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Section B.1 Sample Exercises with Solutions

Here we model one exercise and solution for each learning objective. Your solutions should not look identical to those shown below, but these solutions can give you an idea of the level of detail required for a complete solution.

Example B.1.1. LE1.

Consider the scalar system of equations

3x1+2x2+x4=1x14x2+x37x4=0x2x3=2
  1. Rewrite this system as a vector equation.

  2. Write an augmented matrix corresponding to this system.

Solution.
  1. x1[310]+x2[241]+x3[111]+x4[170]=[102]
  2. [320111417001102]

Example B.1.2. LE2.

  1. For each of the following matrices, explain why it is not in reduced row echelon form.

    A=[404012000000000]B=[012103000000000]C=[144010000000000]
  2. Show step-by-step why

    RREF[031212132411]=[100401010015].
Solution.
    • A=[404012000000000] is not in reduced row echelon form because the pivots are not all 1.

    • B=[012103000000000] is not in reduced row echelon form because the pivots are not descending to the right.

    • C=[144010000000000] is not in reduced row echelon form because not every entry above and below each pivot is zero.

  1. [031212132411][121303122411]Swap Rows 1 and 2[121303120015]Add 2 Row 1 to Row 3[12130113230015]Multiply Row 3 by 13[10531330113230015]Add 2 Row 2 to Row 1[105313301010015]Add 13 Row 3 to Row 2[100401010015]Add 53 Row 3 to Row 1

Example B.1.3. LE3.

Consider each of the following systems of linear equations or vector equations.

  1. 2x1+x2+x3=22x13x23x3=03x1+x2+x3=3
  2. x1[531]+x2[322]+x3[1497]=[104]
  3. x1[011]+x2[144]+x3[243]=[5118]
  • Explain how to find a simpler system or vector equation that has the same solution set for each.

  • Explain whether each solution set has no solutions, one solution, or infinitely-many solutions. If the set is finite, describe it using set notation.

Solution.
  1. RREF[211223303113]=[100001100001]

    This matrix corresponds to the simpler system

    x1=0x2+x3=00=1

    The third equation 0=1 indicates that the system has no solutions. The solution set is .

  2. RREF[5314132901274]=[101201330000]

    This matrix corresponds to the simpler system

    x1x3=2x2+3x3=30=0.

    Since there are three variables and two nontrivial equations, the solution set has infinitely-many solutions.

  3. RREF[0125144111438]=[100301010013]

    This matrix corresponds to the simpler system

    x1=3x2=1x3=3.

    This system has one solution. The solution set is {[313]}.

Example B.1.4. LE4.

Consider the following vector equation.

x1[304]+x2[304]+x3[010]+x4[455]=[11914]
  1. Explain how to find a simpler system or vector equation that has the same solution set.

  2. Explain how to describe this solution set using set notation.

Solution.

First, we compute

RREF[33041100159440514]=[110010010100012].

This corresponds to the simpler system

x1+x2=1x3=1x4=2.

Since the second column is a non-pivot column, we let x2=a. Making this substitution and then solving for x1, x3, and x4 produces the system

x1=1ax2=ax3=1x4=2

Thus, the solution set is {[a+1a12]|aR}.

Example B.1.5. VS1.

Let V be the set of all pairs of numbers (x,y) of real numbers together with the following operations:

(x1,y1)(x2,y2)=(2x1+2x2,2y1+2y2)c(x,y)=(cx,c2y)
  1. Show that scalar multiplication distributes over vector addition:

    c((x1,y1)(x2,y2))=c(x1,y1)c(x2,y2)
  2. Explain why V nonetheless is not a vector space.

Solution.
  1. We compute both sides:

    c((x1,y1)(x2,y2))=c(2x1+2x2,2y1+2y2)=(c(2x1+2x2),c2(2y1+2y2))=(2cx1+2cx2,2c2y1+2c2y2)

    and

    c(x1,y1)c(x2,y2)=(cx1,c2y1)(cx2,c2y2)=(2cx1+2cx2,2c2y1+2c2y2)

    Since these are the same, we have shown that the property holds.

  2. To show V is not a vector space, we must show that it fails one of the 8 defining properties of vector spaces. We will show that scalar multiplication does not distribute over scalar addition, i.e., there are values such that

    (c+d)(x,y)c(x,y)d(x,y)
    • (Solution method 1) First, we compute

      (c+d)(x,y)=((c+d)x,(c+d)2y)=((c+d)x,(c2+2cd+d2)y).

      Then we compute

      c(x,y)d(x,y)=(cx,c2y)(dx,d2y)=(2cx+2dx,2c2y+2d2y).

