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Discover Linear Algebra: A first course in linear algebra

Jeremy Sylvestre

Exercises 10.6 Exercises

Cofactor and adjoint matrices.

For each matrix:
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  1. Compute the associated cofactor matrix.
  2. Compute the associated adjoint matrix.
  3. Compute the product of the matrix and its adjoint.
  4. If the matrix is invertible, use the Inversion by adjoint formula to compute its inverse.
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1.

[−305−3025−15]
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Answer.
  1. [2253−5−40−30−90]
  2. [2−5025−40−93−30]
  3. [−305−3025−15][2−5025−40−93−30]=[900090009]
  4. The determinant of the original matrix is 9, so the matrix is invertible with inverse
    19[2−5025−40−93−30].
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2.

[−350−424425]
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Answer.
  1. [236−16−25−1526201214]
  2. [2−252036−1512−162614]
  3. [−350−424425][2−252036−1512−162614]=[174000174000174]
  4. The determinant of the original matrix is 174, so the matrix is invertible with inverse
    1174[2−252036−1512−162614].
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3.

[5−3813−2−54−9]
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Answer.
  1. [−1919195−5−5−181818]
  2. [−195−1819−51819−518]
  3. [5−3813−2−54−9][−195−1819−51819−518]=0
  4. The determinant of the original matrix is 0, so it is not invertible. Without computing the determinant, we can also see that the adjoint is not invertible because its RREF would have two rows of zeros. As an invertible matrix would have an invertible adjoint, the original matrix cannot be invertible if its adjoint is not invertible.
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4.

[−52−305−2004]
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Answer.
  1. [2000−8−20011−10−25]
  2. [20−8110−20−1000−25]
  3. [−52−305−2004][20−8110−20−1000−25]=[−100000−100000−100]
  4. The determinant of the original matrix is −100, so the matrix is invertible with inverse
    −1100[20−8110−20−1000−25].
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5.

[100−5−4031−5]
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Answer.
  1. [20−2570−5−100−4]
  2. [2000−25−507−1−4]
  3. [100−5−4031−5][2000−25−507−1−4]=[200002000020]
  4. The determinant of the original matrix is 20, so the matrix is invertible with inverse
    120[2000−25−507−1−4].
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6.

[30−1054−142]
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Answer.
  1. [−6−45−45−125−1215]
  2. Notice that the matrix of cofactors is symmetric, hence the adjoint matrix is equal to the matrix of cofactors in this case.
  3. [30−1054−142][−6−45−45−125−1215]=[−23000−23000−23]
  4. The determinant of the original matrix is −23, so the matrix is invertible with inverse
    −123[−6−45−45−125−1215].
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7.

[00011102−2−105001−4]
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Answer.
  1. [−79−4−11−2001−10000−10]
  2. [−71109−2−10−400−1−1000]
  3. [00011102−2−105001−4][−71109−2−10−400−1−1000]=[−10000−10000−10000−1]
  4. The determinant of the original matrix is −1, so the matrix is invertible with inverse
    [7−1−10−921040011000].
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8. Adjoints of special matrices.

For each statement, construct a convincing argument that it is true.
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(a)

The adjoint of a diagonal matrix is always diagonal.
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(b)

The adjoint of an upper triangular matrix is always upper triangular.
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(c)

The adjoint of an lower triangular matrix is always lower triangular.
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(d)

The adjoint of a symmetric matrix is always symmetric.
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Inverses of matrices with variable entries.

For each matrix:
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  1. State a condition on the variable x which must be satisfied in order for the matrix to be invertible.
  2. Under the assumption that your stated invertibility condition on x is satisfied, use the Inversion by adjoint formula to compute its inverse.
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9.

[x+1x−1x0]
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Answer.
  1. Invertibility requires x(1−x)≠0. That is, it requires x≠0,1.
  2. In the case x≠0,1, the inverse matrix is
    1x(1−x)[01−x−xx+1].
    (Note that in this 2×2 case, the Inversion by adjoint formula is really just our old 2×2 inversion formula.)
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10.

[x1xx+1xx+1x+21x]
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Answer.
  1. Invertibility requires x2−x−1≠0. That is, it requires
    x≠1±52.
  2. In the case x2−x−1≠0, the inverse matrix is
    −12(x2−x−1)[x2−x−10−x2+x+12x+2−2x0−x2−x+12x2−x−1].
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11.

[2x+11x−2xx+4x+20−2x+1x+112x−30x00x]
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Answer.
  1. Invertibility requires x(x+1)(x−1)2≠0. That is, it requires x≠0,±1.
  2. In the case x≠0,±1, the inverse matrix is
    [(x+2)(2x−3)(x+1)(x−1)2−1(x+1)(x−1)−x2−4(x+1)(x−1)2−2x3+3x2−9x+1x(x+1)(x−1)2−3(2x−3)(x−1)21x−13(x−2)(x−1)28x2−12x+1x(x−1)2−1x−101x−11x−1−(x+2)(2x−3)(x+1)(x−1)21(x+1)(x−1)x2−4(x+1)(x−1)23x3+2x2−10x+2x(x+1)(x−1)2].
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Cramer’s rule for 2-variable systems.

Use Cramer’s rule to solve each system.
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12.

{5x−12y=12x−5y=1.
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Solution.
Set
A=[5−122−5],b=[11].
Compute detA=−1. Since detA≠0, Cramer’s rule can be applied. Replace one column of A by b and a time and compute determinants.
Ax=[1−121−5]detAx=7Ay=[5121]detAy=3
Finally, apply Cramer’s rule:
x=detAxdetAy=detAydetA=7−1=3−1=−7,=−3.
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13.

