Skip to main content
Logo image

Discover Linear Algebra: A first course in linear algebra

Jeremy Sylvestre

Exercises 4.6 Exercises

1. Rows and columns.

Consider matrix
🔗
A=[8−4−9−9−57426−81−5].
🔗

(a)

Express each row of A as a row vector.
🔗
Answer.
The rows of A are:
Row 1: [8−4−9−9],Row 2: [−5742],Row 3: [6−81−5].
🔗

(b)

Express each column of A as a column vector.
🔗
Answer.
The columns of A are
Column 1:Column 2:Column 3:Column 4:[8−56],[−47−8],[−941],[−92−5].
🔗

(c)

State the dimensions of A.
🔗
Answer.
There are three rows and four columns, so the dimensions of A are 3×4.
🔗
🔗

2. Indexing matrix entries.

Consider matrix
🔗
A=[−2−159−8714−336−5].
🔗

(a)

Identify the value of each of the entries a12,a33,a41.
🔗
Answer.
a12=−1, a33=−3, a41=3.
🔗

(b)

Identify the indices of each of the positive entries of A.
🔗
Answer.
The positive entries are a13=5, a21=9, a23=7, a31=1, a32=4, a41=3, a42=6.
🔗
🔗

3. Matrix dimensions.

State the dimensions of each of the matrices below. Which of these matrices is square?
🔗
[−56−8−4−40712][71−863628−61−2−9000][0810−5−749149−2][209−749−1−7−8−5−1−85937]
🔗
Answer.
The dimensions (in order) are 3×3, 5×3, 3×4, and 4×4. Only the first and last matrices are square.
🔗

4. Identifying the main diagonal in a square matrix.

For those matrices from Exercise 3 that are square, identify the main diagonal.
🔗
Answer.
The first matrix from Exercise 3 (call it A) is square, and has main diagonal a11=−5, a22=−4, and a33=2:
[−56−8−4−40712].
The last matrix from Exercise 3 (call it D) is square, and has main diagonal d11=2, d22=9, d33=−1, and d44=7:
[209−749−1−7−8−5−1−85937].
🔗

(a)

What common pattern do the index pairs for entries along the main diagonal in a square matrix exhibit?
🔗
🔗

(b)

What common pattern do the index pairs for entries above the main diagonal in a square matrix exhibit?
🔗
🔗

(c)

What common pattern do the index pairs for entries below the main diagonal in a square matrix exhibit?
🔗
🔗

5. Matrix entries as a function of row and column indices.

Form the 4×4 matrix A whose entries satisfy the formula aij=i2−j2.
🔗
Answer.
A=[0−3−8−1530−5−12850−7151270]
🔗

6. Restricting matrix entries based on indices.

As usual, let i represent the row index and j represent the column index of each entry in a matrix. Create a 6×6 matrix has the following properties:
🔗
  • All entries with i>j are zero, all other entries are nonzero.
  • Each nonzero entry is positive if i+j is even and negative if i+j is odd.
  • No two nonzero entries have the same absolute value.
🔗
Answer.
Here is one possible matrix that satisfies the stated restrictions:
[1−23−45−607−89−10110012−1314−1500016−1718000019−200000021].
🔗

7. Matrix inequality.

For each example matrix X below, give at least one reason why X is not equal to matrix
🔗
A=[1180−1−7].
🔗

(a)

X=[11800−1−70]
🔗
Answer.
Matrix X is not equal to matrix A because X has a fourth column that A lacks.
🔗

(b)

X=[1180−1−7000]
🔗
Answer.
Matrix X is not equal to matrix A because X has a third row that A lacks.
🔗

(c)

X=[11801−7]
🔗
Answer.
Matrix X is not equal to matrix A because x21=1 while a21=−1.
🔗
🔗

Matrix arithmetic.

In each case, carry out the matrix operations. If the operations cannot be carried out, explain why.
🔗

8.

[21−49]T
🔗
Answer.
[−21−49]
🔗

10.

