1,286} \\
\frac{142}{1,286} & \frac{-18}{1,286} & \frac{462}{1,286} & \frac{110}{1,286} & \frac{590}{1,286} \\
\frac{340}{1,286} & \frac{455}{1,286} & \frac{110}{1,286} & \frac{363}{1,286} & \frac{18}{1,286} \\
\frac{70}{1,286} & \frac{-190}{1,286} & \frac{590}{1,286} & \frac{18}{1,286} & \frac{798}{1,286}
\end{array}\right) \\
& =\left[\left(\frac{1}{1,286}\right)\left(\begin{array}{ccccc}
322 & 412 & 142 & 340 & 70 \\
412 & 627 & -18 & 455 & -190 \\
142 & -18 & 462 & 110 & 590 \\
340 & 455 & 110 & 363 & 18 \\
70 & -190 & 590 & 18 & 798
\end{array}\right)\right] \bullet
{\left[\left(\frac{1}{1,286}\right)\left(\begin{array}{ccccc}
322 & 412 & 142 & 340 & 70 \\
412 & 627 & -18 & 455 & -190 \\
142 & -18 & 462 & 110 & 590 \\
340 & 455 & 110 & 363 & 18 \\
70 & -190 & 590 & 18 & 798
\end{array}\right)\right]} \\
& =\left(\frac{1}{1,286}\right)^{2}\left(\begin{array}{ccccc}
322 & 412 & 142 & 340 & 70 \\
412 & 627 & -18 & 455 & -190 \\
142 & -18 & 462 & 110 & 590 \\
340 & 455 & 110 & 363 & 18 \\
70 & -190 & 590 & 18 & 798
\end{array}\right) \bullet
\left(\begin{array}{ccccc}
322 & 412 & 142 & 340 & 70 \\
412 & 627 & -18 & 455 & -190 \\
142 & -18 & 462 & 110 & 590 \\
340 & 455 & 110 & 363 & 18 \\
70 & -190 & 590 & 18 & 798
\end{array}\right) \\
& =\left(\frac{1}{1,653,796}\right)\left(\begin{array}{ccccc}
322 & 412 & 142 & 340 & 70 \\
412 & 627 & -18 & 455 & -190 \\
142 & -18 & 462 & 110 & 590 \\
340 & 455 & 110 & 363 & 18 \\
70 & -190 & 590 & 18 & 798
\end{array}\right) \bullet \left(\begin{array}{ccccc}
322 & 412 & 142 & 340 & 70 \\
412 & 627 & -18 & 455 & -190 \\
142 & -18 & 462 & 110 & 590 \\
340 & 455 & 110 & 363 & 18 \\
70 & -190 & 590 & 18 & 798
\end{array}\right) .
\end{array}
\end{split}\]
We have
\[\begin{split}
\begin{array}{ll}
& \left(\begin{array}{ccccc}
322 & 412 & 142 & 340 & 70 \\
412 & 627 & -18 & 455 & -190 \\
142 & -18 & 462 & 110 & 590 \\
340 & 455 & 110 & 363 & 18 \\
70 & -190 & 590 & 18 & 798
\end{array}\right)\left(\begin{array}{cccccc}
322 & 412 & 142 & 340 & 70 \\
412 & 627 & -18 & 455 & -190 \\
142 & -18 & 462 & 110 & 590 \\
340 & 455 & 110 & 363 & 18 \\
70 & -190 & 590 & 18 & 798
\end{array}\right) \\
& = \left(\begin{array}{lllll}
d_{11} & d_{12} & d_{13} & d_{14} & d_{15} \\
d_{21} & d_{22} & d_{23} & d_{24} & d_{25} \\
d_{31} & d_{32} & d_{33} & d_{34} & d_{35} \\
d_{41} & d_{42} & d_{43} & d_{44} & d_{45} \\
d_{51} & d_{52} & d_{53} & d_{54} & d_{55}
\end{array}\right),
\end{array}
\end{split}\]
where
\[\begin{split}
\begin{array}{ll}
d_{11} & =(322)(322)+(412)(412)+(142)(142)+(340)(340)+(70)(70) \\
& =103,684+169,744+20,164+115,600+4,900 \\
& =414,092 \\ \\
d_{12} & =(322)(412)+(412)(627)+(142)(-18)+(340)(455)+(70)(-190) \\
