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Chapter 4 · Determinants

Example 19: Use a given product to solve a system

ExamplesMiscellaneous examples · Chapter 4
The question

Use product

[1−1202−33−24][−20192−361−2]\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}

to solve the system of equations

x−y+2z=12y−3z=13x−2y+4z=2\begin{aligned} x - y + 2z &= 1\\ 2y - 3z &= 1\\ 3x - 2y + 4z &= 2 \end{aligned}

About the signs: some PDF viewers do not show the six minus signs inside these two matrices. They are present in the book's text layer, and the book's own solution on the same page prints both matrices with them, exactly as above.

Exam answer

Let

P=[1−1202−33−24],Q=[−20192−361−2]\begin{aligned} P &= \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}, \\ Q &= \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} \end{aligned}

The entries cijc_{ij} of PQPQ (row ii of PP times column jj of QQ⁠):

c11=−2−9+12=1c12=0−2+2=0c13=1+3−4=0c21=0+18−18=0c22=0+4−3=1c23=0−6+6=0c31=−6−18+24=0c32=0−4+4=0c33=3+6−8=1\begin{aligned} c_{11} &= -2 - 9 + 12 = 1 \\ c_{12} &= 0 - 2 + 2 = 0 \\ c_{13} &= 1 + 3 - 4 = 0 \\ c_{21} &= 0 + 18 - 18 = 0 \\ c_{22} &= 0 + 4 - 3 = 1 \\ c_{23} &= 0 - 6 + 6 = 0 \\ c_{31} &= -6 - 18 + 24 = 0 \\ c_{32} &= 0 - 4 + 4 = 0 \\ c_{33} &= 3 + 6 - 8 = 1 \end{aligned} ∴ PQ=[100010001]=I\therefore\ PQ = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I

So PP is invertible and P−1=QP^{-1} = Q⁠.

The given system is PX=BPX = B⁠, where

X=[xyz], B=[112]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\ B = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}

As P−1P^{-1} exists, the system has a unique solution, X=P−1B=QBX = P^{-1}B = QB⁠.

X=[−20192−361−2][112]=[−2+0+29+2−66+1−4]=[053]\begin{aligned} X &= \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \\ &= \begin{bmatrix} -2 + 0 + 2 \\ 9 + 2 - 6 \\ 6 + 1 - 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 5 \\ 3 \end{bmatrix} \end{aligned}

∴ x=0, y=5, z=3\therefore\ x = 0,\ y = 5,\ z = 3⁠.

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

The one idea this problem is built on

If two square matrices multiply to the identity, PQ=IPQ = I⁠, then each is the inverse of the other. Here the first matrix is exactly the coefficient matrix of the system, so the second matrix is its inverse, handed to us ready-made. Solving PX=BPX = B then takes one multiplication, X=QBX = QB⁠, with no determinants or cofactors at all.

Given, to find, and the plan
Given
Two 3×33\times3 matrices, called PP (first) and QQ (second) below, and a system of three equations.
To find
All solutions (x,y,z)(x, y, z) of the system, using the product PQPQ⁠.
Plan
Compute PQPQ entry by entry and find that it is II⁠; explain why that makes Q=P−1Q = P^{-1}⁠; write the system as PX=BPX = B⁠; compute X=QBX = QB⁠; check in all three equations.

Basics you need

1Matrix multiplication

Entry (i,j)(i,j) of a product is row ii of the first matrix times column jj of the second. Two 3×33\times3 matrices give a 3×33\times3 product. Basics: matrix facts the chapter uses

2Inverse and identity

QQ is the inverse of PP when PQ=QP=IPQ = QP = I⁠, where II has 1s on the diagonal and 0s elsewhere; IX=XIX = X for every XX⁠. Basics: transpose and identity

3Determinant of a product

∣PQ∣=∣P∣ ∣Q∣|PQ| = |P|\,|Q| (Theorem 3, Section 4.5), and ∣I∣=1|I| = 1⁠. Basics: why the inverse divides by |A|

4The matrix method

If the coefficient matrix PP is invertible, PX=BPX = B has the unique solution X=P−1BX = P^{-1}B (Section 4.6.1).

Solution

  1. Name the two matrices

    P=[1−1202−33−24],P = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}, Q=[−20192−361−2].Q = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}.

    Both are 3×33\times3⁠, so PQPQ is defined (3 columns of PP⁠, 3 rows of QQ⁠) and is 3×33\times3⁠.

