🔬 Tutorial problems eta \eta

🔬 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

  • \(x^2-y^2=(x-y)(x+y)\)

  • \(x^3-y^3=(x-y)(x^2+xy+y^2)\)