🔬 Tutorial problems theta \theta

🔬 Tutorial problems theta \(\theta\)#

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.

\(\theta\).1#

Find the inverse matrix for the following matrix or show that it does not exist:

\[\begin{split} A=\left(\begin{array}{cc} 3 & -2 \\ 6 & 4 \end{array}\right) \end{split}\]

[Haeussler Jr and Paul, 1987] Section 8.6, Example 3, Part (b)

Note that

\[ \operatorname{det}(A)=(3)(4)-(-2)(6)=12-(-12)=12+12=24 \neq 0 \text {. } \]

Thus we know that \(A\) is non-singular, so that \(A^{-1}\) exists. We will use the adjoint matrix method to find it. Note that

\[\begin{split} \begin{array}{l} C_{11}=(-1)^{1+1} \operatorname{det}((4))=(-1)^{2}(4)=(1)(4)=4, \\ C_{12}=(-1)^{1+2} \operatorname{det}((6))=(-1)^{3}(6)=(-1)(6)=-6, \\ C_{21}=(-1)^{2+1} \operatorname{det}((-2))=(-1)^{3}(-2)=(-1)(-2)=2, \end{array} \end{split}\]

and

\[ C_{22}=(-1)^{2+2} \operatorname{det}((3))=(-1)^{4}(3)=(1)(3)=3 \text {. } \]

Hence the cofactor matrix associated with \(A\) is

\[\begin{split} C_{A}=\left(\begin{array}{cc} 4 & -6 \\ 2 & 3 \end{array}\right) \end{split}\]

This means that the adjoint matrix associated with \(A\) is

\[\begin{split} \operatorname{Adj}(A)=C_{A}^{T}=\left(\begin{array}{cc} 4 & 2 \\ -6 & 3 \end{array}\right) \end{split}\]

Thus the inverse matrix for \(A\) is

\[\begin{split} A^{-1}=\frac{1}{\operatorname{det}(A)} \operatorname{Adj}(A)=\frac{1}{24} A d j(A)=\left(\begin{array}{cc} \frac{1}{6} & \frac{1}{12} \\ \frac{-1}{4} & \frac{1}{8} \end{array}\right) . \end{split}\]

\(\theta\).2#

Find the inverse matrix for the following matrix or show that it does not exist:

\[\begin{split} B=\left(\begin{array}{ccc} 1 & -2 & 1 \\ 2 & -1 & 5 \\ 1 & 1 & 4 \end{array}\right) \text {. } \end{split}\]

[Haeussler Jr and Paul, 1987] Section 8.6, Example 5

Note that

\[\begin{split} \begin{array}{ll} \operatorname{det}(B) &= b_{11} C_{11}+b_{12} C_{12}+b_{13} C_{13} \\ &= 1 C_{11}+(-2) C_{12}+1 C_{13} \\ &= 1(-1)^{1+1}\left|\begin{array}{ll} 1 & 5 \\ 1 & 4 \end{array}\right|-2(1)(-1)^{1+2}\left|\begin{array}{ll} 2 & 5 \\ 1 & 4 \end{array}\right| +1(-1)^{1+3}\left|\begin{array}{ll} 2 & 5 \\ 1 & 4 \end{array}\right| \\ &= 1(-1)^{2}(-4-5)-2(-1)^{3}(8-5)+1(-1)^{4}(2+1) \\ &= 1(1)(-9)-2(-1)(3)+1(1)(3) \\ &= -9+6+3 \\ &= 0 . \end{array} \end{split}\]

Since the determinant of the matrix \(B\) is zero, we know that \(B\) is a singular matrix. This means that it is not invertible, so that \(B^{-1}\) does not exist.

\(\theta\).3#

Apply Gauss-Jordan elimination to an appropriate augmented row matrix to solve the following system of equations:

\[\begin{split} \left\{\begin{array}{ccccc} 3 x+3 y+6 z & = & 12 & \cdots & (\text { Equation } 1) \\ x-3 y+5 z & = & 5 & \cdots & (\text { Equation } 2) \\ 2 x+10 y-3 z & = & 0 & \cdots & (\text { Equation } 3) \end{array}\right\} . \end{split}\]

[Bradley, 2013] Progress Exercises 9.3, Question 1

Consider the following system of three linear equations in three unknown variables:

\[\begin{split} \left\{\begin{array}{cccl} 3 x+3 y+6 z & =12 & \cdots & (\text { Equation } 1) \\ x-3 y+5 z & =5 & \cdots & (\text { Equation } 2) \\ 2 x+10 y-3 z & =0 & \cdots & (\text { Equation } 3) \end{array}\right\} . \end{split}\]