[2−2−30−5341−332−504−2]T
🔗
Answer.
[232−24−5−3100−34−53−2]
🔗

11.

[4121−5−92−93]T
🔗
Answer.
[4121−5−92−93]
🔗

12.

[64−39]+[82−2−8]
🔗
Answer.
[146−51]
🔗

13.

[64−39]+[82−2−875]
🔗
Answer.
Undefined: addition requires matrices of the same dimensions.
🔗

14.

[4−6−8−9−98]−[0−80274]+[4116−3−1]T
🔗
Answer.
[88−7−14−153]
🔗

15.

(a)
[2]+[5]
🔗
Answer.
[7]
Note: that the result is a 1×1 matrix, not simply a scalar (number).
🔗
(b)
[2]+[55]
🔗
Answer.
Undefined: addition requires matrices of the same dimensions.
🔗
🔗

16.

−[7−1−9304]
🔗
Answer.
[−719−30−4]
🔗

18.

3[−72−8812]
🔗
Answer.
[−216−242436]
🔗

19.

−2[−6015]
🔗
Answer.
[120−2−10]
🔗

20.

π[1−32−2]
🔗
Answer.
[π−3π2π−2π]
🔗

23.

4[8]
🔗
Answer.
[32]
Note: that the result is a 1×1 matrix, not simply a scalar (number).
🔗

24.

6[−250]+(−8)[−761]+3[50−3]
🔗
Answer.
[59−18−17]
🔗

25.

2[6−40−6]+3[89−40]+(−2)[52−53]
🔗
Answer.
[2615−2−18]
🔗

26.

(c)
[342−5][13]
🔗
Hint.
Consider the relationship between this matrix multiplication and the two matrix products you have already computed in this exercise.
Answer.
[15−13]
🔗
🔗

27.

(a)
[340][13]
🔗
Answer.
Undefined: the row matrix has more entries than the column matrix, so we cannot carry out the multiply-then-add procedure of matrix multiplication.
🔗
(b)
[34][130]
🔗
Answer.
Undefined: the row matrix has fewer entries than the column matrix, so we cannot carry out the multiply-then-add procedure of matrix multiplication.
🔗
🔗

28.

(a)
[2−5−2][3−32]
🔗
Answer.
[17]
🔗
(b)
[−454][3−32]
🔗
Answer.
[−19]
🔗
(c)
[13−5][3−32]
🔗
Answer.
[−16]
🔗
(d)
[2−5−2−45413−5][3−32]
🔗
Hint.
Consider the relationship between this matrix multiplication task and the three matrix products you have already computed in this exercise.
Answer.
[17−19−16]
🔗
🔗

29.

Note: you should complete Exercise 28 before completing this exercise.
🔗
(a)
[2−5−2][2−1−3]
🔗
Answer.
[15]
🔗
(b)
[−454][2−1−3]
🔗
Answer.
[−25]
🔗
(c)
[13−5][2−1−3]
🔗
Answer.
[14]
🔗
(d)
[2−5−2−45413−5][32−3−12−3]
🔗
Hint.
Consider the relationship between this matrix multiplication task, the final matrix multiplication task in Exercise 28, and the three matrix products you have already computed in this exercise.
Answer.
[1715−19−25−1614]
🔗
🔗

30.

[3−21−2−11][2−2−1−2−30−102]
🔗
Answer.
[90−1−374]
🔗

31.

[2−2−1−2−30−102][3−21−2−11]
🔗
Answer.
Undefined: the first matrix has three columns while the second matrix has only two rows.
🔗

32.

[2−4−10131−2−1][−5211−31610−2]
🔗
Answer.
[100010001]
🔗

33.

[7144−50014240−63647][1411−4−812]
🔗
Answer.
[00000000]
Notice: Here have multiplied two matrices, neither of which is a zero matrix, yet the result is a zero matrix. This confirms that for matrices, result AB=0 does NOT imply that at least one of A or B must be zero.
🔗

34.