& =132,664+258,324-2,556+154,700-13,300 \\
& =529,832, \\ \\
d_{13} & =(322)(142)+(412)(-18)+(142)(462)+(340)(110)+(70)(590) \\
& =45,724-7,416+65,604+37,400+41,300 \\
& =182,612, \\ \\
d_{14} & =(322)(340)+(412)(455)+(142)(110)+(340)(363)+(70)(18) \\
& =109,480+187,460+15,620+123,420+1,260 \\
& =437,240, \\ \\
d_{15} & =(322)(70)+(412)(-190)+(142)(590)+(340)(18)+(70)(798) \\
& =22,540-78,280+83,780+6,120+55,860 \\
& =90,020, \\ \\
d_{21} & =(412)(322)+(627)(412)+(-18)(142)+(455)(340)+(-190)(70) \\
& =132,664+258,324-2,556+154,700-13,300 \\
& =529,832, \\ \\
d_{22} &= (412)(412)+(627)(627)+(-18)(-18)+(455)(455)+(-190)(-190) \\
& =169,744+393,129+324+207,025+36,100 \\
& =806,322 \\ \\
d_{23} &=(412)(142)+(627)(-18)+(-18)(462)+(455)(110)+(-190)(590) \\
& =58,504-11,286-8,316+50,050-112,100 \\
& =-23,148, \\ \\
d_{24}&=(412)(340)+(627)(455)+(-18)(110)+(455)(363)+(-190)(18) \\
& =140,080+285,285-1,980+165,165-3,420 \\
& =585,130 \\ \\
d_{25}&=(412)(70)+(627)(-190)+(-18)(590)+(455)(18)+(-190)(798) \\
& =28,840-119,130-10,620+8,190-151,620 \\
& =-244,340, \\ \\
d_{31}&=(142)(322)+(-18)(412)+(462)(142)+(110)(340)+(590)(70) \\
& =45,724-7,416+65,604+37,400+41,300 \\
& =182,612, \\ \\
d_{32}&=(142)(412)+(-18)(627)+(462)(-18)+(110)(455)+(590)(-190) \\
& =58,504-11,286-8,316+50,050-112,100 \\
& =-23,148, \\ \\
d_{33}&=(142)(142)+(-18)(-18)+(462)(462)+(110)(110)+(590)(590) \\
& =20,164+324+213,444+12,100+348,100 \\
& =594,132, \\ \\
d_{34}&=(142)(340)+(-18)(455)+(462)(110)+(110)(363)+(590)(18) \\
& =48,280-8,190+50,820+39,930+10,620 \\
& =141,460, \\ \\
d_{35}&=(142)(70)+(-18)(-190)+(462)(590)+(110)(18)+(590)(798) \\
& =9,940+3,420+272,580+1,980+470,820 \\
& =758,740, \\ \\
d_{41} &=(340)(322)+(455)(412)+(110)(142)+(363)(340)+(18)(70) \\
& =109,480+187,460+15,620+123,420+1,260 \\
& =437,240 \\ \\
d_{42}&=(340)(412)+(455)(627)+(110)(-18)+(363)(455)+(18)(-190) \\
& =140,080+285,285-1,980+165,165-3,420 \\
& =585,130, \\ \\
d_{43}&=(340)(142)+(455)(-18)+(110)(462)+(363)(110)+(18)(590) \\
& =48,280-8,190+50,820+39,930+10,620 \\
& =141,460, \\ \\
d_{44}&=(340)(340)+(455)(455)+(110)(110)+(363)(363)+(18)(18) \\
& =115,600+207,025+12,100+131,769+324 \\
& =466,818 \\ \\
d_{45}&=(340)(70)+(455)(-190)+(110)(590)+(363)(18)+(18)(798) \\
& =23,800-86,450+64,900+6,534+14,364 \\
& =23,148 \\ \\
d_{51}&=(70)(322)+(-190)(412)+(590)(142)+(18)(340)+(798)(70) \\
& =22,540-78,280+83,780+6,120+55,860 \\
& =90,020 \\ \\
d_{52}&=(70)(412)+(-190)(627)+(590)(-18)+(18)(455)+(798)(-190) \\
& =28,840-119,130-10,620+8,190-151,620 \\
& =-244,340, \\ \\
d_{53}&=(70)(142)+(-190)(-18)+(590)(462)+(18)(110)+(798)(590) \\
& =9,940+3,420+272,580+1,980+470,820 \\