This system of equations can be rewritten as

\[\begin{split} \left\{\begin{array}{ccccc} 3 x+3 y+6 z & = & 12 & \cdots & (\text { Equation } 1) \\ 1 x+(-3) y+5 z & = & 5 & \cdots & (\text { Equation } 2) \\ 2 x+10 y+(-3) z & = & 0 & \cdots & (\text { Equation } 3) \end{array}\right\} . \end{split}\]

The augmented row matrix representation of this system of equations is

\[\begin{split} (A \mid b)=\left(\begin{array}{ccc|c} 3 & 3 & 6 & 12 \\ 1 & -3 & 5 & 5 \\ 2 & 10 & -3 & 0 \end{array}\right) \end{split}\]

Note that

\[\begin{split} \begin{array}{ll} (A \mid b)&=\left(\begin{array}{ccc|c} 3 & 3 & 6 & 12 \\ 1 & -3 & 5 & 5 \\ 2 & 10 & -3 & 0 \end{array}\right) \begin{array}{l} R 1 \rightarrow R 2 \\ R 2 \rightarrow R 1 \\ \\ \end{array} \\ & \longrightarrow\left(\begin{array}{ccc|c} 1 & -3 & 5 & 5 \\ 3 & 3 & 6 & 12 \\ 2 & 10 & -3 & 0 \end{array}\right) \begin{array}{l} \\ R 2 \rightarrow R 2-3 R 1 \\ R 3 \rightarrow R 3-2 R 1 \end{array} \\ & \longrightarrow\left(\begin{array}{ccc|c} 1 & -3 & 5 & 5 \\ 0 & 12 & -9 & -3 \\ 0 & 16 & -13 & -10 \end{array}\right) \quad R 2 \rightarrow\left(\frac{1}{12}\right) R 2 \\ & \longrightarrow\left(\begin{array}{ccc|c} 1 & -3 & 5 & 5 \\ 0 & 1 & \frac{-3}{4} & \frac{-1}{4} \\ 0 & 16 & -13 & -10 \end{array}\right) \begin{array}{l} \\ \\ \quad R 3 \rightarrow R 3-(16) R 2 \\ \end{array} \\ & \longrightarrow\left(\begin{array}{ccc|c} 1 & -3 & 5 & 5 \\ 0 & 1 & \frac{-3}{4} & \frac{-1}{4} \\ 0 & 0 & -1 & -6 \end{array}\right) \begin{array}{l} \\ \\ \quad R 3 \rightarrow(-1) R 3 \\ \end{array} \\ & \longrightarrow\left(\begin{array}{ccc|c} 1 & -3 & 5 & 5 \\ 0 & 1 & \frac{-3}{4} & \frac{-1}{4} \\ 0 & 0 & 1 & 6 \end{array}\right) \begin{array}{c} R 1 \rightarrow R 1-5 R 3 \\ R 2 \rightarrow R 2+\left(\frac{3}{4}\right) R 3 \\ \\ \end{array} \\ & \longrightarrow\left(\begin{array}{ccc|c} 1 & -3 & 0 & -25 \\ 0 & 1 & 0 & \frac{17}{4} \\ 0 & 0 & 1 & 6 \end{array}\right) \begin{array}{c} \quad R 1 \rightarrow R 1+3 R 2 \\ \\ \\ \end{array} \\ & \longrightarrow\left(\begin{array}{ccc|c} 1 & 0 & 0 & \frac{-49}{4} \\ 0 & 1 & 0 & \frac{17}{4} \\ 0 & 0 & 1 & 6 \end{array}\right) \\ & =(I \mid c) . \end{array} \end{split}\]

Thus we can conclude that the unique solution to this system of equations is

\[\begin{split} \left\{\begin{array}{ccc} x & = & \frac{-49}{4} \\ y & = & \frac{17}{4} \\ z & =& 6 \end{array}\right\} . \end{split}\]

\(\theta\).4#

This problem is a part of an application of a linear version of a three market Marshallian cross model.

Solve the following system of linear equations:

\[\begin{split} \left\{\begin{array}{ccccc} P_{1}+4 P_{2}+8 P_{3} & = &26 & \cdots & (\text { Equation } 1) \\ 5 P_{1}+7 P_{2} & = &38 & \cdots & (\text { Equation } 2) \\ 8 P_{1}+12 P_{2}+2 P_{3} & = & 66 & \cdots & (\text { Equation } 3) \end{array}\right\} . \end{split}\]