(a)
[−2][5]
🔗
Answer.
[−10]
Note: that the result is a 1×1 matrix, not simply a scalar (number).
🔗
(b)
[−2][55]
🔗
Answer.
[−10−10]
🔗
(c)
[−2][55]
🔗
Answer.
Undefined: the first matrix has only one column while the second matrix has two rows.
Note: this operation is NOT the same as scalar multiplication.
🔗
🔗

35.

[1−2][−145]
🔗
Hint.
Before you think about how this matrix product should be carried out, first determine the dimensions that the result matrix should have.
Answer.
The first factor in the matrix product is 2×1 and the second factor is 1×3, so the product result should be a 2×3 matrix:
[−1452−8−10].
🔗

36.

[a000b000c][d000e000f]
🔗
Note: see also Exercise 41.
🔗
Answer.
[ad000be000cf]
🔗

38.

[2−2−1−2−30−102]2
🔗
Answer.
[92−42132−425]
🔗

39.

[2−2−1−2−30−102]3
🔗
Answer.
[18−24−17−24−432−17214]
🔗

40.

[3−21−2−11]2
🔗
Answer.
Undefined: cannot take powers of a matrix that is not square as there will be a mismatch between the number of columns of the first factor with the number of rows of the second factor.
🔗

41.

[a000b000c]5
🔗
Note: first see Exercise 36.
🔗
Answer.
[a5000b5000c5]
🔗

43.

[3−21−2−11][3−21−2−11]T
🔗
Answer.
[14−3−36]
🔗

44.

[3−21−2−11]T[3−21−2−11]
🔗
Answer.
[13−41−45−31−32]
🔗

45.

[3−21−2−11][3−21−2−11]T+6[5−36−5−17]
🔗
Answer.
Undefined: we computed that initial matrix product in Exercise 43 and the result had dimensions 2×2, which is not compatible with addition to a 2×3 matrix.
🔗

46.

[1234]2−[14916]
🔗
Answer.
[6666]
Note: that the result is NOT a zero matrix.
🔗

47.

[−232−2]2+[3−11−6][−232−2]−[0−5−25]2
🔗
Solution.
This arithmetic has the algebraic pattern
X2+AX−B2
for
X=[−232−2],A=[3−11−6],B=[0−5−25].
We can reduce the number of matrix products we need to compute by factoring the common X out of the first two terms, but we must be careful to factor it in the correct “direction”:
X2+AX−B2=(X+A)X−B2.
Now compute:
X+A=[−232−2]+[3−11−6]=[123−8],
(X+A)X=[123−8][−232−2]=[2−1−2225],
B2=[0−5−25][0−5−25]=[10−25−1035].
Finally put it all together:
X2+AX−B2=(X+A)X−B2=[2−1−2225]−[10−25−1035]=[−824−12−10].
🔗
🔗

Dimensions of matrix products.

In each case, determine the dimensions of the product matrix AB, or state why the product matrix does not exist, based on the stated dimensions of the individual factor matrices A and B.
🔗

48.

A: 1×4; B: 4×1.
🔗
Answer.
AB: 1×1.
🔗

49.

A: 3×4; B: 4×1.
🔗
Answer.
AB: 3×1.
🔗

50.

A: 2×3; B: 3×4.
🔗
Answer.
AB: 2×4.
🔗

51.

A: 3×4; B: 2×3.
🔗
Answer.
Undefined: the number of columns in A does not match the number of rows in B.
🔗

52.

A: 3×3; B: 3×3.
🔗
Answer.
AB: 3×3.
🔗

53.

A: 3×4; B: 3×4.
🔗
Answer.
Undefined: the number of columns in A does not match the number of rows in B.
🔗

54.

A: 4×3; B: 4×3.
🔗
Answer.
Undefined: the number of columns in A does not match the number of rows in B.
🔗

55.

A: 3×4; B: 4×3.
🔗
Answer.
AB: 3×3.
🔗

56.

A: 4×3; B: 3×4.
🔗
Answer.
AB: 4×4.
🔗

57.

