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Chapter 4 · Determinants

Exercise 4.5 · Q15: Find A−1A^{-1} and use it to solve a system

Exercise 4.5Solving systems AX = B · Section 4.6.1
The question

15. If

A=[2−3532−411−2],A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix},

find A−1A^{-1}⁠. Using A−1A^{-1} solve the system of equations

2x−3y+5z=113x+2y−4z=−5x+y−2z=−3\begin{aligned} 2x - 3y + 5z &= 11\\ 3x + 2y - 4z &= -5\\ x + y - 2z &= -3 \end{aligned}
Exam answer

Expanding along R1R_1⁠:

∣A∣=2∣2−41−2∣−(−3)∣3−41−2∣+5∣3211∣=2(0)+3(−2)+5(1)=−1\begin{aligned} |A| &= 2\begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} - (-3)\begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} \\ &\quad + 5\begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} \\ &= 2(0) + 3(-2) + 5(1) \\ &= -1 \end{aligned}

∣A∣=−1≠0|A| = -1 \ne 0⁠. So A−1A^{-1} exists.

Cofactors:

A11=∣2−41−2∣=0A12=−∣3−41−2∣=2A13=∣3211∣=1\begin{aligned} A_{11} &= \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} = 0 \\ A_{12} &= -\begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} = 2 \\ A_{13} &= \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} = 1 \end{aligned} A21=−∣−351−2∣=−1A22=∣251−2∣=−9A23=−∣2−311∣=−5\begin{aligned} A_{21} &= -\begin{vmatrix} -3 & 5 \\ 1 & -2 \end{vmatrix} = -1 \\ A_{22} &= \begin{vmatrix} 2 & 5 \\ 1 & -2 \end{vmatrix} = -9 \\ A_{23} &= -\begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} = -5 \end{aligned} A31=∣−352−4∣=2A32=−∣253−4∣=23A33=∣2−332∣=13\begin{aligned} A_{31} &= \begin{vmatrix} -3 & 5 \\ 2 & -4 \end{vmatrix} = 2 \\ A_{32} &= -\begin{vmatrix} 2 & 5 \\ 3 & -4 \end{vmatrix} = 23 \\ A_{33} &= \begin{vmatrix} 2 & -3 \\ 3 & 2 \end{vmatrix} = 13 \end{aligned} adj⁡A=[0−122−9231−513]A−1=1∣A∣adj⁡A=−adj⁡A=[01−2−29−23−15−13]\begin{aligned} \operatorname{adj}A &= \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} \\ A^{-1} &= \frac{1}{|A|}\operatorname{adj}A = -\operatorname{adj}A \\ &= \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \end{aligned}

The system is AX=BAX = B⁠, where AA is the given matrix and

X=[xyz], B=[11−5−3]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\ B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}

As A−1A^{-1} exists, the system has a unique solution, X=A−1BX = A^{-1}B⁠.

X=A−1B=[01−2−29−23−15−13][11−5−3]=[0−5+6−22−45+69−11−25+39]=[123]\begin{aligned} X &= A^{-1}B \\ &= \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}\begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \\ &= \begin{bmatrix} 0 - 5 + 6 \\ -22 - 45 + 69 \\ -11 - 25 + 39 \end{bmatrix} \\ &= \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \end{aligned}

∴ x=1, y=2, z=3\therefore\ x = 1,\ y = 2,\ z = 3⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The coefficients of the system are exactly the entries of the given matrix AA⁠, row for row. So the system is AX=BAX = B with this same AA⁠, and once A−1A^{-1} is known the solution is simply X=A−1BX = A^{-1}B: one matrix times a column. Here ∣A∣=−1|A| = -1⁠, so A−1=−adj⁡AA^{-1} = -\operatorname{adj} A⁠, and every entry of the adjoint just changes sign.

Given, to find, and the plan
Given
The 3×33\times3 matrix AA and the system above.
To find
(a) A−1A^{-1}⁠; (b) xx⁠, yy⁠, zz⁠, using A−1A^{-1}⁠.
Plan
(1) ∣A∣|A|⁠, expanding along row 1. (2) All nine minors and cofactors. (3) adj⁡A\operatorname{adj} A and A−1A^{-1}⁠. (4) Check AA−1=IAA^{-1} = I⁠. (5) Write the system as AX=BAX = B and confirm the coefficient matrix is the given AA⁠. (6) X=A−1BX = A^{-1}B⁠. (7) Check in the three equations.