      Since (c+d)x2cx+2dy when c,d,x,y=1, the property fails to hold.

    • (Solution method 2) When we let c,d,x,y=1, we may simplify both sides as follows.

      (c+d)(x,y)=2(1,1)=(21,221)=(2,4)
      c(x,y)d(x,y)=1(1,1)1(1,1)=(11,121)(11,121)=(1,1)(1,1)=(21+21,21+21)=(4,4)

      Since these ordered pairs are different, the property fails to hold.

Example B.1.6. VS2.

  1. Write a statement involving the solutions of a vector equation that's equivalent to each claim below.

    • [13313]is a linear combination of the vectors [101],[202],[303], and [515].

    • [13315]is a linear combination of the vectors [101],[202],[303], and [515].

  2. Use these statements to determine if each vector is or is not a linear combination. If it is, give an example of such a linear combination.

Solution.
  • [13313]is a linear combination of the vectors [101],[202],[303], and [515] exactly when the vector equation

    x1[101]+x2[202]+x3[303]+x4[515]=[13313]

    has a solution. To solve this vector equation, we compute

    RREF[12351300013123513]=[123020001300000].

    We see that this vector equation has solution set {[22a3bab3] | a,bR}, so [13313] is a linear combination; for example, 2[101]+3[515]=[13313]

  • [13315] is a linear combination of the vectors [101],[202],[303], and [515] exactly when the vector equation

    x1[101]+x2[202]+x3[303]+x4[515]=[13315]

    has a solution. To solve this vector equation, we compute

    RREF[12351300013123515]=[123000001000001].

    This vector equation has no solution, so [13315] is not a linear combination.

Example B.1.7. VS3.

  1. Write a statement involving the solutions of a vector equation that's equivalent to each claim below.

    • The set of vectors {[1120],[3233],[107119],[6339]} spans R4.

    • The set of vectors {[1120],[3233],[107119],[6339]} does not span R4.

  2. Explain how to determine which of these statements is true.

Solution.

The set of vectors {[1120],[3233],[107119],[6339]} spans R4 exactly when the vector equation

x1[1120]+x2[3233]+x3[107119]+x4[6339]=v

has a solution for all vR4. If there is some vector vR4 for which this vector equation has no solution, then the set does not span R4. To answer this, we compute

RREF[131061273231130399]=[1013013300000000].

We see that for some vR4, this vector equation will not have a solution, so the set of vectors {[1120],[3233],[107119],[6339]} does not span R4.

Example B.1.8. VS4.

Consider the following two sets of Euclidean vectors.

W={[xyzw]|x+y=3z+2w}U={[xyzw]|x+y=3z+w2}

Explain why one of these sets is a subspace of R3, and why the other is not.

Solution.

To show that W is a subspace, let v=[x1y1z1w1]W and w=[x2y2z2w2]W, so we know that x1+y1=3z1+2w1 and x2+y2=3z2+2w2.

Consider

[x1y1z1w1]+[x2y2z2w2]=[x1+x2y1+y2z1+z2w1+w2].

To see if v+wW, we need to check if (x1+x2)+(y1+y2)=3(z1+z2)+2(w1+w2). We compute

(x1+x2)+(y1+y2)=(x1+y1)+(x2+y2)by regrouping=(3z1+2w1)+(3z2+2w2)since =3(z1+z2)+2(w1+w2)by regrouping.

Thus v+wW, so W is closed under vector addition.

Now consider

cv=[cx1cy1cz1cw1].

Similarly, to check that cvW, we need to check if cx1+cy1=3(cz1)+2(cw1), so we compute

cx1+cy1=c(x1+y1)by factoring=c(3z1+2w1)since =3(cz1)+2(cw1)by regrouping

and we see that cvW, so W is closed under scalar multiplication. Therefore W is a subspace of R3.

Now, to show U is not a subspace, we will show that it is not closed under vector addition.

  • (Solution Method 1) Now let v=[x1y1z1w1]U and w=[x2y2z2w2]U, so we know that x1+y1=3z1+w12 and x2+y2=3z2+w22.

    Consider

    v+w=[x1y1z1w1]+[x2y2z2w2]=[x1+x2y1+y2z1+z2w1+w2].

    To see if v+wU, we need to check if (x1+x2)+(y1+y2)=3(z1+z2)+(w1+w2)2. We compute

    (x1+x2)+(y1+y2)=(x1+y1)+(x2+y2)by regrouping=(3z1+w12)+(3z2+w22)since =3(z1+z2)+(w12+w22)by regrouping

    and thus v+wU \textbf{only when} w12+w22=(w1+w2)2. Since this is not true in general, U is not closed under vector addition, and thus cannot be a subspace.

