🔬 Tutorial problems 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 are for additional practice. The symbol 🍹 indicates additional problems.

\(\eta\).1

Formulation

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

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

This question comes from Haeussler and Paul (1987, pp. 281-282, Section 8.6, Example 3, Part (b))

Solution

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

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

\(\eta\).2

Formulation

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

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

This question comes from Haeussler and Paul (1987, pp. 283-284, Section 8.6, Example 5).

Solution

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.

\(\eta\).3

Formulation

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

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

This question comes from Bradley (2008, p. 501, Progress Exercises 9.3, Question 1).

Solution

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

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

\(\eta\).4

Formulation

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:

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

This example comes from Bradley (2008, p. 502, Progress Exercises 9.3, Question 4).

Solution

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

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

\(\eta\).5

Formulation

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 \(\left.Q_{3}\right)\).

This example comes from Bradley (2008, p. 502, Progress Exercises 9.3, Question 8).

Solution

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.

\(\eta\).6

Formulation

This problem is an example of the Keynesian cross model.

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

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

This example comes from Bradley (2008, p. 510, Progress Exercises 9.4, Question 9).

Solution

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

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

\(\eta\).7 🍹

Formulation

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

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

This question comes from Sydsaeter and Hammond (2006, p. 575, Section 15.6, Problem 2).

Solution

This question is left as an exercise.

\(\eta\).8 🍹

Formulation

This question is an example of the IS-LM macroeconomic model of an economy. The ISLM model consists of two equations. The first equation is the IS equation. It describes the locus of equilibrium points in the goods market. The “IS” term comes from one version of the goods market equilibrium condition, in which investment (\(I\)) must equal savings \((S)\). The second equation is the LM equation. It describes the locus of equilibrium points in the money market. The “LM” term comes from the money market equilibrium condition, in which money demand (\(L\)) must equal money supply (\(M\)). The letter \(L\) stands for liquidity preference, which is one potential source of the demand for money. Note that there is implicitly a third market in this model. This is the bond market. It does not need to be formally considered because Walras’ law ensures that if all but one of the markets in an economy clear, then the remaining market will also clear.

The bond market is implicitly present because there is an opportunity cost involved with holding money balances. This opportunity cost takes the form of the foregone interest which could have been earned if those money balances had been used to purchase bonds. If both the goods market and the money market are simultaneously in equilibrium, then the economy as a whole is in “general equilibrium”. (Further information on the IS-LM model at the intermediate (second year undergraduate) level can be found in Levacic and Rebman (1982), Parkin and Bade (1990), and Wells (1995).)

Assuming that the economy is in a situation of general equilibrium, obtain the equations for the IS curve and LM curve from the equations below and use Cramer’s rule to solve that system of two linear equations in two unknowns to find the equilibrium values of \(Y\) and \(r\). The equations underlying this model are:

Goods Market:

\[\begin{split} \left\{\begin{array}{cccc} Y^{D} & = & C+I & \cdots(G D) \\ C & = & 237.8+0.2 Y & \cdots(C D) \\ I & = & 10-0.4 r & \cdots(I D) \\ Y^{S} & = & Y & \text { (Keynesian GS) } \\ Y^{D} & = & Y^{S} & \text { (GMEquil) } \end{array}\right\} \text{; and} \end{split}\]

Money Market:

\[\begin{split} \left\{\begin{array}{cccc} M^{D} & = & 100+0.1 Y-0.3 r & \cdots(M D) \\ M^{S} & = & 129.225 & (M S) \\ M^{D} & = & M^{S} & (M M E q u i l) \end{array}\right\}. \end{split}\]

This example comes from Bradley (2008, p. 510, Progress Exercises 9.4, Question 10).