{8x+7y=37x+7y=1.
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Solution.
Set
A=[8777],b=[31].
Compute detA=7. Since detA≠0, Cramer’s rule can be applied. Replace one column of A by b and a time and compute determinants.
Ax=[3717]detAx=14Ay=[8371]detAy=−13
Finally, apply Cramer’s rule:
x=detAxdetAy=detAydetA=147=−137.=2,
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14.

{6s+6t=57s+8t=−8.
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Solution.
Set
A=[6678],b=[5−8].
Compute detA=6. Since detA≠0, Cramer’s rule can be applied. Replace one column of A by b and a time and compute determinants.
As=[56−88]detAs=88At=[657−8]detAt=−83
Finally, apply Cramer’s rule:
s=detAsdetAt=detAtdetA=886=−836.=443,
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15.

{6α−9β=−68α+7β=−4.
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Solution.
Set
A=[6−987],b=[−6−4].
Compute detA=114. Since detA≠0, Cramer’s rule can be applied. Replace one column of A by b and a time and compute determinants.
Aα=[−6−9−47]detAα=−78Aβ=[6−68−4]detAβ=24
Finally, apply Cramer’s rule:
α=detAαdetAβ=detAβdetA=−78114=24114=−1319,=419.
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Cramer’s rule for larger systems.

For each system, use Cramer’s rule to determine the value of the specified variable in the solution to the system.
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16.

Variable x in
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{3x2y−9z=6−2x+3z=0x+y−4z=9.
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Solution.
Set
A=[32−9−20311−4],b=[609].
Compute detA=−1. Since detA≠0, Cramer’s rule can be applied. Targeting variable x, replace the first column in A with column b to obtain
Ax=[62−900391−4].
Now compute detAx=36. Therefore, the one unique solution to the system includes value
x=detAxdetA=36−1=−36.
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17.

Variable z in
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{7x−z=6−3x+3y+z=9−4x−1y+z=−11.
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Solution.
Set
\begin{array} ended with \end{bmatrix}
Compute detA=13. Since detA≠0, Cramer’s rule can be applied. Targeting variable z, replace the first third in A with column b to obtain
Az=[706−339−4−1−11].
Now compute detAz=−78. Therefore, the one unique solution to the system includes value
z=detAzdetA=−7813=−6.
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18.

Variable x2 in
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{x1+x2−2x4=−42x1+3x2−6x4=−12x1+5x2+x3−11x4=−52x1+3x2−5x4=4.
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Solution.
Set
A=[110−2230−6251−11230−5],b=[−4−1−54].
Compute detA=1. Since detA≠0, Cramer’s rule can be applied. Targeting variable x2, replace the second column in A with column b to obtain
A2=[1−40−22−10−62−51−11240−5].
Now compute detA2=17. Therefore, the one unique solution to the system includes value
x2=detA2detA=171=17.
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19.

Variable y in
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{3w+3x=1+2x−2y=−4−w+1x−y−3z=23w−3x−2z=0.
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Solution.
Set
A=[330002−20−11−1−33−30−2],b=[1−420].
Compute detA=−120. Since detA≠0, Cramer’s rule can be applied. Targeting variable y, replace the third column in A with column b to obtain
Ay=[331002−40−112−33−30−2].
Now compute detAy=−310. Therefore, the one unique solution to the system includes value
y=detAydetA=−310−120=3112.
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Determinants of products.

For each collection of matrices:
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  1. Compute the determinant of each matrix individually.
  2. Compute the product of the matrices and the determinant of the result.
  3. Verify that the determinant of the product is the product of the determinants.
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20.

A=[707−2],B=[2649]
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Answer.
  1. detA=−14, detB=−6.
  2. AB=[1442624]
    det(AB)=84
  3. (detA)(detB)=(−14)(−6)=84=det(AB)
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21.

A=[14−42],B=[−183−8],C=[−2−2−57]
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Answer.
  1. detA=18, detB=−16, detC=−24.
  2. ABC=[98−190220−356]
    det(ABC)=6912
  3. (detA)(detB)(detC)=(18)(−16)(−24)=6912=det(ABC)
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22.

A=[−10−60−6−646−8],B=[−109−4−70−1−10]
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Answer.
  1. detA=−228, detB=−27.
  2. AB=[76−930480−20−3436]
    det(AB)=6156
  3. (detA)(detB)=(−228)(−27)=6156=det(AB)
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Determinants of inverses.

For each matrix:
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  1. Compute its determinant.
  2. Compute its inverse and the determinant of the result. (You may wish to use Procedure 9.2.2 for the larger matrices.)
  3. Verify that the determinant of the inverse is the reciprocal of the determinant.
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23.

A=[126−9]
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Answer.
  1. detA=−21
  2. A−1=[−121−221−621921]
    det(A−1)=−1/21
  3. det(A−1)=1/(−21)=1/(detA)
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24.

A=[100−220−1−2−2]
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Answer.
  1. detA=−4
  2. A−1=[1001120−32−12−12]
    det(A−1)=−1/4
  3. det(A−1)=1/(−4)=1/(detA)
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25.

A=[11−2001−1−20−10−212−20]
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Answer.
  1. detA=2
  2. A−1=[−2−223−1001−2−112120−12−12]
    det(A−1)=1/2
  3. det(A−1)=1/2=1/(detA)
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Algebra of the determinant.

Let A,B,C represent square matrices of the same size where it is known that
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detA=−3,detB=−1,detC=4.
Use this information to compute the following.
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34.

det(A−1CCTC−1B)
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Answer.
12
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