  • (Solution Method 2) Note that the vector v=[0101] belongs to U since 0+1=3(0)+12. However, the vector 2v=[0202] does not belong to U since 0+23(0)+22. Therefore U is not closed under scalar multiplication, and thus is not a subspace.

Example B.1.9. VS5.

  1. Write a statement involving the solutions of a vector equation that's equivalent to each claim below.

    • The set of vectors {[1344],[1344],[0133]} is linearly independent.

    • The set of vectors {[1344],[1344],[0133]} is linearly dependent.

  2. Explain how to determine which of these statements is true.

Solution.

The set of vectors {[1344],[1344],[0133]} is linearly independent exactly when the vector equation

x1[1344]+x2[1344]+x3[0133]=[0000]

has no non-trivial (i.e. nonzero) solutions. The set is linearly dependent when there exists a nontrivial (i.e. nonzero) solution. We compute

RREF[110331443443]=[110001000000].

Thus, this vector equation has a solution set {[aa0] | aR}. Since there are nontrivial solutions, we conclude that the set of vectors {[1344],[1344],[0133]} is linearly dependent.

Example B.1.10. VS6.

  1. Write a statement involving spanning and independence properties that's equivalent to each claim below.

    • The set of vectors {[1344],[0133],[3111818],[271111]} is a basis of R4.

    • The set of vectors {[1344],[0133],[3111818],[271111]} is not a basis of R4.

  2. Explain how to determine which of these statements is true.

Solution.

The set of vectors {[1344],[0133],[3111818],[271111]} is a basis of R4 exactly when it is linearly independent and the set spans R4. If it is either linearly dependent, or the set does not span R4, then the set is not a basis.

To answer this, we compute

RREF[103231117431811431811]=[1032012100000000].

We see that this set of vectors is linearly dependent, so therefore the set of vectors {[1344],[0133],[3111818],[271111]} is not a basis.

Example B.1.11. VS7.

Consider the subspace

W=span{[1312],[1012],[3612],[1611],[2301]}.
  1. Explain how to find a basis of W.

  2. Explain how to find the dimension of W.

Solution.
  1. Observe that

    RREF[11312306631111022211]=[10201011000001100000]

    If we remove the vectors yielding non-pivot columns, the resulting set will span the same vectors while being linearly independent. Therefore

    {[1312],[1012],[1611]}

    is a basis of W.

  2. Since this (and thus every other) basis has three vectors in it, the dimension of W is 3.

Example B.1.12. VS8.

  1. Given the set

    {x32x2+x+2,2x21,x3+3x2+3x2,x36x2+9x+5}

    write a statement involving the solutions to a polynomial equation that's equivalent to each claim below.

    • The set of polynomials is linearly independent.

    • The set of polynomials is linearly dependent.

  2. Explain how to determine which of these statements is true.

Solution.

The set of polynomials

{x32x2+x+2,2x21,x3+3x2+3x2,x36x2+9x+5}

is linearly independent exactly when the polynomial equation

y1(x32x2+x+2)+y2(2x21)+y3(x3+3x2+3x2)+y4(x36x2+9x+5)=0

has no nontrivial (i.e. nonzero) solutions. The set is linearly dependent when this equation has a nontrivial (i.e. nonzero) solution.

To solve this equation, we distribute and then collect coefficients to obtain

(y1y3+y4)x3+(2y1+2y2+3y36y4)x2+(y1+3y3+9y4)x+(2y1y22y3+5y4)=0.

These polynomials are equal precisely when their coefficients are equal, leading to the system

y1y3+y4=02y1+2y2+3y36y4=0y1++3y3+9y4=02y1y22y3+5y4=0.

To solve this, we compute

RREF[10110223601039021250]=[10030010300012000000]

The system has (infintely many) nontrivial solutions, so we that the set of polynomials is linearly dependent.

Example B.1.13. VS9.

Consider the homogeneous system of equations

x1+x2+3x3+x4+2x5=03x16x3+6x4+3x5=0x1+x2x3+x4=02x12x2+2x3x4+x5=0
  1. Find the solution space of the system.

  2. Find a basis of the solution space.

Solution.
  1. Observe that

    RREF[113120306630111100222110]=[102010011000000110000000]

    Letting x3=a and x5=b (since those correspond to the non-pivot columns), this is equivalent to the system

    x1+2x3+x5=0x2+x3=0x3=ax4+x5=0x5=b

    Thus, the solution set is

    {[2abaabb]|a,bR}.
  2. Since we can write

    [2abaabb]=a[21100]+b[10011],

    a basis for the solution space is

    {[21100],[10011]}.

