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Chapter 4 · Determinants

Exercise 4.5 · Q5: Is 3x−y−2z=23x-y-2z=2⁠, 2y−z=−12y-z=-1⁠, 3x−5y=33x-5y=3 consistent?

Exercise 4.5Solving systems AX = B · Section 4.6
The question

Examine the consistency of the system of equations.

3x−y−2z=22y−z=−13x−5y=3\begin{aligned} 3x-y-2z&=2\\ 2y-z&=-1\\ 3x-5y&=3 \end{aligned}
Exam answer

The system is AX=BAX = B⁠, where

A=[3−1−202−13−50],X=[xyz], B=[2−13]\begin{aligned} A &= \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}, \\ X &= \begin{bmatrix} x \\ y \\ z \end{bmatrix},\ B = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} \end{aligned}

Expanding along R2R_2⁠:

∣A∣=−0∣−1−2−50∣+2∣3−230∣−(−1)∣3−13−5∣=0(−10)+2(6)+1(−12)=0\begin{aligned} |A| &= -0\begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix} + 2\begin{vmatrix} 3 & -2 \\ 3 & 0 \end{vmatrix} \\ &\quad - (-1)\begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix} \\ &= 0(-10) + 2(6) + 1(-12) \\ &= 0 \end{aligned}

∣A∣=0|A| = 0⁠. So AA is singular, and we find (adj⁡A)B(\operatorname{adj}A)B⁠.

Cofactors:

A11=∣2−1−50∣=−5A12=−∣0−130∣=−3A13=∣023−5∣=−6\begin{aligned} A_{11} &= \begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} = -5 \\ A_{12} &= -\begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} = -3 \\ A_{13} &= \begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix} = -6 \end{aligned} A21=−∣−1−2−50∣=10A22=∣3−230∣=6A23=−∣3−13−5∣=12\begin{aligned} A_{21} &= -\begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix} = 10 \\ A_{22} &= \begin{vmatrix} 3 & -2 \\ 3 & 0 \end{vmatrix} = 6 \\ A_{23} &= -\begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix} = 12 \end{aligned} A31=∣−1−22−1∣=5A32=−∣3−20−1∣=3A33=∣3−102∣=6\begin{aligned} A_{31} &= \begin{vmatrix} -1 & -2 \\ 2 & -1 \end{vmatrix} = 5 \\ A_{32} &= -\begin{vmatrix} 3 & -2 \\ 0 & -1 \end{vmatrix} = 3 \\ A_{33} &= \begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix} = 6 \end{aligned} adj⁡A=[−5105−363−6126]\operatorname{adj}A = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix} (adj⁡A)B=[−5105−363−6126][2−13]=[−10−10+15−6−6+9−12−12+18]=[−5−3−6]≠O\begin{aligned} (\operatorname{adj}A)B &= \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} \\ &= \begin{bmatrix} -10 - 10 + 15 \\ -6 - 6 + 9 \\ -12 - 12 + 18 \end{bmatrix} \\ &= \begin{bmatrix} -5 \\ -3 \\ -6 \end{bmatrix} \ne O \end{aligned}

Since ∣A∣=0|A| = 0 and (adj⁡A)B≠O(\operatorname{adj}A)B \ne O⁠, the system is inconsistent.

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

The one idea this problem is built on

The determinant of the coefficient matrix is 0, so the inverse does not exist and we use the second test of Section 4.6.1: compute (adj⁡A)B(\operatorname{adj}A)B⁠. It is not the zero column, and that rules out every solution, because a solution X0X_0 would force (adj⁡A)B=∣A∣X0=O(\operatorname{adj}A)B = |A|X_0 = O⁠. Behind it is a hidden relation: the first equation minus twice the second gives 3x−5y=43x - 5y = 4⁠, while the third says 3x−5y=33x - 5y = 3⁠.

Given, to find, and the plan
Given
Three linear equations in x,y,zx, y, z⁠. The second has no xx term and the third has no zz term; those coefficients are 0.
To find
Whether the system is consistent or inconsistent.
Plan
(1) Write AX=BAX = B⁠, with zeros for the missing terms. (2) Expand ∣A∣|A| along a row containing a zero; it is 0. (3) Find all nine cofactors and adj⁡A\operatorname{adj}A⁠. (4) Compute (adj⁡A)B(\operatorname{adj}A)B and conclude, with the reason. (5) Confirm by combining the equations.