& =758,740, \\ \\
d_{54}&=(70)(340)+(-190)(455)+(590)(110)+(18)(363)+(798)(18) \\
& =23,800-86,450+64,900+6,534+14,364 \\
& =23,148, \\ \\
d_{55} & =(70)(70)+(-190)(-190)+(590)(590)+(18)(18)+(798)(798) \\
& =4,900+36,100+348,100+324+636,804 \\
& =1,026,228 .
\end{array}
\end{split}\]
This means that
\[\begin{split}
\begin{array}{ll}
& \left(\begin{array}{ccccc}
322 & 412 & 142 & 340 & 70 \\
412 & 627 & -18 & 455 & -190 \\
142 & -18 & 462 & 110 & 590 \\
340 & 455 & 110 & 363 & 118 \\
70 & -190 & 590 & 18 & 798
\end{array}\right)\left(\begin{array}{ccccc}
322 & 412 & 142 & 340 & 70 \\
412 & 627 & -18 & 455 & -190 \\
142 & -18 & 462 & 110 & 590 \\
340 & 455 & 110 & 363 & 118 \\
70 & -190 & 590 & 18 & 798
\end{array}\right) \\
= & \left(\begin{array}{ccccc}
414,092 & 529,832 & 182,612 & 437,240 & 90,020 \\
529,832 & 806,322 & -23,148 & 585,130 & -244,340 \\
182,612 & -23,148 & 594,132 & 141,460 & 758,740 \\
437,240 & 585,130 & 141,460 & 466,818 & 23,148 \\
90,020 & -244,340 & 758,740 & 23,148 & 1,026,228
\end{array}\right) .
\end{array}
\end{split}\]
As such, we know that
\[\begin{split}
\begin{array}{ll}
& P P=\left(\frac{1}{1,653,796}\right)\left(\begin{array}{ccccc}
414,092 & 529,832 & 182,612 & 437,240 & 90,020 \\
529,832 & 806,322 & -23,148 & 585,130 & -244,340 \\
182,612 & -23,148 & 594,132 & 141,460 & 758,740 \\
437,240 & 585,130 & 141,460 & 466,818 & 23,148 \\
90,020 & -244,340 & 759,010 & 758,740 & 1,026,228
\end{array}\right) \\
& =\left(\frac{1}{1,286}\right)^{2}\left(\begin{array}{ccccc}
414,092 & 529,832 & 182,612 & 437,240 & 90,020 \\
529,832 & 806,322 & -23,148 & 585,130 & -244,340 \\
182,612 & -23,148 & 594,132 & 141,460 & 758,740 \\
437,240 & 585,130 & 141,460 & 466,818 & 23,148 \\
90,020 & -244,340 & 758,740 & 23,148 & 1,026,228
\end{array}\right) \\
& =\left(\frac{1}{1,286}\right)\left(\begin{array}{ccccc}
322 & 412 & 142 & 340 & 70 \\
412 & 627 & -18 & 455 & 190 \\
142 & -18 & 462 & 110 & 590 \\
340 & 455 & 110 & 363 & 18 \\
70 & 190 & 590 & 18 & 798
\end{array}\right) \\
& =\left(\begin{array}{lllll}
\frac{322}{1,286} & \frac{412}{1,286} & \frac{142}{1,286} & \frac{340}{1,286} & \frac{70}{1,286} \\
\frac{412}{1,286} & \frac{627}{1,286} & \frac{-18}{1,286} & \frac{455}{1,286} & \frac{-190}{1,286} \\
\frac{142}{1,286} & \frac{-18}{1,286} & \frac{462}{1,286} & \frac{110}{1,286} & \frac{590}{1,286} \\
\frac{340}{1,286} & \frac{455}{1,286} & \frac{110}{1,286} & \frac{363}{1,286} & \frac{18}{1,286} \\
\frac{70}{1,286} & \frac{-190}{1,286} & \frac{590}{1,286} & \frac{18}{1,286} & \frac{798}{1,286}
\end{array}\right) \\
& =P \text {. }
\end{array}
\end{split}\]
Thus we can conclude that the hat matrix is idempotent.