    Names make the argument readable. The signs matter: the question only works with the six minus signs shown. Without them the (1,1)(1,1) entry of the product would be 1⋅2+1⋅9+2⋅6=231\cdot2 + 1\cdot9 + 2\cdot6 = 23⁠, not 1, and the product would be useless. With them, Step 2 gives exactly II⁠, which is what the book's solution finds.
    Order of a product, Chapter 3, Section 3.4.5. Basics: rows, columns and positions

  2. Compute PQPQ⁠, entry by entry

    Row ii of PP times column jj of QQ⁠:

    c11=1(−2)+(−1)9+2⋅6=1,c12=1⋅0+(−1)2+2⋅1=0,c13=1⋅1+(−1)(−3)+2(−2)=0,c21=0(−2)+2⋅9+(−3)6=0,c22=0⋅0+2⋅2+(−3)1=1,c23=0⋅1+2(−3)+(−3)(−2)=0,c31=3(−2)+(−2)9+4⋅6=0,c32=3⋅0+(−2)2+4⋅1=0,c33=3⋅1+(−2)(−3)+4(−2)=1.\begin{aligned} c_{11} &= 1(-2) + (-1)9 + 2\cdot6 = 1,\\ c_{12} &= 1\cdot0 + (-1)2 + 2\cdot1 = 0,\\ c_{13} &= 1\cdot1 + (-1)(-3) + 2(-2) = 0,\\ c_{21} &= 0(-2) + 2\cdot9 + (-3)6 = 0,\\ c_{22} &= 0\cdot0 + 2\cdot2 + (-3)1 = 1,\\ c_{23} &= 0\cdot1 + 2(-3) + (-3)(-2) = 0,\\ c_{31} &= 3(-2) + (-2)9 + 4\cdot6 = 0,\\ c_{32} &= 3\cdot0 + (-2)2 + 4\cdot1 = 0,\\ c_{33} &= 3\cdot1 + (-2)(-3) + 4(-2) = 1. \end{aligned} PQ=[100010001]=I.PQ = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I.
    These are the textbook's sums: for example c11=−2−9+12c_{11} = -2 - 9 + 12⁠, c22=0+4−3c_{22} = 0 + 4 - 3⁠, c33=3+6−8c_{33} = 3 + 6 - 8⁠. Every diagonal entry is 1 and every other entry is 0, so the product is the 3×33\times3 identity.
    Multiplication of matrices, Chapter 3, Section 3.4.5. Basics: matrix facts the chapter uses

  3. Why PQ=IPQ = I means Q=P−1Q = P^{-1}

    Take determinants of PQ=IPQ = I⁠: ∣P∣ ∣Q∣=∣I∣=1|P|\,|Q| = |I| = 1⁠. So ∣P∣≠0|P| \ne 0⁠, PP is non-singular, and P−1P^{-1} exists (Theorem 4). Now pre-multiply PQ=IPQ = I by P−1P^{-1}⁠:

    P−1(PQ)=P−1I(P−1P)Q=P−1Q=P−1.\begin{aligned} P^{-1}(PQ) &= P^{-1}I\\ (P^{-1}P)Q &= P^{-1}\\ Q &= P^{-1}. \end{aligned}

    So

    [1−1202−33−24]−1=[−20192−361−2].\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}.
    The definition of an inverse asks for PQ=IPQ = I and QP=IQP = I⁠. The textbook writes "Hence" after checking only PQPQ⁠; the argument above fills that gap: for square matrices, a one-sided inverse is automatically the inverse. (The Check box below also confirms QP=IQP = I directly.)
    Theorem 3 (∣AB∣=∣A∣ ∣B∣|AB| = |A|\,|B|⁠) and Theorem 4, Section 4.5; associative law, Chapter 3, Section 3.4.6. Lesson: inverse of a matrix

  4. Write the system as PX=BPX = B

    Reading the coefficients of x,y,zx, y, z from each equation (x−y+2zx - y + 2z⁠: 1,−1,21, -1, 2⁠; 2y−3z2y - 3z⁠: 0,2,−30, 2, -3⁠; 3x−2y+4z3x - 2y + 4z⁠: 3,−2,43, -2, 4⁠):

    [1−1202−33−24][xyz]=[112].\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}.

    The coefficient matrix is exactly PP⁠, so the system is PX=BPX = B with XX the column of x,y,zx, y, z and BB the column of 1,1,21, 1, 2⁠.

    This is why the question supplies PQPQ⁠: PP is the coefficient matrix, so its inverse QQ is exactly what the matrix method needs. The second equation has no xx⁠, so its coefficient is 0.
    Section 4.6.1. Lesson: solving systems AX = B

  5. Solve: X=P−1B=QBX = P^{-1}B = QB

    Pre-multiplying PX=BPX = B by P−1P^{-1} gives X=P−1BX = P^{-1}B⁠, and P−1=QP^{-1} = Q by Step 3:

    [xyz]=[−20192−361−2][112].\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}.

    Row times column:

    row 1: (−2)1+0⋅1+1⋅2=0,row 2: 9⋅1+2⋅1+(−3)2=5,row 3: 6⋅1+1⋅1+(−2)2=3.\begin{aligned} \text{row 1}&: \ (-2)1 + 0\cdot1 + 1\cdot2 = 0,\\ \text{row 2}&: \ 9\cdot1 + 2\cdot1 + (-3)2 = 5,\\ \text{row 3}&: \ 6\cdot1 + 1\cdot1 + (-2)2 = 3. \end{aligned} [xyz]=[053].\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 5 \\ 3 \end{bmatrix}.