[Bradley, 2013] Progress Exercises 9.3, Question 4

Consider the following system of three linear equations in three unknown variables:

\[\begin{split} \left\{\begin{array}{ccccc} P_{1}+4 P_{2}+8 P_{3} & = & 26 & \cdots & (\text { Equation } 1) \\ 5 P_{1}+7 P_{2} & = & 38 & \cdots & (\text { Equation } 2) \\ 8 P_{1}+12 P_{2}+2 P_{3} & = & 66 & \cdots & (\text { Equation } 3) \end{array}\right\} . \end{split}\]

This system of equations can be rewritten as

\[\begin{split} \left\{\begin{array}{cccl} 1 P_{1}+4 P_{2}+8 P_{3} & =26 & \cdots & (\text { Equation } 1) \\ 5 P_{1}+7 P_{2}+0 P_{3} & =38 & \cdots & (\text { Equation } 2) \\ 8 P_{1}+12 P_{2}+2 P_{3} & =66 & \cdots & (\text { Equation } 3) \end{array}\right\} . \end{split}\]

The augmented row matrix representation of this system of equations is

\[\begin{split} (A \mid b)=\left(\begin{array}{ccc|c} 1 & 4 & 8 & 26 \\ 5 & 7 & 0 & 38 \\ 8 & 12 & 2 & 66 \end{array}\right) \end{split}\]

Note that

\[\begin{split} \begin{array}{ll} (A \mid b) &=\left(\begin{array}{ccc|c} 1 & 4 & 8 & 26 \\ 5 & 7 & 0 & 38 \\ 8 & 12 & 2 & 66 \end{array}\right) \begin{array}{l} R 2 \rightarrow R 2-5 R 1 \\ R 3 \rightarrow R 3-8 R 1 \end{array} \\ & \longrightarrow\left(\begin{array}{ccc|c} 1 & 4 & 8 & 26 \\ 0 & -13 & -40 & -92 \\ 0 & -20 & -62 & -142 \end{array}\right) \quad R 2 \rightarrow\left(\frac{-1}{13}\right) R 2 \\ & \longrightarrow\left(\begin{array}{ccc|c} 1 & 4 & 8 & 26 \\ 0 & 1 & \frac{40}{13} & \frac{92}{13} \\ 0 & -20 & -62 & -142 \end{array}\right) \quad R 3 \rightarrow R 3+(20) R 2 \\ & \longrightarrow\left(\begin{array}{ccc|c} 1 & 4 & 8 & 26 \\ 0 & 1 & \frac{40}{13} & \frac{92}{13} \\ 0 & 0 & \frac{-6}{13} & \frac{-6}{13} \end{array}\right) \quad R 3 \rightarrow\left(\frac{-13}{6}\right) R 3 \\ & \longrightarrow\left(\begin{array}{ccc|c} 1 & 4 & 8 & 26 \\ 0 & 1 & \frac{40}{13} & \frac{92}{13} \\ 0 & 0 & 1 & 1 \end{array}\right) \begin{array}{c} R 1 \rightarrow R 1-8 R 3 \\ R 2 \rightarrow R 2-\left(\frac{40}{13}\right) R 3 \end{array} \\ & \longrightarrow\left(\begin{array}{ccc|c} 1 & 4 & 0 & 18 \\ 0 & 1 & 0 & \frac{52}{13} \\ 0 & 0 & 1 & 1 \end{array}\right) \quad R 1 \rightarrow R 1-4 R 2 \\ & \longrightarrow\left(\begin{array}{lll|c} 1 & 0 & 0 & \frac{26}{13} \\ 0 & 1 & 0 & \frac{52}{13} \\ 0 & 0 & 1 & 1 \end{array}\right) \\ & =\left(\begin{array}{lll|l} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 1 \end{array}\right) \\ & =(I \mid c) . \end{array} \end{split}\]

Thus we can conclude that the unique solution to this system of equations is

\[\begin{split} \left\{\begin{array}{l} P_{1}=2 \\ P_{2}=4 \\ P_{3}=1 \end{array}\right\} . \end{split}\]

\(\theta\).5#

This problem is a linear version of a three market Marshallian Cross model.

Assuming that all three markets are in equilibrium, solve the following system of linear equations.

\[\begin{split} \left\{\begin{array}{cccccc} Q^D_1 & = & 50 - 2P_1 + 5P_2 - 3P_3 & & & \cdots(D1) \\ Q^S_1 & = & 8P_1 - 5 & & & \cdots(S1) \\ Q^D_1 & = & Q^S_1 & = & Q_1 & \cdots(E1) \\ Q^D_2 & = & 22 + 7P_1 - 2P_2 + 5P_3 & & & \cdots(D2) \\ Q^S_2 & = & 12P_2 - 5 & & & \cdots(S2) \\ Q^D_2 & = & Q^S_2 & = & Q_2 & \cdots(E2) \\ Q^D_3 & = & 17 + P_1 + 5P_2 - 3P_3 & & & \cdots(D3) \\ Q^S_3 & = & 4P_3 - 1 & & & \cdots(S3) \\ Q^D_3 & = & Q^S_3 & = & Q_3 & \cdots(E3) \\ \end{array}\right\} . \end{split}\]

Hint: There are two ways that you might proceed here.