A: 3×3; B: 3×4.
🔗
Answer.
AB: 3×4.
🔗

58.

A: 3×4; B: 4×4.
🔗
Answer.
AB: 3×4.
🔗

59.

A: 1×1; B: 2×2.
🔗
Answer.
Undefined: the number of columns in A does not match the number of rows in B.
🔗

60.

A: 1×1; B: 1×3.
🔗
Answer.
AB: 1×3.
🔗

61.

A: 4×1; B: 1×3.
🔗
Answer.
AB: 4×3.
🔗
🔗

Interpreting matrix equality as a system of equations.

In each case, use the definition of equal matrices to determine the variable values that make the matrix equality true. In any equality where 0 appears, assume it represents a zero matrix of appropriate dimensions.
🔗

62.

[b+3cc+3b−2aa+4ca−4b−3c]=[−53−90]
🔗
Solution.
Comparing corresponding entries on each side of the matrix equality leads to system of equations
{b+3c=−5−2a+3b+c=3a+4c=−9a−4b−3c=0.
Turn this system into an augmented matrix and reduce:
[013−5−2313104−92−4−30]→reducerow[100−10101001−20000].
This system is consistent with one unique solution a=−1, b=1, c=−2.
🔗

63.

k1[321]+k2[731]=[−3−7−5]
🔗
Solution.
First compute the left-hand side to a single column matrix:
k1[321]+k2[731]=[3k12k1k1]+[7k23k2k2]=[3k1+7k22k1+3k2k1+k2].
Our simplified matrix equality is now
[3k1+7k22k1+3k2k1+k2]=[−3−7−5].
Comparing corresponding entries on each side of the matrix equality leads to system of equations
{3k1+7k2=−32k1+3k2=−7k1+k2=−5.
Turn this system into an augmented matrix and reduce:
[37−323−711−5]→reducerow[10−8013000].
This system is consistent with one unique solution k1=−8, b=3. You may test that this is the correct solution by recomputing the matrix algebra on the left-hand side of the original matrix equality with those values in place of k1 and k2.
🔗

64.

c1[0−11]+c2[1−2−1]=[−5115]
🔗
Solution.
First compute the left-hand side to a single column matrix:
c1[0−11]+c2[1−2−1]=[0−c1c1]+[c2−2c2−c2]=[c12k1+3k2k1+k2].
Our simplified matrix equality is now
[c2−c1−2c2c1−c2]=[−5115].
Comparing corresponding entries on each side of the matrix equality leads to system of equations
{c2=−5−c1−2c2=11c1−c2=5.
Turn this system into an augmented matrix and reduce:
[01−5−1−2111−15]→reducerow[100010001].
This system is inconsistent, so there is no combination of values for c1 and c2 that make the original matrix equality true.
Careful: It is tempting to attempt to solve the system of equations directly, without using an augmented matrix, as we clearly have c2=−5 from the first equation, and then we can use the third equation to solve c1=0. But if you substitute those two values into the left-hand side of the original matrix equality and recompute the matrix algebra, you will find that the matrix equality is not true for these values. The reason for this is that the original matrix equality leads to a system of three equations, and we cannot ignore any one equation in that system.
🔗

65.