(g)
The residual-making matrix is given by
\[\begin{split}
\begin{array}{ll}
M&=I-P \\
& =\left(\begin{array}{ccccc}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1
\end{array}\right)-\left(\begin{array}{ccccc}
\frac{322}{1,286} & \frac{412}{1,286} & \frac{142}{1,286} & \frac{340}{1,286} & \frac{70}{1,286} \\
\frac{412}{1,286} & \frac{627}{1,286} & \frac{-18}{1,286} & \frac{455}{1,286} & \frac{-190}{1,286} \\
\frac{142}{1,286} & \frac{-18}{1,286} & \frac{462}{1,286} & \frac{110}{1,286} & \frac{590}{1,286} \\
\frac{340}{1,286} & \frac{455}{1,286} & \frac{110}{1,286} & \frac{363}{1,286} & \frac{18}{1,286} \\
\frac{70}{1,286} & \frac{-190}{1,286} & \frac{590}{1,286} & \frac{18}{1,286} & \frac{798}{1,286}
\end{array}\right) \\
& =\left(\begin{array}{ccccc}
\frac{1,286}{1,286} & 0 & 0 & 0 & 0 \\
0 & \frac{1,286}{1,286} & 0 & 0 & 0 \\
0 & 0 & \frac{1,286}{1,286} & 0 & 0 \\
0 & 0 & 0 & \frac{1,286}{1,286} & 0 \\
0 & 0 & 0 & 0 & \frac{1,286}{1,286}
\end{array}\right)-\left(\begin{array}{cccccc}
\frac{322}{1,286} & \frac{412}{1,286} & \frac{142}{1,286} & \frac{340}{1,286} & \frac{70}{1,286} \\
\frac{412}{1,286} & \frac{627}{1,286} & \frac{-18}{1,286} & \frac{455}{1,286} & \frac{-190}{1,286} \\
\frac{142}{1,286} & \frac{-18}{1,286} & \frac{462}{1,286} & \frac{110}{1,286} & \frac{590}{1,286} \\
\frac{340}{1,286} & \frac{455}{1,286} & \frac{110}{1,286} & \frac{363}{1,286} & \frac{18}{1,286} \\
\frac{70}{1,286} & \frac{-190}{1,286} & \frac{590}{1,286} & \frac{18}{1,286} & \frac{798}{1,286}
\end{array}\right) \\
& =\left(\begin{array}{ccccc}
\frac{1,286}{1,286}-\frac{322}{1,286} & 0-\frac{412}{1,286} & 0-\frac{142}{1,1286} & 0-\frac{340}{1,286} & 0-\frac{70}{1,286} \\
0-\frac{412}{1,286} & \frac{1,286}{1,286}-\frac{627}{1,286} & 0-\left(\frac{-18}{1,286}\right) & 0-\frac{455}{1,286} & 0-\left(\frac{-190}{1,286}\right) \\
0-\frac{142}{1,286} & 0-\left(\frac{-18}{1,286}\right) & \frac{1,286}{1,286}-\frac{462}{1,286} & 0-\frac{110}{1,286} & 0-\frac{590}{1,286} \\
0-\frac{340}{1,286} & 0-\frac{455}{1,286} & 0-\frac{110}{1,286} & \frac{1,286}{1,286}-\frac{363}{1,286} & 0-\frac{18}{1,286} \\
0-\frac{70}{1,286} & 0-\left(\frac{-190}{1,286}\right) & 0-\frac{590}{1,286} & 0-\frac{18}{1,286} & \frac{1,286}{1,286}-\frac{798}{1,286}