You could use the equilibrium (market clearing) conditions for the three markets to construct a system of six linear equations in six unknown variables. In this approach, the unknown variables are the three equilibrium quantities \(\left(Q_{1}, Q_{2}\right.\) and \(\left.Q_{3}\right)\) and the three equilibrium prices \(\left(P_{1}, P_{2}\right.\) and \(P_{3}\) ).

Alternatively, you could use use the equilibrium (market clearing) conditions for the three markets to construct a system of three linear equations in three unknown variables. In this case, the unknown variables are the three equilibrium prices \(\left(P_{1}, P_{2}\right.\) and \(\left.P_{3}\right)\). If you employ this approach, you will need to use some of the original equations and the equilibrium values for the prices to obtain the three equilibrium quantities \(\left(Q_{1}, Q_{2}\right.\) and \(\left.Q_{3}\right)\).

[Bradley, 2013] Progress Exercises 9.3, Question 8

Approach One

Consider the following three market linear Marshallian cross model:

\[\begin{split} \left\{\begin{array}{cccccc} Q^D_1 & = & 50 - 2P_1 + 5P_2 - 3P_3 & & & \cdots(D1) \\ Q^S_1 & = & 8P_1 - 5 & & & \cdots(S1) \\ Q^D_1 & = & Q^S_1 & = & Q_1 & \cdots(E1) \\ Q^D_2 & = & 22 + 7P_1 - 2P_2 + 5P_3 & & & \cdots(D2) \\ Q^S_2 & = & 12P_2 - 5 & & & \cdots(S2) \\ Q^D_2 & = & Q^S_2 & = & Q_2 & \cdots(E2) \\ Q^D_3 & = & 17 + P_1 + 5P_2 - 3P_3 & & & \cdots(D3) \\ Q^S_3 & = & 4P_3 - 1 & & & \cdots(S3) \\ Q^D_3 & = & Q^S_3 & = & Q_3 & \cdots(E3) \\ \end{array}\right\} . \end{split}\]

Using the three equilibrium (market clearing) conditions, we can rewrite this model as a system of six linear equations in six unknown variables as follows:

\[\begin{split} \left\{\begin{array}{cccc} Q_{1}+2 P_{1}-5 P_{2}+3 P_{3} & = & 50 & \cdots\left(D 1^{\prime}\right) \\ Q_{1}-8 P_{1} & = & -5 & \cdots\left(S 1^{\prime}\right) \\ Q_{2}-7 P_{1}+2 P_{2}-5 P_{3} & = & 22 & \cdots\left(D 2^{\prime}\right) \\ Q_{2}-12 P_{2} & = & -5 & \cdots\left(S 2^{\prime}\right) \\ Q_{3}-P_{1}-5 P_{2}+3 P_{3} & = & 17 & \cdots\left(D 3^{\prime}\right) \\ Q_{3}-4 P_{3} & = & -1 & \cdots\left(S 3^{\prime}\right) \end{array}\right\} . \end{split}\]

The augmented row matrix representation of this system is

\[\begin{split} (A \mid b)=\left(\begin{array}{cccccc|c} 1 & 0 & 0 & 2 & -5 & 3 & 50 \\ 1 & 0 & 0 & -8 & 0 & 0 & -5 \\ 0 & 1 & 0 & -7 & 2 & -5 & 22 \\ 0 & 1 & 0 & 0 & -12 & 0 & -5 \\ 0 & 0 & 1 & -1 & -5 & 3 & 17 \\ 0 & 0 & 1 & 0 & 0 & -4 & -1 \end{array}\right) \end{split}\]