k1[02−21]+k2[1−232]+k3[2−235]+k4[1043]=0
🔗
Solution.
First compute the left-hand side to a single matrix:
k1[02−21]+k2[1−232]+k3[2−235]+k4[1043]=[02k1−2k1k1]+[k2−2k23k22k2]+[2k3−2k33k35k3]+[k404k43k4]=[k2+2k3+k42k1−2k2−2k3−2k1+3k2+3k3+4k4k1+2k2+5k3+3k4].
In the context of the original matrix equality, the 0 symbol on the right-hand side should represent the 2×2 zero matrix, and so our simplified matrix equality is now
[k2+2k3+k42k1−2k2−2k3−2k1+3k2+3k3+4k4k1+2k2+5k3+3k4]=[0000].
Comparing corresponding entries on each side of the matrix equality leads to homogeneous system of equations
{k2+2k3+k4=02k1−2k2−2k3=0−2k1+3k2+3k3+4k4=0k1+2k2+5k3+3k4=0.
Since the system is homogeneous, we reduce only with the coefficient matrix:
[01212−2−20−23341253]→reducerow[10040107001−30000].
This system is consistent (as all homogeneous systems are) but the solution requires a parameter, so in fact there are an infinite number of combinations of scalar values for k1,k2,k3,k4 that make the original matrix equality true. In particular, if we assign parameter k4=t, the general solution is
k1=−4t,k2=−7t,k3=3t,k4=t.
Careful: We may have been misled by the zero matrix on the right-hand side of the original matrix equality, and decided that the solution must be to set each of k1,k2,k3,k4 to 0. While that is one possible solution to the matrix equation (which we would call the trivial solution), by re-interpreting the matrix equation as a system of (ordinary) equations, we were able to arrive at the complete collection of all possible solutions to the matrix equation. So while it is possible to make the matrix equality true by setting each of k1,k2,k3,k4 to 0, it is also possible, for example, by setting
k1=−4,k2=−7,k3=3,k4=1.
🔗

66.

c1[1123−20]+c2[121311]+c3[131331]=0
🔗
Solution.
First compute the left-hand side to a single matrix:
c1[1123−20]+c2[121311]+c3[131331]=[c1c12c13c1−2c10]+[c22c2c23c2c2c2]+[c33c3c33c33c3c3]=[c1+c2+c3c1+2c2+3c32c1+c2+c33c1+3c2+3c3−2c1+c2+3c3c2+c3].
In the context of the original matrix equality, the 0 symbol on the right-hand side should represent the 3×2 zero matrix, and so our simplified matrix equality is now
[c1+c2+c3c1+2c2+3c32c1+c2+c33c1+3c2+3c3−2c1+c2+3c3c2+c3]=[000000].
Comparing corresponding entries on each side of the matrix equality leads to homogeneous system of equations
{c1+c2+c3=0c1+2c2+3c3=02c1+c2+c3=03c1+3c2+3c3=0−2c1+c2+3c3=0c2+c3=0.
Since the system is homogeneous, we reduce only with the coefficient matrix:
[111123211333−213011]→reducerow[100010001000000000].
This system is consistent (as all homogeneous systems are) with one unique solution, the trivial one: c1=0, c2=0, and c3=0.
Careful: We may have been misled by the zero matrix on the right-hand side of the original matrix equality, and prematurely jumped to the conclusion that the only solution must be to set each of c1,c2,c3 to 0 (the trivial solution). While that turned out to be the case this time, it is important that we went through the full process of re-interpreting the matrix equation as a system of (ordinary) equations so that we could confirm that the trivial solution was indeed the only solution.
🔗
🔗

Isolating via matrix algebra.

In each case, use matrix algebra to isolate and determine the unknown variable matrix.
🔗

67.

[−870−3]−2X=[50−68]
🔗
Solution.
It’s less messy to perform algebraic manipulations with letters, so assign
A=[−870−3],B=[50−68].
Then we have
A−2X=BA−B=2XX=12(A−B)
and we may calculate
A−B=[−870−3]−[50−68]=[−1376−11],
X=12(A−B)=12[−1376−11]=[−132723−112].
🔗

68.

5XT+[−3−736−1−1]=0
🔗
Solution.
It’s less messy to perform algebraic manipulations with letters, so assign
A=[−3−736−1−1].
Then we have
5XT+A=05XT=−AXT=−15A.
The final algebraic step is to “undo” the transpose that has been applied to X, but by Rule 5.a of Proposition 4.5.1, transpose reverses itself:
XT=−15A(XT)T=(−15A)TX=−15AT,
where in the last step we have also applied Rule 5.c of Proposition 4.5.1.
The above algebra now tells us what to calculate:
AT=[−3−736−1−1]T=[−33−1−76−1],
X=−15AT=−15[−33−1−76−1]=[35−351575−6515].
🔗
🔗

Systems as matrix equations.

