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Discover Linear Algebra: A first course in linear algebra

Jeremy Sylvestre

Exercises 2.6 Exercises

Identifying leading/constrained/free variables.

Assume that each matrix below represents the augmented matrix of a linear system of equations in the variables x1,x2,…,xn (where n is the number of variable columns in the matrix).
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For each matrix, identify the leading variables. Then decide which variables are free and which are constrained.
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1.

[05989−2]
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Answer.
Both variables x1 and x2 are leading:
[05989−2].
The boxed entries could be converted into leading ones through row operations. Therefore, both variables are constrained and there are no free variables. (This implies the underlying system has one unique solution.)
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2.

[7−3140058]
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Answer.
Variables x1 and x3 are leading:
[7−3140058].
The boxed entries could be converted into leading ones through row operations. These two variables are constrained by the respective equations in which they are the leading variables, which leaves x2 as a free variable. (This implies the underlying system has an infinite number of solutions.)
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3.

[000085000000010300000−4−4−7]
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Answer.
Variables x2,x4,x5 are leading:
[000085000000010300000−4−4−7].
Variable x2 already has a leading one in its column, and the other two boxed entries could be converted into leading ones through row operations. These three variables are constrained by the respective equations in which they are the leading variables, which leaves x1 and x3 as free variables. (This implies the underlying system has an infinite number of solutions.)
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Recognizing echelon forms.

For each matrix, decide whether it is in row echelon form (REF). For those that are, decide whether it is in reduced row echelon form (RREF).
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4.

[10−90−50105−90010700015]
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Answer.
This matrix is in REF but not RREF because there are columns that contain both a leading one and another nonzero entry.
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5.

[10002000−1]
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Answer.
This matrix is not in REF because the first nonzero entry in each of the second and third rows is not one.
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6.

[10040100001−20000]
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Answer.
This matrix is in RREF.
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7.

[1030000001000001]
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Answer.
This matrix is in not in REF because is a zero row that appears above some of the nonzero rows.
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8.

[100100000]
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Answer.
This matrix is not in REF because the leading one in the second row does not appear to the right of the leading one in the first column.
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9.

[010000100001]
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Answer.
This matrix is in RREF.
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Choosing row operations.

For each matrix, choose a single row operation that makes incremental progress according to Procedure 2.3.2.
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10.

[6−2703312692−1−2−3−5]
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Answer.
The leftmost column that is not all zeros is the first column, so the immediate goal here should be to obtain a leading one in the first column (Step 1.a of Procedure 2.3.2). Any one of the operations 16R1, 13R2, 12R3, or R2−R3 may be employed to do so. However, it is usually preferable (though not necessary) to avoid introducing fractions in the matrix, so R2−R3 may be the best choice:
[6−2703312692−1−2−3−5]→R2−R3[6−27031249192−1−2−3−5].
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11.

[9−23−6−3−2−141−4−702−4−9−8]
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Answer.
The leftmost column that is not all zeros is the first column, but we already have a leading one in the first column, so the immediate goal should be to move it to the first row (Step 1.b of Procedure 2.3.2). Therefore, our choice of operation should be R1↔R3:
[9−23−6−3−2−141−4−702−4−9−8]→R1↔R3[1−4−70−3−2−149−23−62−4−9−8].
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12.

[1−41440−20−205−584−8−32−4−45]
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Answer.
The leftmost column that is not all zeros is the first column, but we already have a leading one in the first column, and that leading one is already in the first row. So the immediate goal should be to use that leading one to eliminate the other nonzero entries in the first column (Step 1.c of Procedure 2.3.2). Focusing on the 5 in the first column, combining it with −5 times the leading one will cancel to zero. So our choice of operation should be R3−5R1:
[1−41440−20−205−584−8−32−4−45]→R3−5R1[1−41440−20−200153−16−28−32−4−45].
Note: It is acceptable (and advisable!) to combine multiple operations in one calculation step when carrying out an “elimination” step in Procedure 2.3.2 (in this case, Step 1.c). That is, normally we would simultaneously perform our chosen operation R3−5R1 as well as the operation R4+3R1 in order to eliminate both of the other nonzero entries in the first column.
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13.

[1−2−3903710−674]
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Answer.
This matrix already exhibits all the goals of Step 1 of Procedure 2.3.2, so we move on to Step 2. Ignoring the first row completely, the leftmost column that is not all zeros is the second column, and so the immediate goal should be to obtain a leading one in that column (Step 2.a of Procedure 2.3.2). We can do so using either operation 13R2 or operation −16R3. Since 13 is a simpler fraction and creates the leading one in the second row instead of the third, we’ll choose that option:
[1−2−3903710−674]→13R2[1−2−390173130−674].
Warning. It may be tempting to attempt to avoid fractions altogether by using the −2 above that leading 3 to obtain the leading one. But notice what happens if we instead choose operation R2+R1:
[1−2−3903710−674]→R2+R1[1−2−39114100−674].
Our goal was to obtain a leading one in the second column. While we have succeeded in obtaining a one in the second column, it is not a leading one, because a row operation must be carried through the entire row, including any “leading” zeros. The operation R2+R1 would reverse progress already made in reducing this matrix: in particular, it reverses the progress of Step 1.c of Procedure 2.3.2 by “un-eliminating” an entry in the first column below that leftmost leading one. It is precisely for this reason that Step 2.a of Procedure 2.3.2 advises not to use the first row to attempt to obtain a second leading one.
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14.