Note that

\[\begin{split} \begin{array}{ll} (A \mid b)&=\left(\begin{array}{cccccc|c} 1 & 0 & 0 & 2 & -5 & 3 & 50 \\ 1 & 0 & 0 & -8 & 0 & 0 & -5 \\ 0 & 1 & 0 & -7 & 2 & -5 & 22 \\ 0 & 1 & 0 & 0 & -12 & 0 & -5 \\ 0 & 0 & 1 & -1 & -5 & 3 & 17 \\ 0 & 0 & 1 & 0 & 0 & -4 & -1 \end{array}\right) \begin{array}{l} R 1 \rightarrow R 4 \\ R 2 \rightarrow R 1 \\ R 3 \rightarrow R 5 \\ R 4 \rightarrow R 2 \\ R 5 \rightarrow R 6 \\ R 6 \rightarrow R 3 \end{array} \\ & \longrightarrow\left(\begin{array}{cccccc|c} 1 & 0 & 0 & -8 & 0 & 0 & -5 \\ 0 & 1 & 0 & 0 & -12 & 0 & -5 \\ 0 & 0 & 1 & 0 & 0 & -4 & -1 \\ 1 & 0 & 0 & 2 & -5 & 3 & 50 \\ 0 & 1 & 0 & -7 & 2 & -5 & 22 \\ 0 & 0 & 1 & -1 & -5 & 3 & 17 \end{array}\right) \begin{array}{l} \\ \\ \\ R 4 \rightarrow R 4-R 1 \\ R 5 \rightarrow R 5-R 2 \\ R 6 \rightarrow R 6-R 3 \end{array} \\ & \longrightarrow\left(\begin{array}{cccccc|c} 1 & 0 & 0 & -8 & 0 & 0 & -5 \\ 0 & 1 & 0 & 0 & -12 & 0 & -5 \\ 0 & 0 & 1 & 0 & 0 & -4 & -1 \\ 0 & 0 & 0 & 10 & -5 & 3 & 55 \\ 0 & 0 & 0 & -7 & 14 & -5 & 27 \\ 0 & 0 & 0 & -1 & -5 & 7 & 18 \end{array}\right) \begin{array}{l} \\ \\ \\ R 4 \rightarrow R 6 \\ \\ R 6 \rightarrow R 4 \end{array} \\ & \longrightarrow\left(\begin{array}{cccccc|c} 1 & 0 & 0 & -8 & 0 & 0 & -5 \\ 0 & 1 & 0 & 0 & -12 & 0 & -5 \\ 0 & 0 & 1 & 0 & 0 & -4 & -1 \\ 0 & 0 & 0 & -1 & -5 & 7 & 18 \\ 0 & 0 & 0 & -7 & 14 & -5 & 27 \\ 0 & 0 & 0 & 10 & -5 & 3 & 55 \end{array}\right) \begin{array}{c} \\ \\ \\ R 4 \rightarrow(-1) R 4 \\ R 5 \rightarrow R 5-7 R 4 \\ R 6 \rightarrow R 6+(10) R 4 \end{array} \\ & \longrightarrow\left(\begin{array}{cccccc|c} 1 & 0 & 0 & -8 & 0 & 0 & -5 \\ 0 & 1 & 0 & 0 & -12 & 0 & -5 \\ 0 & 0 & 1 & 0 & 0 & -4 & -1 \\ 0 & 0 & 0 & 1 & 5 & -7 & -18 \\ 0 & 0 & 0 & 0 & 49 & -54 & -99 \\ 0 & 0 & 0 & 0 & -55 & 73 & 235 \end{array}\right) \begin{array}{l} \\ \\ \\ \\ R 5 \rightarrow\left(\frac{1}{49}\right) R 5 \\ \end{array} \\ & \longrightarrow\left(\begin{array}{cccccc|c} 1 & 0 & 0 & -8 & 0 & 0 & -5 \\ 0 & 1 & 0 & 0 & -12 & 0 & -5 \\ 0 & 0 & 1 & 0 & 0 & -4 & -1 \\ 0 & 0 & 0 & 1 & 5 & -7 & -18 \\ 0 & 0 & 0 & 0 & 1 & \frac{-54}{49} & \frac{-99}{49} \\ 0 & 0 & 0 & 0 & -55 & 73 & 235 \end{array}\right) \begin{array}{l} \\ \\ \\ \\ \\ R 6 \rightarrow R 6+(55) R 5 \end{array} \end{array} \end{split}\]
\[\begin{split} \begin{array}{ll} & \longrightarrow\left(\begin{array}{cccccc|c} 1 & 0 & 0 & -8 & 0 & 0 & -5 \\ 0 & 1 & 0 & 0 & -12 & 0 & -5 \\ 0 & 0 & 1 & 0 & 0 & -4 & -1 \\ 0 & 0 & 0 & 1 & 5 & -7 & -18 \\ 0 & 0 & 0 & 0 & 1 & \frac{-54}{49} & \frac{-99}{49} \\ 0 & 0 & 0 & 0 & 0 & \frac{607}{49} & \frac{6,070}{49} \end{array}\right) \begin{array}{l} \\ \\ \\ \\ \\ R 6 \rightarrow\left(\frac{49}{607}\right) R 6 \end{array} \\ & \longrightarrow\left(\begin{array}{cccccc|c} 1 & 0 & 0 & -8 & 0 & 0 & -5 \\ 0 & 1 & 0 & 0 & -12 & 0 & -5 \\ 0 & 0 & 1 & 0 & 0 & -4 & -1 \\ 0 & 0 & 0 & 1 & 5 & -7 & -18 \\ 0 & 0 & 0 & 0 & 1 & \frac{-54}{49} & \frac{-99}{49} \\ 0 & 0 & 0 & 0 & 0 & 1 & 10 \end{array}\right) \\ & \longrightarrow\left(\begin{array}{cccccc|c} 1 & 0 & 0 & -8 & 0 & 0 & -5 \\ 0 & 1 & 0 & 0 & -12 & 0 & -5 \\ 0 & 0 & 1 & 0 & 0 & -4 & -1 \\ 0 & 0 & 0 & 1 & 5 & -7 & -18 \\ 0 & 0 & 0 & 0 & 1 & \frac{-54}{49} & \frac{-99}{49} \\ 0 & 0 & 0 & 0 & 0 & 1 & 10 \end{array}\right) \begin{array}{c} \\ \\ R 3 \rightarrow R 3+4 R 6 \\ R 4 \rightarrow R 4 + 7 R 6 \\ R 5 \rightarrow R 5+\left(\frac{54}{49}\right) R 6 \\ \end{array} \\ & \longrightarrow\left(\begin{array}{cccccc|c} 1 & 0 & 0 & -8 & 0 & 0 & -5 \\ 0 & 1 & 0 & 0 & -12 & 0 & -5 \\ 0 & 0 & 1 & 0 & 0 & 0 & 39 \\ 0 & 0 & 0 & 1 & 5 & 0 & 52 \\ 0 & 0 & 0 & 0 & 1 & 0 & 9 \\ 0 & 0 & 0 & 0 & 0 & 1 & 10 \end{array}\right) \begin{array}{c} \\ R 2 \rightarrow R 2+(12) R 5 \\ \\ R 4 \rightarrow R 4 - 5 R 5 \\ \\ \end{array} \\ & \longrightarrow\left(\begin{array}{cccccc|c} 1 & 0 & 0 & -8 & 0 & 0 & -5 \\ 0 & 1 & 0 & 0 & 0 & 0 & 103 \\ 0 & 0 & 1 & 0 & 0 & 0 & 39 \\ 0 & 0 & 0 & 1 & 0 & 0 & 7 \\ 0 & 0 & 0 & 0 & 1 & 0 & 9 \\ 0 & 0 & 0 & 0 & 0 & 1 & 10 \end{array}\right) \begin{array}{c} R 1 \rightarrow R1 + 8 R 4 \\ \\ \\ \\ \\ \end{array} \\ & \longrightarrow\left(\begin{array}{cccccc|c} 1 & 0 & 0 & 0 & 0 & 0 & 51 \\ 0 & 1 & 0 & 0 & 0 & 0 & 103 \\ 0 & 0 & 1 & 0 & 0 & 0 & 39 \\ 0 & 0 & 0 & 1 & 0 & 0 & 7 \\ 0 & 0 & 0 & 0 & 1 & 0 & 9 \\ 0 & 0 & 0 & 0 & 0 & 1 & 10 \end{array}\right) \\ & =(I \mid c) . \end{array} \end{split}\]