Express each system of equations as a single matrix equation Ax=b.
🔗

69.

{x−y=−62x−3y=−14
🔗
Answer.
[1−12−3][xy]=[−6−14]
🔗

70.

{2s−t=08s−2t=−14s−8t=0
🔗
Answer.
[2−18−24−8][st]=[0−10]
🔗

71.

{2x+5y−4z=9x+3y−3z=53x+7y−5z=13
🔗
Answer.
[25−413−337−5][xyz]=[9513]
🔗

72.

{2x16x2−x3+3x4=−12x1+3x2−x3−x4=−9
🔗
Answer.
[26−1313−1−1][x1x2x3x4]=[−12−9]
🔗

73.

{3p1+5p2−2p3=0p1+2p2−2p3=04p1+5p2+4p3=02p1+1p2+8p3=0
🔗
Answer.
[35−212−2454218][p1p2p3]=[0000]
🔗
🔗

Testing system solutions.

In each case, use matrix multiplication to determine whether the variable values constitute a solution to the specified system of equations.
🔗

74.

System from Exercise 69; potential solution of x=−4,y=2.
🔗
Solution.
As in the answer to Exercise 69, we have coefficient matrix
A=[1−12−3].
The potential solution can be expressed as column vector
x=[−42].
Multiply these together:
Ax=[1−12−3][−42]=[−6−14].
This result matches the column of constants from the answer to Exercise 69, which verifies that the proposed solution is an actual solution.
🔗

75.

System from Exercise 70; potential solution of s=3,t=−3.
🔗
Solution.
As in the answer to Exercise 70, we have coefficient matrix
A=[2−18−24−8].
The potential solution can be expressed as column vector
x=[3−3].
Multiply these together:
Ax=[2−18−24−8][3−3]=[93036].
This result does not match the column of constants
b=[0−10]
from the answer to Exercise 70, so the proposed solution is not an actual solution.
🔗

76.

System from Exercise 71:
🔗
  1. Potential solution of x=2,y=1,z=0.
  2. Potential solution of x=−1,y=3,z=1.
🔗
Solution.
As in the answer to Exercise 71, we have coefficient matrix
A=[25−413−337−5].
The two potential solutions can be expressed as column vectors
x1=[210],x2=[−131].
Multiply each of these by the coefficient matrix:
Ax1=[25−413−337−5][210]Ax2=[25−413−337−5][−131]=[9513],=[9513].
In both cases, the multiplication result matches the column of constants from the answer to Exercise 71, which verifies that each of the proposed solutions is an actual solution.
🔗

77.

System from Exercise 72:
🔗
  1. Potential solution of x1=−3,x2=0,x3=6,x4=0.
  2. Potential solution of x1=−1,x2=−2,x3=1,x4=1.
  3. Potential solution of x1=1,x2=1,x3=1,x4=1.
🔗
Solution.
As in the answer to Exercise 72, we have coefficient matrix
A=[26−1313−1−1].
The three potential solutions can be expressed as column vectors
x1=[−3060],x2=[−1−211],x3=[1111].
Multiply each of these by the coefficient matrix in turn:
Ax1=[26−1313−1−1][−3060]=[−12−9]Ax2=[26−1313−1−1][−1−211]=[−12−9]Ax3=[26−1313−1−1][1111]=[102].
Each of the first two results matches the column of constants from the answer to Exercise 72, which verifies that these two proposed solution are both actual solutions. However, the third result does not match that column of constants, and so the third proposed solution is not actually a solution.
🔗

78.

System from Exercise 73; potential solution of p1=−6,p2=4,p3=1.
🔗
Solution.
As in the answer to Exercise 73, we have coefficient matrix
A=[35−212−2454218].
The potential solution can be expressed as column vector
p=[−641].
Multiply these together:
Ap=[35−212−2454218][−641]=[0000].
This result matches the (homogeneous) column of constants from the answer to Exercise 73, which verifies that the proposed solution is an actual (and in this case, nontrivial) solution.
🔗
🔗

Expressing system solutions in vector form.