Basics you need

1Expanding a 3×33\times3 determinant

Along row 1: ∣A∣=a11M11−a12M12+a13M13|A| = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13}⁠. Basics: the + − + sign pattern

2Minor, cofactor, adjoint

MijM_{ij}⁠: delete row ii⁠, column jj⁠. Aij=(−1)i+jMijA_{ij} = (-1)^{i+j}M_{ij}⁠. adj⁡A\operatorname{adj} A = transpose of [Aij][A_{ij}]⁠. Basics: where the determinant comes from

3The inverse

If ∣A∣≠0|A| \ne 0⁠, A−1=1∣A∣adj⁡AA^{-1} = \frac{1}{|A|}\operatorname{adj} A⁠. With ∣A∣=−1|A| = -1⁠, that is −adj⁡A-\operatorname{adj} A⁠. Basics: why the inverse divides by |A|

4A system is a matrix product

Row ii of the coefficient matrix times XX is the left side of equation ii⁠. Basics: matrix facts the chapter uses

Solution

  1. Compute ∣A∣|A| by expanding along row 1

    ∣A∣=2∣2−41−2∣−(−3)∣3−41−2∣+5∣3211∣=2(0)−(−3)(−2)+5(1)=0−6+5=−1.\begin{aligned} |A| &= 2\begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} - (-3)\begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix}\\ &\qquad + 5\begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix}\\ &= 2(0) - (-3)(-2) + 5(1)\\ &= 0 - 6 + 5 = -1. \end{aligned}

    (The minors 00⁠, −2-2⁠, 11 are worked out in Step 2.) ∣A∣=−1≠0|A| = -1 \ne 0⁠, so AA is nonsingular and A−1A^{-1} exists.

    In the middle term three signs meet: the position sign −-⁠, the entry −3-3⁠, and the minor −2-2⁠. Their product is −(−3)(−2)=−6-(-3)(-2) = -6⁠. A negative determinant is perfectly fine; only ∣A∣=0|A| = 0 would mean no inverse.
    Section 4.2.3; Theorem 4, Section 4.5 (invertible exactly when nonsingular). Lesson: expanding a 3 × 3 determinant

  2. Find all nine minors and cofactors

    Row 1 (entries 2,−3,52, -3, 5⁠):

    M11=∣2−41−2∣=(2)(−2)−(−4)(1)=0,A11=(−1)1+1(0)=0,M12=∣3−41−2∣=(3)(−2)−(−4)(1)=−2,A12=(−1)1+2(−2)=2,M13=∣3211∣=(3)(1)−(2)(1)=1,A13=(−1)1+3(1)=1.\begin{aligned} M_{11} &= \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix}\\ &= (2)(-2) - (-4)(1) = 0,\\ A_{11} &= (-1)^{1+1}(0) = 0,\\ M_{12} &= \begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix}\\ &= (3)(-2) - (-4)(1) = -2,\\ A_{12} &= (-1)^{1+2}(-2) = 2,\\ M_{13} &= \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix}\\ &= (3)(1) - (2)(1) = 1,\\ A_{13} &= (-1)^{1+3}(1) = 1. \end{aligned}

    Row 2 (entries 3,2,−43, 2, -4⁠):

    M21=∣−351−2∣=(−3)(−2)−(5)(1)=1,A21=(−1)2+1(1)=−1,M22=∣251−2∣=(2)(−2)−(5)(1)=−9,A22=(−1)2+2(−9)=−9,M23=∣2−311∣=(2)(1)−(−3)(1)=5,A23=(−1)2+3(5)=−5.\begin{aligned} M_{21} &= \begin{vmatrix} -3 & 5 \\ 1 & -2 \end{vmatrix}\\ &= (-3)(-2) - (5)(1) = 1,\\ A_{21} &= (-1)^{2+1}(1) = -1,\\ M_{22} &= \begin{vmatrix} 2 & 5 \\ 1 & -2 \end{vmatrix}\\ &= (2)(-2) - (5)(1) = -9,\\ A_{22} &= (-1)^{2+2}(-9) = -9,\\ M_{23} &= \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix}\\ &= (2)(1) - (-3)(1) = 5,\\ A_{23} &= (-1)^{2+3}(5) = -5. \end{aligned}

