Solutions

Exercise 1.1 1. Given: [pic] Find: (a) [pic] [pic]
(b) [pic] [pic] (c) [pic] [pic] (d) [pic] [pic] (e) [pic] (where [pic] denotes the transpose of [pic]) [pic] (f) [pic] [pic] (g) [pic] [pic]
2. Given: [pic] Find [pic] and [pic] [pic] [pic] (Note, a scalar) 3. Given: [pic] Find (if they exist) [pic], [pic], [pic], [pic], [pic], and [pic]. In each case, explain why the matrix product does (or does not) exist. [pic] [pic] does not exist because number of elements in a row of AT (=1)
does not equal number of elements in a column of B (=3). (Or,
equivalently, because number of columns in AT (=1) does not equal
number of rows in B (=3) [pic] [pic] does not exist because number of elements in a row of BT (=3)
does not equal number of elements in a column of A (=1). (Or,
equivalently, because number of columns in BT (=3) does not equal
number of rows in A (=1) [pic] (Note, same as BA) [pic] (Note, same as AB)
4. Given: [pic] Find (if they exist) [pic], [pic], [pic], [pic], [pic], and [pic]. AB does not exist because number of elements in a row of A (=2) does
not equal number of elements in a column of B (=3) [pic][pic] Exercise 1.2
1. Find the inverses (if they exist) of: [pic] The inverses are: [pic]
C-1 does not exist, because each element of row 2 equals 3 times the
corresponding element of row 1. (Equally, each element of column 2
equals 4 times the corresponding element of column 1.) [pic] See section 19.10 of the book. 2. Using your answers from question 1 above, solve where possible the
following sets of simultaneous equations. If you find that no solution
is possible, explain why. (a) x + 3y = 5 -x + 2y = 2 As explained in section 19.11 of the book, we can write this pair of
simultaneous equations in matrix form as Av = b where [pic] ; [pic] ; and [pic] Therefore v = A-1b. From (1) above, we have [pic], so [pic]; that is, x = 0.8, y = 1.4. (You can check that these values are correct by substituting them back
into the given simultaneous equations.) (b) 4y = (7 [pic] Following the same steps as in (a) above, and using [pic], from
question (1), we find [pic]. (Don't forget to check by substitution). (c) x + 4y = 1 [pic] In this case the relevant matrix in question (1) is C, which has no
inverse. Therefore the simultaneous equations in this case do not have
a solution. (This is because they are not independent of one another;
for example, the second equation is 3 times the first.) (d) x + 3y = 1 [pic] Following the same steps as in (a) and (b) above, and using [pic], from
question (1), we find [pic]. (Don't forget to check by substitution).
3. Find the determinant, all minors and cofactors, and the inverse of
each of the following matrices: (a) [pic] Determinant = 20 Matrix of minors: [pic] Matrix of cofactors (= signed minors): [pic] Inverse matrix: [pic]
(b) [pic] Determinant: -4 Matrix of minors: [pic] Matrix of cofactors (=signed minors) [pic] Inverse matrix: [pic] (c) [pic] Determinant: -4 Minors [pic] Cofactors (= signed minors): [pic] Inverse matrix: [pic] (d) [pic] Determinant: 4 Minors: [pic] Cofactors: [pic] inverse matrix: [pic] Exercise 1.3
1. Use Cramer's rule to find x in the equation systems below. (a) [pic] [pic] [pic] As explained in section 19.11 of the book, we can write this set of
simultaneous equations in matrix form as Av = b where [pic] ; [pic] ; and [pic] Cramer's rule tells us that we can solve for each variable as follows. (i) To solve for x, we form a new matrix, A1, obtained by replacing
the first column of A with the vector b. Thus [pic] The solution value of the variable x is then given by [pic]; that is,
the ratio of the determinants of the two matrices. In this example,
the two determinants are [pic], so [pic] (ii) To solve for y, we form a new matrix, A2, obtained by replacing
the second column of A with the vector b. Thus [pic] The solution value of the variable y is then given by [pic]; that is,
the ratio of the determinants of the two matrices. In this example,
the two determinants are [pic], so [pic] (iii) To solve for z, we form a new matrix, A3, obtained by replacing
the third column of A with the vector b. Thus [pic] The solution value of the variable y is then given by [pic]; that is,
the ratio of the determinants of the two matrices. In this example,
the two determinants are [pic], so [pic]. So our solutions are: x = -7, y = 9, z = 7 (You can check that these values are correct by substituting them back
into the given simultaneous equations.) (b) [pic] [pic] [pic] Using the method of (a) above: (i) Solution for x. We have [pic] ; [pic] ; [pic] So [pic] (ii) Solution for y. We have [pic] ; [pic] ; [pic] So [pic] (iii) Solution for z. We have [pic] ; [pic] ; [pic] So [pic] So our solutions are: x = -2, y = 2.75, z = 5 (You can check that these values are correct by substituting them back
into the given simultaneous equations.)