In each case, express the general solution to the system Ax=b in vector form. (See Subsubsection 4.4.4.2.) Then use your vector form of the solution to determine the specific solution obtained by setting all parameters to 0, as well as all possible specific solutions obtained by setting a single parameter to 1 while setting all other parameters to 0.
🔗

79.

A=[1236]; b=[412].
🔗
Solution.
Form the augmented matrix and reduce:
[1243612]→reducerow[124000].
Using x,y as the variables, variable y is free and we assign it a parameter: y=t. Interpreting the first row in the reduced matrix as an equation, we can isolate
x=4−2t.
Express the general solution in vector form:
x=[xy]=[4−2tt]=[40]+t[−21].
Finally, the requested specific solutions are
x0=[40],x1=[21].
🔗

80.

A=[1236]; b=0.
🔗
Solution.
This is the homogeneous version of the system in Exercise 79, so we have general solution
x=t[−21]
and requested specific solution
x0=0,x1=[−21].
🔗

81.

A=[1−8]; b=[7].
🔗
Solution.
Form the augmented matrix:
[1−87].
This matrix is already reduced. Using x,y as the variables, variable y is free and we assign it a parameter: y=t. Interpreting the row in the augmented matrix as an equation, we can isolate
x=7+8t.
Express the general solution in vector form:
x=[xy]=[7+8tt]=[70]+t[81].
Finally, the requested specific solutions are
x0=[70],x1=[151].
🔗

82.

A=[0141−2−12−3428]; b=[2−48].
🔗
Solution.
Form the augmented matrix and reduce:
[01421−2−12−4−34288]→reducerow[10−4001420000].
Using x,y,z as the variables, variable z is free and we assign it a parameter: z=t. Interpreting the nonzero rows of the reduced matrix as equations, we can isolate
x=4t,y=2−4t.
Express the general solution in vector form:
x=[xyz]=[4t2−4tt]=[020]+t[4−41].
Finally, the requested specific solutions are
x0=[020],x1=[4−21].
🔗

83.

A=[0141−2−12−3428]; b=0.
🔗
Solution.
This is the homogeneous version of the system in Exercise 82, so we have general solution
x=t[4−41]
and the requested specific solutions are
x0=0,x1=[4−41].
🔗

84.

A=[13−4]; b=0.
🔗
Solution.
This is a homogeneous system with a coefficient matrix that is already reduced. Using x,y,z as the variables, variables y and z are free and we assign them parameters: y=s, z=t. Interpreting the row in A as coefficients in a homogeneous equation, we can isolate
x=−3s+4t.
Express the general solution in vector form:
x=[xyz]=[−3s+4tst]=s[−310]+t[401].
Finally, the requested specific solutions are
x0=0,xs=[−310],xt=[401].
🔗

85.

A=[1−21103000142−12−422061]; b=[−2−101].
🔗
Solution.
Form the augmented matrix and reduce:
[1−211030−200142−1−102−4220611]→reducerow[1−206103001420−50000015].
Using x1,x2,x3,x4,x5,x6 as the variables, variables x2,x4,x5 are free and we assign parameters to them:
x2=r,x4=s,x5=t.
Interpreting the rows of the reduced matrix as equations, we can isolate
x1=3+2r−6s−t,x3=−5−4s−2t,x6=5.
Express the general solution in vector form:
x=[x1x2x3x4x5x6]=[3+2r−6s−tr−5−4s−2tst5]=[30−5005]+r[210000]+s[−60−4100]+t[−10−2010].
Finally, the requested specific solutions are
x0=[30−5005],x1=[51−5005],x2=[−30−9105],x3=[20−7015].
🔗

86.