    Row 3 (entries 1,1,−21, 1, -2⁠):

    M31=∣−352−4∣=(−3)(−4)−(5)(2)=2,A31=(−1)3+1(2)=2,M32=∣253−4∣=(2)(−4)−(5)(3)=−23,A32=(−1)3+2(−23)=23,M33=∣2−332∣=(2)(2)−(−3)(3)=13,A33=(−1)3+3(13)=13.\begin{aligned} M_{31} &= \begin{vmatrix} -3 & 5 \\ 2 & -4 \end{vmatrix}\\ &= (-3)(-4) - (5)(2) = 2,\\ A_{31} &= (-1)^{3+1}(2) = 2,\\ M_{32} &= \begin{vmatrix} 2 & 5 \\ 3 & -4 \end{vmatrix}\\ &= (2)(-4) - (5)(3) = -23,\\ A_{32} &= (-1)^{3+2}(-23) = 23,\\ M_{33} &= \begin{vmatrix} 2 & -3 \\ 3 & 2 \end{vmatrix}\\ &= (2)(2) - (-3)(3) = 13,\\ A_{33} &= (-1)^{3+3}(13) = 13. \end{aligned}
    M11=0M_{11} = 0 because the rows (2,−4)(2, -4) and (1,−2)(1, -2) are proportional (the first is twice the second). A zero cofactor is allowed; it just puts a 0 into the adjoint. Each sign (−1)i+j(-1)^{i+j} is written explicitly so none is forgotten.
    Definitions 1 and 2, Section 4.4. Lesson: minors and cofactors

  3. Build adj⁡A\operatorname{adj} A and A−1A^{-1}

    Matrix of cofactors:

    [021−1−9−522313]\begin{bmatrix} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{bmatrix}

    Transpose it:

    adj⁡A=[0−122−9231−513].\operatorname{adj} A = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}.

    Since ∣A∣=−1|A| = -1⁠, dividing by ∣A∣|A| changes every sign:

    A−1=1−1adj⁡A=[01−2−29−23−15−13].A^{-1} = \frac{1}{-1}\operatorname{adj} A = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}.

    This answers the first part of the question.

    The adjoint is the transpose of the cofactor matrix, so the cofactors of row 1 of AA (0,2,10, 2, 1⁠) fill column 1. Then 1∣A∣=1−1=−1\frac{1}{|A|} = \frac{1}{-1} = -1 multiplies every entry, so the inverse has whole-number entries. This happens whenever ∣A∣=±1|A| = \pm1⁠.
    Definition 3, Section 4.5.1; Theorem 4, Section 4.5. Lesson: the adjoint

  4. Check: AA−1=IAA^{-1} = I

    Row ii of AA times column jj of A−1A^{-1}⁠:

    (1,1): 2(0)+(−3)(−2)+5(−1)=0+6−5=1,(1,2): 2(1)+(−3)(9)+5(5)=2−27+25=0,(1,3): 2(−2)+(−3)(−23)+5(−13)=−4+69−65=0,(2,1): 3(0)+2(−2)+(−4)(−1)=0−4+4=0,(2,2): 3(1)+2(9)+(−4)(5)=3+18−20=1,(2,3): 3(−2)+2(−23)+(−4)(−13)=−6−46+52=0,(3,1): 1(0)+1(−2)+(−2)(−1)=0−2+2=0,(3,2): 1(1)+1(9)+(−2)(5)=1+9−10=0,(3,3): 1(−2)+1(−23)+(−2)(−13)=−2−23+26=1.\begin{aligned} (1,1)&:\ 2(0) + (-3)(-2) + 5(-1)\\ &= 0 + 6 - 5 = 1,\\ (1,2)&:\ 2(1) + (-3)(9) + 5(5)\\ &= 2 - 27 + 25 = 0,\\ (1,3)&:\ 2(-2) + (-3)(-23) + 5(-13)\\ &= -4 + 69 - 65 = 0,\\ (2,1)&:\ 3(0) + 2(-2) + (-4)(-1)\\ &= 0 - 4 + 4 = 0,\\ (2,2)&:\ 3(1) + 2(9) + (-4)(5)\\ &= 3 + 18 - 20 = 1,\\ (2,3)&:\ 3(-2) + 2(-23) + (-4)(-13)\\ &= -6 - 46 + 52 = 0,\\ (3,1)&:\ 1(0) + 1(-2) + (-2)(-1)\\ &= 0 - 2 + 2 = 0,\\ (3,2)&:\ 1(1) + 1(9) + (-2)(5)\\ &= 1 + 9 - 10 = 0,\\ (3,3)&:\ 1(-2) + 1(-23) + (-2)(-13)\\ &= -2 - 23 + 26 = 1. \end{aligned}