\end{array}\right) \\
& =\left(\begin{array}{ccccc}
\frac{964}{1,286} & \frac{-412}{1,286} & \frac{-142}{1,286} & \frac{-340}{1,286} & \frac{-70}{1,286} \\
\frac{1-4212}{1,286} & \frac{659}{1,286} & \frac{18}{1,286} & \frac{-455}{1,286} & \frac{190}{1,286} \\
\frac{142}{1,286} & \frac{18}{1,286} & \frac{324}{1,286} & \frac{-110}{1,286} & \frac{-590}{1,286} \\
\frac{-340}{1,286} & \frac{-455}{1,286} & \frac{-110}{1,286} & \frac{923}{1,286} & \frac{-18}{1,286} \\
\frac{-70}{1,286} & \frac{190}{1,286} & \frac{-590}{1,286} & \frac{-18}{1,286} & \frac{488}{1,286}
\end{array}\right) .
\end{array}
\end{split}\]
(h)
Note that
\[\begin{split}
\begin{array}{ll}
M^{T} & =\left(\begin{array}{ccccc}
\frac{964}{1,286} & \frac{-412}{1,286} & \frac{-142}{1,286} & \frac{-340}{1,286} & \frac{-70}{1,286} \\
\frac{-412}{1,286} & \frac{659}{1,286} & \frac{18}{1,286} & \frac{-455}{1,286} & \frac{190}{1,286} \\
\frac{-142}{1,286} & \frac{18}{1,286} & \frac{824}{1,286} & \frac{-110}{1,286} & \frac{-590}{1,286} \\
\frac{-340}{1,286} & \frac{-455}{1,286} & \frac{-110}{1,286} & \frac{993}{1,286} & \frac{-18}{1,286} \\
\frac{-70}{1,286} & \frac{190}{1,286} & \frac{-590}{1,286} & \frac{-18}{1,286} & \frac{488}{1,286}
\end{array}\right)^{T} \\
& =\left(\begin{array}{llllll}
\frac{964}{1,286} & \frac{-412}{1,286} & \frac{-142}{1,286} & \frac{-340}{1,286} & \frac{-70}{1,286} \\
\frac{-412}{1,286} & \frac{659}{1,286} & \frac{18}{1,286} & \frac{-455}{1,286} & \frac{190}{1,286} \\
\frac{-142}{1,286} & \frac{18}{1,286} & \frac{824}{1,286} & \frac{-110}{1,286} & \frac{-590}{1,286} \\
\frac{-340}{1,286} & \frac{-455}{1,286} & \frac{-110}{1,286} & \frac{993}{1,286} & \frac{-18}{1,286} \\
\frac{-70}{1,286} & \frac{190}{1,286} & \frac{-590}{1,286} & \frac{-18}{1,286} & \frac{488}{1,286}
\end{array}\right) \\
& =M .
\end{array}
\end{split}\]
Thus we can conclude that the residual making matrix is symmetric.
(i)
This is left as an exercise. While it is tedious, the practice that you will gain in matrix multiplication by completing this exercise might be useful. You should find that \(M M=M\), so that the residual-making matrix is indeed idempotent.