Thus we can conclude that the equilibrium quantities and prices are

\[\begin{split} \begin{array}{c} Q_{1}=51 \text { units, } \\ Q_{2}=103 \text { units, } \\ Q_{3}=39 \text { units, } \\ P_{1}=\$ 7.00 \text { per unit, } \\ P_{2}=\$ 9.00 \text { per unit, } \\ P_{3}=\$ 10.00 \text { per unit. } \end{array} \end{split}\]

Approach Two

Note that

\(Q_{1}^{D}=Q_{1}^{S} \Longleftrightarrow 50-2 P_{1}+5 P_{2}-3 P_{3}=8 P_{1}-5 \Longleftrightarrow 10 P_{1}-5 P_{2}+3 P_{3}=55\),

\(Q_{2}^{D}=Q_{2}^{S} \Longleftrightarrow 22+7 P_{1}-2 P_{2}+5 P_{3}=12 P_{2}-5 \Longleftrightarrow-7 P_{1}+14 P_{2}-5 P_{3}=27\), and

\(Q_{3}^{D}=Q_{3}^{S} \Longleftrightarrow 17+1 P_{1}+5 P_{2}-3 P_{3}=4 P_{3}-1 \Longleftrightarrow-1 P_{1}-5 P_{2}+7 P_{3}=18\).