2. Solve the following system of linear equations using (a) matrix
inversion, and (b) Cramer's rule. [pic] (a) Solution by matrix inversion. We can write this set of simultaneous equations in matrix form as Av = b where [pic] ; [pic] ; and [pic] The solution is v = A-1b. We find A-1 as [pic]. So [pic] So our solutions are: x = 4, y = 0, z = 1 (You can check that these values are correct by substituting them back
into the given simultaneous equations.)
(b) Solution by Cramer's rule. Using the method of question 1 above: (i) Solution for x. We have [pic] ; [pic] ; [pic] So [pic] (ii) Solution for y. We have [pic] ; [pic] ; [pic] So [pic] (iii) Solution for z. We have [pic] ; [pic] ; [pic] So [pic] So our solutions are: x = 4, y = 0, z = 1, as before.
3. In the economy of the Kingdom of Monomania, households spend their
incomes on domestically produced goods, Cd, and imported goods, M.
This spending is observed to follow the relationship [pic] (1) where a is a parameter and Y = households' income. Imports are
observed to be related to household consumption by the relationship [pic] (where m is a parameter)
(2) The equilibrium condition for this economy is that aggregate output, Y,
must equal the demand for output for domestic consumption, Cd, and
domestic investment, I, plus demand for exports by foreigners, X.
Therefore the equilibrium condition is [pic] (3) (where I and X are assumed to be exogenous). Note that household
income is necessarily equal to output because households earn their
incomes (wages, salaries and profits) by producing output. (a) Show that, after suitable rearrangement, the set of
simultaneous equations 1 to 3 above can be written as Ax = b, where
A is a matrix containing the parameters of the model (a and m); x
is a vector containing the endogenous variables (Y, Cd and M) and b
is a vector containing the exogenous variables (I and X). We can present the equations of this model in table form as: |Y |-Cd | |= |I + X|(from equation 3) |
| |-mCd |+M |= |0 |(from equation 2) |
|[pic] |-Cd | |= |0 |(from equations 1 & |
| | | | | |2) | This helps us to see that the equations can be expressed in matrix
form. For example, the coefficients of Y in column 1 are 1, 0 and
[pic]. This gives us the first column of the coefficient matrix.
Similarly the coefficients of Cd in the second column are -1, -m,
and -1. This gives us the second column of the coefficient matrix.
After some fiddling around, we find that the equations can be
expressed in matrix form as follows: [pic] or in matrix notation, Ax = b. The key point here is that we have now the endogenous variables Y,
Cd and M on the left hand side (the vector x). These endogenous
variables are determined by the values of the exogenous variables I
and X (the vector b), together with the values of the parameters
(the matrix A.) (b) By calculating A-1, find the equilibrium values of the
endogenous variables in terms of the values of the parameters and
exogenous variables. Given [pic], we want to find [pic]. To find A-1, we follow the procedure explained in section 19.13 of
the book. The matrix of transposed cofactors, [pic], is [pic] The determinant of [pic]is [p