\(\eta\).3#
Compute the following determinants
(a) \(\mathrm{det} \left( \begin{array}{cc} 5,& 1 \\ 0,& 1 \end{array} \right)\)
(b) \(\mathrm{det} \left( \begin{array}{cc} 2,& 1 \\ 1,& 2 \end{array} \right)\)
(c) \(\mathrm{det} \left( \begin{array}{ccc} 1,& 5,& 8 \\ 0,& 2,& 1 \\ 0,& -1,& 2 \end{array} \right)\)
(d) \(\mathrm{det} \left( \begin{array}{ccc} 1,& 0,& 3 \\ 1,& 1,& 0 \\ 0,& 0,& 8 \end{array} \right)\)
(e) \(\mathrm{det} \left( \begin{array}{cccc} 1,& 5,& 8,& 17 \\ 0,& -2,& 13,& 0 \\ 0,& 0,& 1,& 2 \\ 0,& 0,& 0,& 2 \end{array} \right)\)
(f) \(\mathrm{det} \left( \begin{array}{cccc} 2,& 1,& 0,& 0 \\ 1,& 2,& 0,& 0 \\ 0,& 0,& 2,& 0 \\ 0,& 0,& 0,& 2 \end{array} \right)\)
Solution for the selected problems
(a) By definition of a \((2 \times 2)\) determinant, we have
\[\begin{split}
\mathrm{det}
\left(
\begin{array}{cc}
5,& 1 \\
0,& 1
\end{array}
\right)
= 5\cdot 1 - 0\cdot 1 = 5
\end{split}\]
(d) By recursive definition of a \((3 \times 3)\) determinant, using the last raw for the recursive expansion (it has most zeros!), we have
\[\begin{split}
\det
\left(
\begin{array}{ccc}
1,& 0,& 3 \\
1,& 1,& 0 \\
0,& 0,& 8
\end{array}
\right)
=
8 (-1)^{3+3}
\det
\left(
\begin{array}{cc}
1,& 0 \\
1,& 1 \\
\end{array}
\right)
= 8 (1-0) = 8
\end{split}\]
(f) By recursive definition of a \((4 \times 4)\) determinant, using the last raw twice for the recursive expansion (it has most zeros!), we have
\[\begin{split}\det
\left(
\begin{array}{cccc}
2,& 1,& 0,& 0 \\
1,& 2,& 0,& 0 \\
0,& 0,& 2,& 0 \\
0,& 0,& 0,& 2
\end{array}
\right)
=
2 \cdot 2 \cdot
\det
\left(
\begin{array}{cc}
2,& 1 \\
1,& 2 \\
\end{array}
\right)
= 4 (4-1) = 12
\end{split}\]
\(\eta\).4#
Consider an \((n \times n)\) Vandermonde matrix [this one can be named :)] of the form
\[\begin{split}
V =
\begin{bmatrix}
1 & 1 & \cdots & 1 \\
x_1 & x_2 & \cdots & x_n \\
x_1^2 & x_2^2 & \cdots & x_n^2 \\
\vdots & \vdots & \ddots & \vdots \\
x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1}
\end{bmatrix}
\end{split}\]
Show that the determinant of \(V\) is given by
\[
\det(V) = \Pi_{j<i \leqslant n}(x_i-x_j)
\]
for the cases \(n=2\), \(n=3\) and \(n=4\)
Properties of the determinants help in finding an elegant solution. In particular, you may find useful that the determinant does not changes when a row/column of a matrix is added/subtracted from the other one, see for example here for explanation.