Thus we have the following system of three linear equations in three unknown variables:

Equation (1): \(10 P_{1}-5 P_{2}+3 P_{3}=55\),

Equation (2): \(-7 P_{1}+14 P_{2}-5 P_{3}=27\),

and

Equation (3): \(-1 P_{1}-5 P_{2}+7 P_{3}=18\).

The augmented row-matrix representation of this three equation system is

\[\begin{split} (A \mid b)=\left(\begin{array}{ccc|c} 10 & -5 & 3 & 55 \\ -7 & 14 & -5 & 27 \\ -1 & -5 & 7 & 18 \end{array}\right) \end{split}\]

We now apply the Gauss-Jordan elimination procedure to this augmented row matrix.

\[\begin{split} \begin{array}{ll} (A \mid b) &\rightarrow\left(\begin{array}{ccc|c} 10 & -5 & 3 & 55 \\ -7 & 14 & -5 & 27 \\ -1 & -5 & 7 & 18 \end{array}\right) \begin{array}{c} R 1 \rightarrow R 1+(R 2+2 R 3) \\ \\ \\ \end{array} \\ & \rightarrow\left(\begin{array}{ccc|c} 1 & -1 & 12 & 118 \\ -7 & 14 & -5 & 27 \\ -1 & -5 & 7 & 18 \end{array}\right) \begin{array}{c} \\ R 2 \rightarrow R 2+7 R 1 \\ R 3 \rightarrow R 3+R 1 \end{array} \\ & \rightarrow\left(\begin{array}{ccc|c} 1 & -1 & 12 & 118 \\ 0 & 7 & 79 & 853 \\ 0 & -6 & 19 & 136 \end{array}\right) \quad R 2 \rightarrow R 2+R 3 \\ & \rightarrow\left(\begin{array}{ccc|c} 1 & -1 & 12 & 118 \\ 0 & 1 & 98 & 989 \\ 0 & -6 & 19 & 136 \end{array}\right) \begin{array}{c} \\ \\ R 3 \rightarrow R 3+6 R 2 \end{array} \\ & \rightarrow\left(\begin{array}{ccc|c} 1 & -1 & 12 & 118 \\ 0 & 1 & 98 & 989 \\ 0 & 0 & 607 & 6,070 \end{array}\right) \begin{array}{c} \\ \\ R 3 \rightarrow R 3 \times\left(\frac{1}{607}\right) \end{array} \\ & \begin{array}{l} \rightarrow\left(\begin{array}{ccc|c} 1 & -1 & 12 & 118 \\ 0 & 1 & 98 & 989 \\ 0 & 0 & 1 & \frac{6,070}{607} \end{array}\right) \end{array} \begin{array}{l} \\ \\ \text { Simplify the fraction in the constant column. } \end{array} \\ &\rightarrow\left(\begin{array}{ccc|c} 1 & -1 & 12 & 118 \\ 0 & 1 & 98 & 989 \\ 0 & 0 & 1 & 10 \end{array}\right) \begin{array}{l} R 1 \rightarrow R 1-12 R 3 \\ R 2 \rightarrow R 2-98 R 3 \\ \\ \end{array} \\ & \rightarrow\left(\begin{array}{ccc|c} 1 & -1 & 0 & -2 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & 10 \end{array}\right) \begin{array}{l} R 1 \rightarrow R 1+R 2 \\ \\ \\ \end{array} \\ & \rightarrow\left(\begin{array}{ccc|c} 1 & 0 & 0 & 7 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & 10 \end{array}\right) \\ & \rightarrow \quad(I \mid c) . \end{array} \end{split}\]

Thus we can conclude that the equilibrium prices are \(P_{1}^{*}=\$ 7\) per unit, \(P_{2}^{*}=\$ 9\) per unit, and \(P_{3}^{*}=\$ 10\) per unit.

Upon substituting \(P_{1}=\$ 7\) into the supply equation for commodity one, we obtain

\[ Q_{1}^{*}=8 P_{1}^{*}-5=8(7)-5=56-5=51 \text { units. } \]

Upon substituting \(P_{2}=\$ 9\) into the supply equation for commodity two, we obtain

\[ Q_{2}^{*}=12 P_{2}^{*}-5=12(9)-5=108-5=103 \text { units. } \]

Upon substituting \(P_{3}=\$ 10\) into the supply equation for commodity three, we obtain

\[ Q_{3}^{*}=4 P_{3}^{*}-3=4(10)-1=40-1=39 \text { units. } \]

Thus we can conclude that the equilibrium quantities are \(Q_{1}^{*}=51\) units, \(Q_{2}^{*}=103\) units, and \(Q_{3}^{*}=39\) units.

\(\theta\).6#

This problem is an example of the Keynesian cross model.