Solution

Suppose that the goods market is described by the following equations:

\[\begin{split} \left\{\begin{array}{cccc} Y^{D} & = & C+I & \cdots(G D) \\ C & = & 237.8+0.2 Y & \cdots(C D) \\ I & = & 10-0.4 r & \cdots(I D) \\ Y^{S} & = & Y & (\text {Keynesian } G S) \\ Y^{D} & = & Y^{S} & (G M E) \end{array}\right\} \end{split}\]

Note that

\[\begin{split} \begin{array}{lccc} & Y^{S} & = & Y^{D} \\ \Longleftrightarrow & Y & = & C+I \\ \Longleftrightarrow & Y & = & (237.8+0.2 Y)+(10-0.4 r) \\ \Longleftrightarrow & Y & = & 237.8+0.2 Y+10-0.4 r \\ \Longleftrightarrow & Y & = & 247.8+0.2 Y-0.4 r \\ \Longleftrightarrow & (1-0.2) Y+0.4 r & = & 247.8 \\ \Longleftrightarrow & 0.8 Y+0.4 r & = & 247.8 . \end{array} \end{split}\]

Thus the equation of the IS curve for this economy is

\[ 0.8 Y+0.4 r=247.8 \]

Suppose that the money market is described by the following equations:

\[\begin{split} \left\{ \begin{array}{cccc} M^{D} & = & 100+0.1 Y-0.3 r & \cdots(M D) \\ M^{S} & = & 129.225 & (M S) \\ M^{D} & = & M^{S} & (M M E) \end{array} \right\} \end{split}\]

Note that

\[\begin{split} \begin{array}{cccc} & M^{D} & = & M^{S} \\ \Longleftrightarrow & 100+0.1 Y-0.3 r & = & 129.225 \\ \Longleftrightarrow & 0.1 Y-0.3 r &= &29.225 . \end{array} \end{split}\]

Thus the equation of the LM curve for this economy is

\[ 0.1 Y-0.3 r=29.225 \]

The entire economy is in equilibrium when both the goods market and the money market are in equilibrium. This will occur at points where the IS curve and the LM curve intersect. As such, the equilibrium values of income (or output) and the interest rate are given by the solution to the following system of two linear equations in these two unknown variables:

\[\begin{split} \left\{ \begin{array}{cccc} 0.8 Y+0.4 r & = & 247.8 & ...(IS) \\ 0.1 Y-0.3 r & = & 29.225 & ...(LM) \end{array} \right\} \end{split}\]

This system of equations can be represented as a matrix equation of the form

\[ A x=b \]

where

\[\begin{split} \begin{array}{ll} A &=\left(\begin{array}{cc} 0.8 & 0.4 \\ 0.1 & -0.3 \end{array}\right), \\ x &=\left(\begin{array}{c} Y \\ r \end{array}\right) \end{array} \end{split}\]

and

\[\begin{split} b=\left(\begin{array}{c} 247.8 \\ 29.225 \end{array}\right) \end{split}\]

Note that

\[\begin{split} \bar{A}_{Y}=\left(\begin{array}{cc} 247.8 & 0.4 \\ 29.225 & -0.3 \end{array}\right) \end{split}\]

and

\[\begin{split} \bar{A}_{r}=\left(\begin{array}{cc} 0.8 & 247.8 \\ 0.1 & 29.225 \end{array}\right) . \end{split}\]

We have

\[\begin{split} \begin{array}{ll} \operatorname{det}(A) & =(0.8)(-0.3)-(0.4)(0.1) \\ & =-0.24-0.04 \\ & =-0.28 \\ & \neq 0, \\ \operatorname{det}\left(\bar{A}_{Y}\right) & =(247.8)(-0.3)-(0.4)(29.225) \\ & =-74.34-11.69 \\ & =-86.03 \end{array} \end{split}\]

and

\[\begin{split} \begin{array}{ll} \operatorname{det}\left(\bar{A}_{r}\right) & =(0.8)(29.225)-(247.8)(0.1) \\ & =23.38-24.78 \\ & =-1.4 \end{array} \end{split}\]

Thus we have

\[ Y=\frac{\operatorname{det}\left(\bar{A}_{Y}\right)}{\operatorname{det}(A)}=\frac{-86.03}{-0.28}=\frac{86.03}{0.28} \approx 307.25 \]

and

\[ r=\frac{\operatorname{det}\left(\bar{A}_{r}\right)}{\operatorname{det}(A)}=\frac{-1.4}{-0.28}=\frac{1.4}{0.28}=5 . \]

\(\eta\).9 🍹

Formulation

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 \(\frac{\partial x_{2}}{\partial p_{1}}\).

Solution

This question is left as an exercise.