Using the hint we perform the following derivation:
\[\begin{split}
\det
\begin{pmatrix}
1 & 1 & \cdots & 1 \\
x_1 & x_2 & \cdots & x_n \\
x_1^2 & x_2^2 & \cdots & x_n^2 \\
\vdots & \vdots & \ddots & \vdots \\
x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1}
\end{pmatrix}
=
\det
\begin{pmatrix}
1 & 0 & \cdots & 0 \\
x_1 & x_2-x_1 & \cdots & x_n-x_1 \\
x_1^2 & x_2^2-x_1^2 & \cdots & x_n^2-x_1^2 \\
\vdots & \vdots & \ddots & \vdots \\
x_1^{n-1} & x_2^{n-1}-x_1^{n-1} & \cdots & x_n^{n-1}-x_1^{n-1}
\end{pmatrix}
=
\end{split}\]
(expand along the first row)
\[\begin{split}
=
\det
\begin{pmatrix}
x_2-x_1 & x_3-x_1 & \cdots & x_n-x_1 \\
x_2^2-x_1^2 & x_3^2-x_1^2 & \cdots & x_n^2-x_1^2 \\
\vdots & \vdots & \ddots & \vdots \\
x_2^{n-1}-x_1^{n-1} & x_3^{n-1}-x_1^{n-1} & \cdots & x_n^{n-1}-x_1^{n-1}
\end{pmatrix}
=
\end{split}\]
(take out common factor from each row)
\[\begin{split}
=
(x_2-x_1)(x_3-x_1)\dots(x_n-x_1)
\det
\begin{pmatrix}
1 & 1 & \cdots & 1 \\
\frac{x_2^2-x_1^2}{x_2-x_1} & \frac{x_3^2-x_1^2}{x_3-x_1} & \cdots & \frac{x_n^2-x_1^2}{x_n-x_1} \\
\vdots & \vdots & \ddots & \vdots \\
\frac{x_2^{n-1}-x_1^{n-1}}{x_2-x_1} & \frac{x_3^{n-1}-x_1^{n-1}}{x_3-x_1} & \cdots & \frac{x_n^{n-1}-x_1^{n-1}}{x_n-x_1} \\
\end{pmatrix}
\end{split}\]
It is not hard to show (by long division of polynomials and mathematical induction) that
\[
\frac{a^{k}-b^{k}}{a-b}
=a^{k-1}+a^{k-2}b + \dots + ab^{k-2} + b^{k-1}
=\sum_{i=1}^{k} a^{k-i}b^{i-1}
\]
Continuing the derivation:
\[\begin{split}
=
\prod_{i=2}^n (x_i-x_1)
\det
\begin{pmatrix}
1 & \cdots & 1 \\
x_2+x_1 & \cdots & x_n+x_1 \\
\vdots & \ddots & \vdots \\
\sum_{i=1}^{n-1} x_2^{n-1-i}x_1^{i-1} &
\cdots &
\sum_{i=1}^{n-1} x_n^{n-1-i}x_1^{i-1}
\end{pmatrix}
=
\end{split}\]
We can then repeat the same procedure:
subtract the first row from all other
notice regularities in the polynomials
expand the determinant along the first row
take out common factors from each row
It is then obvious how to prove the main formula by mathematical induction.
\(n=2\)
\[\begin{split}
\det
\begin{pmatrix}
1 & 1 \\
x & y
\end{pmatrix}
=y-x
\end{split}\]
\(n=3\)
\[\begin{split}
\det
\begin{pmatrix}
1 & 1 & 1 \\
x & y & z \\
x^2 & y^2 & z^2
\end{pmatrix}
=
\det
\begin{pmatrix}
1 & 0 & 0 \\
x & y-x & z-x \\
x^2 & y^2-x^2 & z^2-x^2
\end{pmatrix}
=
\end{split}\]
\[\begin{split}
=
\det
\begin{pmatrix}
y-x & z-x \\
y^2-x^2 & z^2-x^2
\end{pmatrix}
=
\end{split}\]
\[\begin{split}
=
(y-x)(z-x)
\det
\begin{pmatrix}
1 & 1 \\
x+y & x+z
\end{pmatrix}
=
\end{split}\]
\[
=
(y-x)(z-x)(x+z-x-y)=(y-x)(z-x)(z-y)
\]
\(n=4\)
Follow the same scheme, using
\(x^2-y^2=(x-y)(x+y)\)
\(x^3-y^3=(x-y)(x^2+xy+y^2)\)