    So

    AA−1=[100010001]=I.AA^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I.
    Because A−1A^{-1} has whole-number entries, we can check AA−1=IAA^{-1} = I directly. Each 1 on the diagonal is 1∣A∣\frac{1}{|A|} times "a row times its own cofactors" =1−1(−1)= \frac{1}{-1}(-1)⁠; each 0 is "a row times another row's cofactors".
    Theorem 1, Section 4.5; Note in Section 4.4. Basics: why the inverse divides by |A|

  5. Write the system as AX=BAX = B

    Read the coefficients of x,y,zx, y, z row by row: (2,−3,5)(2, -3, 5)⁠, (3,2,−4)(3, 2, -4)⁠, (1,1,−2)(1, 1, -2)⁠. These are exactly the rows of the given AA⁠. So the system is AX=BAX = B with

    X=[xyz],B=[11−5−3],X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\qquad B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}, AX=[2x−3y+5z3x+2y−4zx+y−2z].AX = \begin{bmatrix} 2x - 3y + 5z \\ 3x + 2y - 4z \\ x + y - 2z \end{bmatrix}.

    AA is 3×33\times3 and XX is 3×13\times1⁠, so AXAX is 3×13\times1⁠, the order of BB⁠.

    The phrase "using A−1A^{-1}" only makes sense if the coefficient matrix of the system is the given AA⁠, so we confirm it entry by entry. Had the coefficients been arranged differently (for example as the columns of AA⁠), we would need the inverse of A′A' instead.
    Section 4.6.1 (matrix form of a system). Lesson: solving systems AX = B

  6. Compute X=A−1BX = A^{-1}B

    A−1A^{-1} is 3×33\times3⁠, BB is 3×13\times1⁠, so A−1BA^{-1}B is 3×13\times1⁠:

    row 1: 0(11)+1(−5)+(−2)(−3)=0−5+6=1,row 2: (−2)(11)+9(−5)+(−23)(−3)=−22−45+69=2,row 3: (−1)(11)+5(−5)+(−13)(−3)=−11−25+39=3.\begin{aligned} \text{row 1}&:\ 0(11) + 1(-5) + (-2)(-3)\\ &= 0 - 5 + 6 = 1,\\ \text{row 2}&:\ (-2)(11) + 9(-5) + (-23)(-3)\\ &= -22 - 45 + 69 = 2,\\ \text{row 3}&:\ (-1)(11) + 5(-5) + (-13)(-3)\\ &= -11 - 25 + 39 = 3. \end{aligned} X=[xyz]=[123].X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}.

    So x=1x = 1⁠, y=2y = 2⁠, z=3z = 3.

    Since AA is nonsingular, AX=BAX = B gives A−1(AX)=A−1BA^{-1}(AX) = A^{-1}B⁠, that is X=A−1BX = A^{-1}B (associativity and A−1A=IA^{-1}A = I⁠). So this is the unique solution.
    Section 4.6.1, Case I; Section 3.4.6 (associativity). Lesson: inverse of a matrix

  7. Check in all three original equations

    2x−3y+5z=2(1)−3(2)+5(3)=2−6+15=11,3x+2y−4z=3(1)+2(2)−4(3)=3+4−12=−5,x+y−2z=1+2−2(3)=−3.\begin{aligned} 2x - 3y + 5z &= 2(1) - 3(2) + 5(3)\\ &= 2 - 6 + 15 = 11,\\ 3x + 2y - 4z &= 3(1) + 2(2) - 4(3)\\ &= 3 + 4 - 12 = -5,\\ x + y - 2z &= 1 + 2 - 2(3) = -3. \end{aligned}

    All three hold.