[1−59−7400−5−9−300−275]
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Answer.
This matrix already exhibits all the goals of Step 1 of Procedure 2.3.2, so we move on to Step 2. Ignoring the first row completely, the leftmost column that is not all zeros is the third column, and we may obtain a leading one in that column (Step 2.a of Procedure 2.3.2) with either operation −15R2 or −12R3. We choose the latter operation as 12 is the simpler fraction:
[1−59−7400−5−9−300−275]→−12R3[1−59−7400−5−9−3001−72−52].
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15.

[1828068501570−8−70]
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Answer.
This matrix already exhibits all the goals of Step 1 of Procedure 2.3.2, so we move on to Step 2. Ignoring the first row completely, the leftmost column that is not all zeros is the second column. We already have a leading one in that column but it’s in the third row, so we move it up to the second row (Step 2.b of Procedure 2.3.2) with operation R2↔R3:
[1828068501570−8−70]→R2↔R3[1828015706850−8−70].
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16.

[13−35−9001−150003−3002−7−6]
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Answer.
This matrix already exhibits all the goals of Step 1 of Procedure 2.3.2, so we move on to Step 2. Ignoring the first row completely, the leftmost column that is not all zeros is the third column. We already have a leading one in that column and it is already in the second row, so we move on to using that leading one to eliminate the other entries in its column (Step 2.c of Procedure 2.3.2). Working from top-to-bottom in that column, we have a −3 entry in the first row that can be eliminated by combining it with +3 times the leading one in the second row:
[13−35−9001−150003−3002−7−6]→R1+3R2[13026001−150003−3002−7−6].
Note: It is acceptable (and advisable!) to combine multiple operations in one calculation step when carrying out an “elimination” step in Procedure 2.3.2 (in this case, Step 2.c). That is, normally we would simultaneously perform our chosen operation R1+3R2 as well as the operation R4−2R2 in order to eliminate both of the other nonzero entries in the third column.
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17.

[10−71−60126−7000−7100023]
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Answer.
This matrix already exhibits all the goals of both Step 1 and Step 2 of Procedure 2.3.2, so we move on to Step 3. Ignoring the first two rows completely, the leftmost column that is not all zeros is the fourth column. so our goal is to obtain a leading one in that column (Step 3.a of Procedure 2.3.2). We could do this by introducing fractions (using either −17R3 or 12R4), but another option avoiding fractions is R3+4R4:
[10−71−60126−7000−7100023]→R3+4R4[10−71−60126−700011300023].
Warning. Step 3.a of Procedure 2.3.2 advises us to avoid attempting to use the first two rows to obtain the next leading one, for good reason. In this matrix, we should not try to use the 1 in the first row, fourth column, neither to obtain a leading one in the third or fourth rows, nor to attempt to eliminate the leading entries in the third or fourth rows.
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18.

[1−80000014−50000100000]
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Answer.
This matrix is already quite simplified. In fact, it is only one operation away from RREF, as all that remains is to use the leading one in the fifth column to eliminate the entry above it:
[1−80000014−50000100000]→R2+5R3[1−8000001400000100000].
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Determining general solutions.

The matrices below are all already in RREF. For each, determine whether the associated system of equations is consistent. If it is, carry out the following additional tasks.
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  1. Write out that system in variables of your choosing.
  2. Determine the general solution to the system, expressed in parametric form if necessary.
  3. If the system has an infinite number of solutions, use your expression of the general solution to determine at least three different specific solutions.
Note: if the matrix does not have a vertical line separating the last column, interpret the matrix to be the coefficient matrix of a homogeneous system.
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19.

[100060100−70010000012]
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Solution.
The system is consistent but there are no free variables, hence there is one unique solution. The associated system of equations is identical to that solution; using variables x1,x2,x3,x4 we have
x1=6,x2=−7,x3=0,x4=2.
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20.

[10050105001−30000]
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Solution.
The system is consistent but there are no free variables, hence there is one unique solution. The associated system of equations is identical to that solution if we ignore the irrelevant 0=0 equation; using variables x1,x2,x3 we have
x1=5,x2=5,x3=−3,
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21.