A=[1−21103000142−12−422061]; b=0.
🔗
Solution.
This is the homogeneous version of the system in Exercise 85, so we have general solution
x=r[210000]+s[−60−4100]+t[−10−2010]
and the requested specific solutions are
x0=0,xr=[210000],xs=[−60−4100],xt=[−10−2010].
🔗
🔗

Matrix arithmetic entry patterns.

For each matrix operation, write out the relationship between the entries of the “input” matrices and the entries of the resulting “output” matrix.
🔗

87. Addition.

For C=A+B, where A and B are both of size m×n, express the general entry cij of the m×n sum matrix C as a formula in the entries of A and B.
🔗
🔗

88. Subtraction.

For C=A−B, where A and B are both of size m×n, express the general entry cij of the m×n difference matrix C as a formula in the entries of A and B.
🔗
🔗

89. Negation.

For C=−A, where A is of size m×n, express the general entry cij of the m×n negative matrix C as a formula in the entries of A.
🔗
🔗

90. Scalar multiplication.

For C=kA, where A is of size m×n and k is a scalar, express the general entry cij of the m×n scaled matrix C as a formula in the entries of A.
🔗
🔗

91. Transpose.

For C=AT, where A is of size m×n, express the general entry cij of the m×n transposed matrix C as a formula in the entries of A.
🔗
🔗

93. Square.

For C=A2, where A is of square size n×n, express the general entry cij of the n×n power matrix C as a formula in the entries of A.
🔗
🔗
🔗

94. Extend the pattern.

In Exercise 37 we found that by multiplying two matrices whose entries follow a particular pattern, we can effectively “isolate” the addition part of the multiply-then-add procedure of matrix multiplication. Or, put another way, this pattern reveals that there is a way to “recast” addition of scalars (numbers) as a form of multiplication of 2×2 matrices.
🔗

(a)

Can you create another, similar “form” of 2×2 matrix that does the same?
🔗
🔗

(b)

Can you create a similar “form” of 3×3 matrix that does the same?
🔗
🔗

(c)

Can you create a “form” of 2×2 matrix that “recasts” multiplication of scalars as a form of matrix multiplication?
🔗
🔗
🔗

Verifying matrix algebra rules.

Practise verifying matrix algebra rules. Use the provided proofs of Rule 1.b and Rule 2.c from Proposition 4.5.1 as templates/guides to carrying out your own verifications. You may also find your answers to Exercises 4.6.87–4.6.93 helpful. (Recall: that 0 represents a zero matrix of appropriate dimensions, not the scalar 0.)
🔗

96.

Rule: A+A=2A for every matrix A.
🔗
🔗

97.

Rule: 1A=A for every matrix A.
🔗
🔗

101.

Rule: if matrix A has a row that is all zeros, then so does the product AB for every matrix B for which the product is defined.
🔗
🔗

102.

Rule: if matrix B has a column that is all zeros, then so does the product AB for every matrix A for which the product is defined.
🔗
🔗
🔗

106. Combinations of solutions.

(a)

Suppose x0 is a column vector that represents a specific solution to the homogeneous system Ax=0, and let k represent an arbitrary scalar value. Verify that the scaled vector y=kx0 also represents a specific solution to that same homogeneous system.
🔗
🔗

(b)

Suppose x1 and x2 are column vectors of the same dimensions, each of which represents a specific solution to the homogeneous system Ax=0. Verify that the sum vector y=x1+x2 also represents a specific solution to that same homogeneous system.
🔗
🔗

(c)

Suppose x1,x2,…,xm are all column vectors of the same dimensions, each of which represents a specific solution to the homogeneous system Ax=0, and let each of k1,k2,…,km represent an arbitrary scalar value. Verify that the linear combination vector
🔗
y=k1x1+k2x2+…+kmxm
also represents a specific solution to that same homogeneous system.
🔗
🔗
🔗

(d)

Do the patterns verified in this exercise hold true for scales/sums/combinations of solutions to a nonhomogeneous system?
🔗
🔗
🔗
🔗
PrevTopNext