Example B.1.14. AT1.

Consider the following maps of polynomials S:PP and T:PP defined by

S(f(x))=3xf(x) and T(f(x))=3f(x)f(x).

Explain why one of these maps is a linear transformation, and why the other map is not.

Solution.

To show S is a linear transformation, we must show two things:

S(f(x)+g(x))=S(f(x))+s(g(x))
S(cf(x))=cS(f(x))

To show S respects addition, we compute

S(f(x)+g(x))=3x(f(x)+g(x))by definition of =3xf(x)+3xg(x)by distributing

But note that S(f(x))=3xf(x) and S(g(x))=3xg(x), so we have S(f(x)+g(x))=S(f(x))+S(g(x)).

For the second part, we compute

S(cf(x))=3x(cf(x))by definition of =3cxf(x)rewriting the multiplication.

But note that cS(f(x))=c(3xf(x))=3cxf(x) as well, so we have S(cf(x))=cS(f(x)). Now, since S respects both addition and scalar multiplication, we can conclude S is a linear transformation.

  • (Solution method 1) As for T, we compute

    T(f(x)+g(x))=3(f(x)+g(x))(f(x)+g(x))by definition of =3(f(x)+g(x))(f(x)+g(x))since the derivative is linear=3f(x)f(x)+3f(x)g(x)+3f(x)g(x)+3g(x)g(x)by distributing

    However, note that T(f(x))+T(g(x))=3f(x)f(x)+3g(x)g(x), which is not always the same polynomial (for example, when f(x)=g(x)=x). So we see that T(f(x)+g(x))T(f(x))+T(g(x)), so T does not respect addition and is therefore not a linear transformation.

  • (Solution method 2) As for T, we may choose the polynomial f(x)=x and scalar c=2. Then

    T(cf(x))=T(2x)=3(2x)(2x)=3(2)(2x)=12x.

    But on the other hand,

    cT(f(x))=2T(x)=2(3)(x)(x)=2(3)(1)(x)=6x.

    Since this isn't the same polynomial, T does not preserve multiplication and is therefore not a linear transformation.

Example B.1.15. AT2.

  1. Find the standard matrix for the linear transformation T:R3R4 given by

    T([xyz])=[x+yx+3yz7x+y+3z0].
  2. Let S:R4R3 be the linear transformation given by the standard matrix

    [234101113224].

    Compute S([2132]).

Solution.
  1. Since

    T([100])=[1170]T([010])=[1310]T([001])=[0130],

    the standard matrix for T is [110131713000].

  2. S([2132])=2S(e1)+S(e2)+3S(e3)+2S(e4)
    =2[203]+[312]+3[412]+2[114]=[1346].

Example B.1.16. AT3.

Let T:R4R3 be the linear transformation given by

T([xyzw])=[x+3y+2z3w2x+4y+6z10wx+6yz+3w]
  1. Explain how to find the image of T and the kernel of T.

  2. Explain how to find a basis of the image of T and a basis of the kernel of T.

  3. Explain how to find the rank and nullity of T, and why the rank-nullity theorem holds for T.

Solution.
  1. To find the image we compute

    Im(T)=T(span{e1,e2,e3,e4})
    =span{T(e1),T(e2),T(e3),T(e4)}
    =span{[121],[346],[261],[3103]}.
  2. The kernel is the solution set of the corresponding homogeneous system of equations, i.e.

    x+3y+2z3w=02x+4y+6z10w=0x+6yz+3w=0.

    So we compute

    RREF[1323024610016130]=[105900112000000].

    Then, letting z=a and w=b we have

    kerT={[5a+9ba2bab]|a,bR}.
  3. Since Im(T)=span{[121],[346],[261],[3103]}, we simply need to find a linearly independent subset of these four spanning vectors. So we compute

    RREF[1323246101613]=[105901120000].

    Since the first two columns are pivot columns, they form a linearly independent spanning set, so a basis for ImT is {[121],[346]}.

    To find a basis for the kernel, note that

    kerT={[5a+9ba2bab]|a,bR}
    ={a[5110]+b[9201]|a,bR}
    =span{[5110],[9201]}.

    so a basis for the kernel is

    {[5110],[9201]}.
  4. The dimension of the image (the rank) is 2, the dimension of the kernel (the nullity) is 2, and the dimension of the domain of T is 4, so we see 2+2=4, which verifies that the sum of the rank and nullity of T is the dimension of the domain of T.

Example B.1.17. AT4.

Let T:R4R3 be the linear transformation given by the standard matrix [1323246101613].