    So x=0x = 0⁠, y=5y = 5⁠, z=3z = 3. Since PP is invertible, this is the only solution.

    P−1(PX)=(P−1P)X=IX=XP^{-1}(PX) = (P^{-1}P)X = IX = X⁠, so pre-multiplying isolates XX⁠. The textbook's sums are −2+0+2-2 + 0 + 2⁠, 9+2−69 + 2 - 6 and 6+1−46 + 1 - 4⁠.
    Case I, Section 4.6.1. Lesson: solving systems AX = B

  6. Check in every original equation

    x−y+2z=0−5+6=1,2y−3z=10−9=1,3x−2y+4z=0−10+12=2.\begin{aligned} x - y + 2z &= 0 - 5 + 6 = 1,\\ 2y - 3z &= 10 - 9 = 1,\\ 3x - 2y + 4z &= 0 - 10 + 12 = 2. \end{aligned}

    All three right-hand sides 1,1,21, 1, 2 are reproduced.

    Substituting into the given equations confirms both the product of Step 2 and the set-up of Step 4. A solution with x=0x = 0 is perfectly normal: 0 is a number like any other.
Result

The product of the two given matrices is II⁠, so the second is the inverse of the first, which is the coefficient matrix of the system. Hence

X=[−20192−361−2][112]=[053],X = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 5 \\ 3 \end{bmatrix},

that is x=0x = 0⁠, y=5y = 5⁠, z=3z = 3. (This agrees with the textbook's own solution.)

Check it with numbers

The other order, QPQP⁠, should also be II⁠. Row ii of QQ times column jj of PP⁠:

d11=−2+0+3=1,d12=2+0−2=0,d13=−4+0+4=0,d21=9+0−9=0,d22=−9+4+6=1,d23=18−6−12=0,d31=6+0−6=0,d32=−6+2+4=0,d33=12−3−8=1.\begin{aligned} d_{11} &= -2 + 0 + 3 = 1, & d_{12} &= 2 + 0 - 2 = 0,\\ d_{13} &= -4 + 0 + 4 = 0, & d_{21} &= 9 + 0 - 9 = 0,\\ d_{22} &= -9 + 4 + 6 = 1, & d_{23} &= 18 - 6 - 12 = 0,\\ d_{31} &= 6 + 0 - 6 = 0, & d_{32} &= -6 + 2 + 4 = 0,\\ d_{33} &= 12 - 3 - 8 = 1. \end{aligned}

So QP=IQP = I too. Also, expanding along column 1 of PP (it contains a 0): ∣P∣=1(8−6)−0+3(3−4)=−1|P| = 1(8 - 6) - 0 + 3(3 - 4) = -1⁠, and expanding along row 1 of QQ (it contains a 0): ∣Q∣=−2(−4+3)−0+1(9−12)=−1|Q| = -2(-4 + 3) - 0 + 1(9 - 12) = -1⁠. Then ∣P∣ ∣Q∣=1=∣I∣|P|\,|Q| = 1 = |I|⁠, as Step 3 requires.

Common mistakes
  1. Reading the matrices without their minus signs. Some PDF viewers hide them; the unsigned product is not II (its first entry is 23).
  2. Writing X=BQX = BQ⁠. BB is 3×13\times1⁠, so BQBQ is not defined; the inverse goes on the left: X=QBX = QB⁠.
  3. Using PP instead of QQ⁠. PBPB would just compute the left sides of the equations at x,y,z=1,1,2x, y, z = 1, 1, 2⁠; it does not solve anything.
  4. Computing the inverse from scratch. Not wrong, but the question gives Q=P−1Q = P^{-1} for free; using it is the point.
What to take away
  1. For square matrices, PQ=IPQ = I already implies Q=P−1Q = P^{-1} (and QP=IQP = I⁠).
  2. If the coefficient matrix's inverse is known, the solution of PX=BPX = B is one multiplication: X=P−1BX = P^{-1}B⁠.
  3. Always check the solution in the original equations.

Every fact used, and where it comes from

FactUsed inSource
Product of matrices; order of the productSteps 1, 2, 5Chapter 3, Section 3.4.5 (Basics: matrix facts the chapter uses)
∣AB∣=∣A∣ ∣B∣|AB| = |A|\,|B|⁠; ∣I∣=1|I| = 1Step 3Theorem 3, Section 4.5
∣A∣≠0  ⟺  A|A| \ne 0 \iff A invertibleStep 3Theorem 4, Section 4.5 (Lesson: inverse of a matrix)
Associative law; IX=XIX = XSteps 3, 5Chapter 3, Section 3.4.6
Writing a system as AX=BAX = B⁠; X=A−1BX = A^{-1}BSteps 4, 5Section 4.6.1, Case I (Lesson: solving systems AX = B)
Expansion along a row or columnCheckSection 4.2.3 (Lesson: expanding a 3 × 3 determinant)
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