[10−4−801270000]
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Solution.
The system is consistent.
  1. Ignoring the irrelevant 0=0 equation, the associated system of equations in variables x,y,z is
    {x−4z=−8y+2z=7.
  2. There is no leading one in the third column of the matrix, hence variable z is free. Assigning parameter z=t and isolating the two constrained variables in the equations above yields general solution
    x=−8+4t,y=7−2t,z=t.
  3. We may determine specific solutions by choosing arbitrary values for the parameter. For example:
    • Parameter value t=0 leads to specific solution
      x=−8,y=7,z=0.
    • Parameter value t=1 leads to specific solution
      x=−4,y=5,z=1.
    • Parameter value t=−3 leads to specific solution
      x=−20,y=13,z=−3.
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22.

[180−10013]
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Solution.
The system is consistent.
  1. The associated system of equations in variables x,y,z is
    {x+8y=−1z=3.
  2. There is no leading one in the second column of the matrix, hence variable y is free. Assigning parameter y=t and isolating the two constrained variables in the equations above yields general solution
    x=−1−8t,y=t,z=3.
  3. We may determine specific solutions by choosing arbitrary values for the parameter. For example:
    • Parameter value t=0 leads to specific solution
      x=−1,y=0,z=3.
    • Parameter value t=1 leads to specific solution
      x=−9,y=1,z=3.
    • Parameter value t=2 leads to specific solution
      x=−1−82,y=2,z=3.
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23.

[107−60019−100000100000]
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Solution.
The system is inconsistent due to the appearance of a leading one in the “equals” column.
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24.

[107−60019−100000100000]
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Solution.
We assume this to be the coefficient matrix of a homogeneous system, which is always consistent.
  1. The associated system of equations using variables x1,x2,x3,x4,x5 is
    {x1+7x3−6x4=0x2+9x3−x4=0x5=0.
    (We ignore the irrelevant 0=0 equation.)
  2. The third and fourth columns in the matrix do not contain leading ones, hence variables x3 and x4 are free. Assigning parameters x3=s and x4=t and isolating for the three constrained variables in the equations above leads to general solution
    x1=−7s+6t,x2=−9s+t,x3=s,x4=t,x5=0.
  3. We may determine specific solutions by choosing different combinations of arbitrary values for the parameters. For example:
    • Parameter values s=0 and t=0 lead to the trivial solution
      x1=0,x2=0,x3=0,x4=0,x5=0.
    • Parameter values s=1 and t=0 lead to solution
      x1=−7,x2=−9,x3=1,x4=0,x5=0.
    • Parameter values s=0 and t=1 lead to solution
      x1=6,x2=1,x3=0,x4=1,x5=0.
    • Parameter values s=−2 and t=3 lead to solution
      x1=32,x2=21,x3=−2,x4=3,x5=0.
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Solving systems.

For each system:
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  1. Form the associated augmented matrix. (If it is a homogeneous system, form the coefficient matrix instead.)
  2. Use Procedure 2.3.2 to reduce the matrix to RREF.
  3. Determine whether the system is consistent. If it is, determine the general solution to the system, expressed in parametric form if necessary.
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25.

{4x+3y=45x+4y=6
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Solution.
  1. The augmented matrix is
    [434546].
  2. The matrix can be reduced to RREF by Procedure 2.3.2 using the following sequence of operations.
    1. R2−R1
    2. R1↔R2
    3. R2−4R1
    4. −R2
    5. R1−R2
    [434546]→reducerow[10−2014]
  3. The system is consistent and has one unique solution of x=−2,y=4.
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26.

{4x−36y=12−3x+27y=−7
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Solution.
  1. The augmented matrix is
    [4−3612−327−7].
  2. The matrix can be reduced to RREF by Procedure 2.3.2 using the following sequence of operations.
    1. R1+R2
    2. R2+3R1
    3. 18R2
    4. R1−5R2
    [4−3612−327−7]→reducerow[1−90001]
  3. The system is inconsistent.
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27.

{7x−z=6−3x+3y+z=9−4x−1y+z=−1114x−1y−2z=7
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Solution.
  1. The augmented matrix is
    [70−16−3319−4−11−1114−1−27].
  2. The matrix can be reduced to RREF by Procedure 2.3.2 using the following sequence of operations.
    1. R2−R3
    2. R1↔R2
    3. R2−7R1; R3+4R1; R4−14R1
    4. R4−2R2
    5. −R4
    6. R2↔R4
    7. R1−4R2; R3−15R2; R4+28R2
    8. R4+R3
    [70−16−3319−4−11−1114−1−27]→reducerow[10000105001−60000]
  3. The system is consistent and has one unique solution of x=0,y=5,z=−6.
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28.