  1. Explain why T is or is not injective.

  2. Explain why T is or is not surjective.

Solution.

Compute

RREF[1323246101613]=[105901120000].
  1. Note that the third and fourth columns are non-pivot columns, which means kerT contains infinitely many vectors, so T is not injective.

  2. Since there are only two pivots, the image (i.e. the span of the columns) is a 2-dimensional subspace (and thus does not equal R3), so T is not surjective.

Example B.1.18. MX1.

Of the following three matrices, only two may be multiplied.

A=[1301]B=[412]C=[013125]

Explain which two may be multiplied and why. Then show how to find their product.

Solution.

AC is the only one that can be computed, since C corresponds to a linear transformation R3R2 and A corresponds to a linear transfromation R2R2. Thus the composition AC corresponds to a linear transformation R3R2 with a 2×3 standard matrix. We compute

AC(e1)=A([01])=0[10]+1[31]=[31]AC(e2)=A([12])=1[10]2[31]=[72]AC(e3)=A([35])=3[10]+5[31]=[125].

Thus

AC=[3712125].

Example B.1.19. MX2.

Let A be a 4×4 matrix.

  1. Give a 4×4 matrix P that may be used to perform the row operation R3R3+4R1.

  2. Give a 4×4 matrix Q that may be used to perform the row operation R14R1.

  3. Use matrix multiplication to describe the matrix obtained by applying R34R1+R3 and then R14R1 to A (note the order).

Solution.
  1. P=[1000010040100001]

  2. Q=[4000010000100001]

  3. QPA

Example B.1.20. MX3.

Explain why each of the following matrices is or is not invertible by disussing its corresponding linear transformation. If the matrix is invertible, explain how to find its inverse.

D=[1102255423204435]N=[391113921339315412216]
Solution.

We compute

RREF(D)=[1000010000100001].

We see D is bijective, and therefore invertible. To compute the inverse, we solve Dx=e1 by computing

RREF[11021255402320044350]=[10002101003800103600018].

Similarly, we solve Dx=e2 by computing

RREF[11020255412320044350]=[1000801001400101300013].

Similarly, we solve Dx=e3 by computing

RREF[11020255402320144350]=[10002301004100103900019].

Similarly, we solve Dx=e4 by computing

RREF[11020255402320044351]=[10002010040010400011].

Combining these, we obtain

D1=[218232381441436133948391].

We compute

RREF(N)=[1303001200000000].

We see N is not bijective and thus is not invertible.

Example B.1.21. GT1.

Let A be a 4×4 matrix with determinant 7.

  1. Let B be the matrix obtained from A by applying the row operation R3R3+3R4. What is det(B)?

  2. Let C be the matrix obtained from A by applying the row operation R23R2. What is det(C)?

  3. Let D be the matrix obtained from A by applying the row operation R3R4. What is det(D)?

Solution.
  1. Adding a multiple of one row to another row does not change the determinant, so det(B)=det(A)=7.

  2. Scaling a row scales the determinant by the same factor, so so det(B)=3det(A)=3(7)=21.

  3. Swaping rows changes the sign of the determinant, so det(B)=det(A)=7.

Example B.1.22. GT2.

Show how to compute the determinant of the matrix

A=[1301112411133125]
Solution.

Here is one possible solution, first applying a single row operation, and then performing Laplace/cofactor expansions to reduce the determinant to a linear combination of 2×2 determinants:

det[1301112411133125]=det[1301001111133125]=(1)det[131113315]+(1)det[130111312]=(1)((1)det[1315](1)det[3115]+(3)det[3113])+==(1)((1)det[1112](3)det[1132])=(1)(8+1430)+(1)(115)=10

Here is another possible solution, using row and column operations to first reduce the determinant to a 3×3 matrix and then applying a formula:

det[1301112411133125]=det[1301001111133125]=det[1301001011123127]=det[1301111200103127]=det[131112317]=((7181)(3+221))=10

Example B.1.23. GT3.

Explain how to find the eigenvalues of the matrix [22107].

Solution.

Compute the characteristic polynomial:

det(AλI)=det[2λ2107λ]
=(2λ)(7λ)+20=λ25λ+6=(λ2)(λ3)

The eigenvalues are the roots of the characteristic polynomial, namely 2 and 3.

Example B.1.24. GT4.

Explain how to find a basis for the eigenspace associated to the eigenvalue 3 in the matrix

[78289113252].
Solution.

The eigenspace associated to 3 is the kernel of A3I, so we compute

RREF(A3I)=RREF[73828931132523]=
RREF[108286113251]=[1010132000].

Thus we see the kernel is

{[a32aa]|aR}

which has a basis of {[1321]}.