Use Cramer’s rule to solve the following system of linear equations:

\[\begin{split} \left\{\begin{array}{lll} Y=C+I_{0} & \cdots(E 1) \\ C=C_{0}+b Y & \cdots(C 1) \end{array}\right\} \end{split}\]

[Bradley, 2013] Progress Exercises 9.4, Question 9

Consider the following Keynesian cross model of a closed economy in which there is no government sector:

\[\begin{split} \left\{\begin{array}{lll} Y & =C+I_{0} & \cdots(E 1) \\ C & =C_{0}+b Y & \cdots(C 1) \end{array}\right\} \end{split}\]

The equilibrium condition is given by equation (E1). The consumption function is given by equation (C1). Note that \(b, C_{0}\) and \(I_{0}\) are exogenous parameters. Note also that \(0<b<1\). Recall that \(b\) is the marginal propensity to consume, \(C_{0}\) is autonomous consumption and \(I_{0}\) is autonomous investment. This model can rewritten as

\[\begin{split} \left\{\begin{array}{cccc} Y-C & = & I_{0} & \cdots\left(E 1^{\prime}\right) \\ -b Y+C & = & C_{0} & \cdots\left(C 1^{\prime}\right) \end{array}\right\} \end{split}\]

This system of two linear equations in two unknown variables can be expressed as the matrix equation

\[ A x=b, \]

where

\[\begin{split} A=\left(\begin{array}{cc} 1 & -1 \\ -b & 1 \end{array}\right), \\ x=\left(\begin{array}{l} Y \\ C \end{array}\right) \end{split}\]

and

\[\begin{split} b=\left(\begin{array}{c} I_{0} \\ C_{0} \end{array}\right) \end{split}\]

Note that

\[\begin{split} \bar{A}_{Y}=\left(\begin{array}{cc} I_{0} & -1 \\ C_{0} & 1 \end{array}\right) \end{split}\]

and

\[\begin{split} \bar{A}_{C}=\left(\begin{array}{cc} 1 & I_{0} \\ -b & C_{0} \end{array}\right) \end{split}\]

We have

\[\begin{split} \begin{array}{ll} \operatorname{det}(A) & =(1)(1)-(-1)(-b) \\ & =1-b \\ & >0, \\ \\ \operatorname{det}\left(\bar{A}_{Y}\right) & =\left(I_{0}\right)(1)-(-1)\left(C_{0}\right) \\ & =I_{0}+C_{0} \end{array} \end{split}\]

and

\[\begin{split} \begin{array}{ll} \operatorname{det}\left(\bar{A}_{C}\right) & =(1)\left(C_{0}\right)-\left(I_{0}\right)(-b) \\ & =C_{0}+b I_{0} . \end{array} \end{split}\]

Thus we have

\[ Y=\frac{\operatorname{det}\left(\bar{A}_{Y}\right)}{\operatorname{det}(A)}=\left(\frac{I_{0}+C_{0}}{1-b}\right) \]

and

\[ C=\frac{\operatorname{det}\left(\bar{A}_{C}\right)}{\operatorname{det}(A)}=\left(\frac{C_{0}+b I_{0}}{1-b}\right) . \]

\(\theta\).7#

Employ Gauss-Jordan elimination to determine the possible solutions of the following system of linear equations for different values of \(a\) and \(b\) :

\[\begin{split} \left\{\begin{array}{ccccc} x+y-z & = & 1 & \cdots & (\text { Equation } 1) \\ x-y+2 z & = & 2 & \cdots & (\text { Equation } 2) \\ x+2 y-a z & = & b & \cdots & (\text { Equation } 3) \end{array}\right\} . \end{split}\]

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Section 15.6, Problem 2

This question is left as an exercise.

\(\theta\).8#

Consider the matrix equation \(A x=b\) where

\[\begin{split} A=\left(\begin{array}{ccc} 0 & -p_{1} & -p_{2} \\ -p_{1} & U_{11} & U_{12} \\ -p_{2} & U_{21} & U_{22} \end{array}\right) \end{split}\]

\(x=\left(\frac{\partial \lambda}{\partial p_{1}}, \frac{\partial x_{1}}{\partial p_{1}}, \frac{\partial x_{2}}{\partial p_{1}}\right)^{T}\) and \(b=\left(x_{1}, \lambda, 0\right)^{T}\). This equation might have been derived by conducting a comparative static exercise on the first-order conditions of a consumer’s budget-constrained utility maximisation problem. Use Cramer’s rule to find the comparative static effects \(\frac{\partial x_{1}}{\partial p_{1}}\) and \(\frac{\partial x_{2}}{\partial p_{1}}\).

This question is left as an exercise.

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