{5x−8y+35z=362x−3y+13z=14
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Solution.
  1. The augmented matrix is
    [5−835362−31314].
  2. The matrix can be reduced to RREF by Procedure 2.3.2 using the following sequence of operations.
    1. R1−2R2
    2. R2−2R1
    3. R1+2R2
    [5−835362−31314]→reducerow[10−1401−5−2]
  3. The system is consistent and has one free variable, z. The simplified system of equations is
    {x−z=4y−5z=−2
    Assigning parameter z=t and isolating the two constrained variables in the equations above leads to general solution
    x=4+ty=−2+5tz=t.
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29.

{6x1−13x2−17x3−6x4=07x1−15x2−20x3−7x4=05x1−11x2−14x3−4x4=0
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Solution.
  1. The system is homogeneous with coefficient matrix
    [6−13−17−67−15−20−75−11−14−4].
  2. The matrix can be reduced to RREF by Procedure 2.3.2 using the following sequence of operations.
    1. R1−R3
    2. R2−7R1; R3−5R1
    3. −R2
    4. R1+2R2; R3+R2
    5. −R3
    6. R1+16R3; R2+7R3
    [6−13−17−67−15−20−75−11−14−4]→reducerow[10−5001−100001]
  3. The system is consistent and has one free variable, x3. Remember that the matrices above are only the coefficient matrices of a homogeneous system; the corresponding augmented matrix is
    [10−50001−10000010]
    The simplified system of equations is
    {x1−5x3=0x2−x3=0x4=0
    Assigning parameter x3=t and isolating the constrained variables in the first two equations above leads to general solution
    x1=5t,x2=t,x3=t,x4=0.
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30.

{2x1−4x2−24x3−3x4+18x5=9−3x1+4x2+26x3+8x4−7x5=−17−1x1+2x2+12x3+2x4−7x5=−53x1−5x2−31x3−6x4+18x5=15
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Solution.
  1. The augmented matrix is
    [2−4−24−3189−34268−7−17−12122−7−53−5−31−61815].
  2. The matrix can be reduced to RREF by Procedure 2.3.2 using the following sequence of operations.
    1. −R3
    2. R1↔R3
    3. R2+3R1; R3−2R1; R4−3R1
    4. R2↔R4
    5. R1+2R2; R4+2R2
    6. R1+2R3; R4−2R3
    [2−4−24−3189−34268−7−17−12122−7−53−5−31−61815]→reducerow[10−20930150−3000014−1000000]
  3. The system is consistent and has two free variables, x3 amd x5. The simplified system of equations is
    {x1−2x3+9x5=3x2+5x3−3x5=0x4+4x5=−1
    Assigning parameters x3=s and x5=t and isolating the constrained variables in the equations above leads to general solution
    x1=3+2s−9tx2=−5s+3tx3=sx4=−1−4tx5=t.
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31. Patterns of RREF.

(a)

Determine all possible matrices that have two rows, two columns, and are in RREF. (In your example matrices, every entry will be one of must be one or must be zero or could be any value. Write an asterisk ∗ for entries that could be any value.)
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(b)

Determine all possible matrices that have three rows, three columns, and are in RREF.
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(c)

Without writing all the possibilities down, can you determine how many different possible forms of RREF matrix with four rows and four columns there are?
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(d)

Based on your answers so far, can you come up with a pattern for the number of different possible forms of RREF matrix with n rows and n columns? How does the pattern change for matrices with more rows than columns? Or for matrices with fewer rows than columns?
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32. Existence of a unique solution for two variables.

(a)

Suppose a,b,c,d are constant values so that ad−bc is not equal to zero. Use Proposition 2.5.6 to demonstrate that every system that has coefficient matrix
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[abcd]
has one unique solution.
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(b)

Now consider the geometric interpretation of Task a. Suppose we have system of equations
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{ax+by=k1cx+dy=k2.
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(i)
What are the slopes of the lines involved in the system? (For simplicity, assume both b≠0 and d≠0.)
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(ii)
Suppose the system has no solution. How must the slopes of the two lines be related?
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(iii)
Suppose the system has an infinite number of solutions. How must the slopes of the two lines be related?
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(iv)
The two cases of Task b.ii and Task b.iii can be summarized into one case as “not the case of a single unique solution.” In each of those two cases, what does the relationship between the slopes of the two lines imply about the value of ad−bc?
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33. A special type of coefficient matrix.

A matrix is called upper triangular when it has the same number of columns as rows and its entries follow the pattern:
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  • The first entry in the second row is zero.
  • The first two entries in the third row are zero.
  • The first three entries in the fourth row are zero.
  • And so on.
For example, the matrices
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[ab0c][abc0de00f][abcd0efg00hi000j]
are all upper triangular.
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Using Proposition 2.5.6 for inspiration, determine a simple condition or set of conditions by which one can predict, without performing any row operations, whether a homogeneous system of equations with an upper triangular coefficient matrix will have one unique solution or will have an infinite number of solutions.
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