    The check uses the equations as printed, so it also confirms Step 5's reading of BB⁠, signs included.
Result

∣A∣=−1|A| = -1 and

A−1=[01−2−29−23−15−13].A^{-1} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}.

The system has the unique solution x=1x = 1⁠, y=2y = 2⁠, z=3z = 3.

(Textbook answer: A−1A^{-1} as above, x=1x = 1⁠, y=2y = 2⁠, z=3z = 3⁠.)

Check it with numbers

∣A∣|A| along row 3. Signs + − ++\,-\,+⁠, entries 1,1,−21, 1, -2⁠, minors 2,−23,132, -23, 13⁠:

∣A∣=1(2)−1(−23)+(−2)(13)=2+23−26=−1,\begin{aligned} |A| &= 1(2) - 1(-23) + (-2)(13)\\ &= 2 + 23 - 26 = -1, \end{aligned}

the same as along row 1.

A−1A=IA^{-1}A = I too. For instance, row 2 of A−1A^{-1} times column 1 of AA⁠, and row 3 of A−1A^{-1} times column 3 of AA⁠:

(−2)(2)+9(3)+(−23)(1)=−4+27−23=0,(−1)(5)+5(−4)+(−13)(−2)=−5−20+26=1.\begin{aligned} &(-2)(2) + 9(3) + (-23)(1)\\ &= -4 + 27 - 23 = 0,\\ &(-1)(5) + 5(-4) + (-13)(-2)\\ &= -5 - 20 + 26 = 1. \end{aligned}
Common mistakes
  1. Forgetting the sign of ∣A∣|A|⁠. With ∣A∣=−1|A| = -1⁠, A−1=−adj⁡AA^{-1} = -\operatorname{adj} A⁠. Using adj⁡A\operatorname{adj} A itself as the inverse gives x=−1x = -1⁠, y=−2y = -2⁠, z=−3z = -3⁠, which fail every equation.
  2. Getting ∣A∣=11|A| = 11⁠. The middle term is −(−3)(−2)=−6-(-3)(-2) = -6⁠, not +6+6⁠; with +6+6 the determinant would be 1111⁠.
  3. Skipping the transpose. The cofactor matrix is not symmetric here, so using it untransposed gives a wrong inverse (its first row would be 0,−2,−10, -2, -1 after the sign change, instead of 0,1,−20, 1, -2⁠).
What to take away
  1. When the question gives AA and then a system, check that the system's coefficient matrix is that AA⁠; then one inverse serves both parts.
  2. If ∣A∣=±1|A| = \pm1⁠, the inverse has whole-number entries and AA−1=IAA^{-1} = I can be checked directly.
  3. X=A−1BX = A^{-1}B⁠: inverse on the left, column of constants on the right.

Every fact used, and where it comes from

FactUsed inSource
Expansion along a row, signs (−1)i+j(-1)^{i+j}Step 1Section 4.2.3 (Lesson: expanding a 3 × 3 determinant)
Minor and cofactorStep 2Definitions 1 and 2, Section 4.4 (Lesson: minors and cofactors)
adj⁡A\operatorname{adj} A = transpose of [Aij][A_{ij}]Step 3Definition 3, Section 4.5.1 (Lesson: the adjoint)
Invertible iff nonsingular; A−1=1∣A∣adj⁡AA^{-1} = \frac{1}{|A|}\operatorname{adj} ASteps 1, 3Theorem 4, Section 4.5 (Lesson: inverse of a matrix)
A(adj⁡A)=∣A∣ IA(\operatorname{adj} A) = |A|\,I⁠; row times other cofactors is 0Step 4Theorem 1, Section 4.5; Note in Section 4.4
Matrix product, row times columnSteps 4, 5, 6Section 3.4.5 (Chapter 3)
A system written as AX=BAX = B⁠; X=A−1BX = A^{-1}B uniqueSteps 5, 6Section 4.6.1, Case I (Lesson: solving systems)
Associativity of multiplicationStep 6Section 3.4.6 (Chapter 3)
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