Basics you need

1Missing terms are zero coefficients

2y−z=−12y - z = -1 is 0x+2y+(−1)z=−10x + 2y + (-1)z = -1⁠, and 3x−5y=33x - 5y = 3 is 3x+(−5)y+0z=33x + (-5)y + 0z = 3⁠. Every row of AA needs all three coefficients. Basics: algebra the chapter leans on

2Expanding a 3×33\times3 determinant

Along any row, ∣A∣=|A| = (each entry) ×\times (its cofactor), added up. The signs (−1)i+j(-1)^{i+j} follow

+−+−+−+−+\begin{matrix}+&-&+\\-&+&-\\+&-&+\end{matrix}

A row with a zero saves one minor. Basics: the + − + sign pattern

3The test when ∣A∣=0|A| = 0

Section 4.6.1, Case II: if (adj⁡A)B≠O(\operatorname{adj}A)B \ne O⁠, the system has no solution. If (adj⁡A)B=O(\operatorname{adj}A)B = O⁠, it may or may not have solutions. Basics: why the inverse divides by |A|

Solution

  1. Write the system as AX=BAX = B

    A=[3−1−202−13−50]X=[xyz],B=[2−13]\begin{gathered} A=\begin{bmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{bmatrix}\\ X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}2\\-1\\3\end{bmatrix} \end{gathered}

    AA is 3×33\times3 and XX is 3×13\times1⁠, so AXAX is 3×13\times1⁠. Row by row:

    3x+(−1)y+(−2)z=3x−y−2z0x+2y+(−1)z=2y−z3x+(−5)y+0z=3x−5y\begin{aligned} 3x+(-1)y+(-2)z&=3x-y-2z\\ 0x+2y+(-1)z&=2y-z\\ 3x+(-5)y+0z&=3x-5y \end{aligned}

    These are the three left sides, so AX=BAX = B is the given system.

    The zeros matter: they keep each coefficient in the column of its own unknown. Leaving them out would shift entries into the wrong columns.
    Basics: matrix facts the chapter uses

  2. Expand ∣A∣|A| along row 2

    Row 2 is (0,2,−1)(0, 2, -1)⁠: its first entry is 0, so its first term vanishes whatever its minor is. The row 2 signs are − + −-\,+\,-⁠:

    ∣A∣=−0∣−1−2−50∣+2∣3−230∣−(−1)∣3−13−5∣\begin{aligned} |A|&=-0\begin{vmatrix}-1&-2\\-5&0\end{vmatrix}+2\begin{vmatrix}3&-2\\3&0\end{vmatrix}\\ &\quad-(-1)\begin{vmatrix}3&-1\\3&-5\end{vmatrix} \end{aligned}

    The two minors that matter:

    ∣3−230∣=3(0)−3(−2)=6∣3−13−5∣=3(−5)−3(−1)=−12\begin{aligned} \begin{vmatrix}3&-2\\3&0\end{vmatrix}&=3(0)-3(-2)=6\\ \begin{vmatrix}3&-1\\3&-5\end{vmatrix}&=3(-5)-3(-1)=-12 \end{aligned} ∣A∣=0+2(6)+1(−12)=12−12=0|A|=0+2(6)+1(-12)=12-12=0

    So ∣A∣=0|A| = 0⁠: AA is singular and has no inverse.

    The third term is −(−1)×(−12)-(-1)\times(-12)⁠: the pattern sign −-⁠, the entry −1-1⁠, and the minor −12-12⁠. Two minus signs make +1+1⁠, times −12-12 gives −12-12⁠. With ∣A∣=0|A| = 0⁠, Case I is closed and we need Case II of Section 4.6.1.
    Expansion along a row, Section 4.2.3. Lesson: expanding a 3 × 3 determinant

  3. All nine cofactors

    A11=(−1)1+1∣2−1−50∣=0−5=−5A12=(−1)1+2∣0−130∣=−(0+3)=−3A13=(−1)1+3∣023−5∣=0−6=−6A21=(−1)2+1∣−1−2−50∣=−(0−10)=10A22=(−1)2+2∣3−230∣=0+6=6A23=(−1)2+3∣3−13−5∣=−(−15+3)=12A31=(−1)3+1∣−1−22−1∣=1+4=5A32=(−1)3+2∣3−20−1∣=−(−3−0)=3A33=(−1)3+3∣3−102∣=6−0=6\begin{aligned} A_{11}&=(-1)^{1+1}\begin{vmatrix}2&-1\\-5&0\end{vmatrix}\\ &=0-5=-5\\ A_{12}&=(-1)^{1+2}\begin{vmatrix}0&-1\\3&0\end{vmatrix}\\ &=-(0+3)=-3\\ A_{13}&=(-1)^{1+3}\begin{vmatrix}0&2\\3&-5\end{vmatrix}\\ &=0-6=-6\\ A_{21}&=(-1)^{2+1}\begin{vmatrix}-1&-2\\-5&0\end{vmatrix}\\ &=-(0-10)=10\\ A_{22}&=(-1)^{2+2}\begin{vmatrix}3&-2\\3&0\end{vmatrix}\\ &=0+6=6\\ A_{23}&=(-1)^{2+3}\begin{vmatrix}3&-1\\3&-5\end{vmatrix}\\ &=-(-15+3)=12\\ A_{31}&=(-1)^{3+1}\begin{vmatrix}-1&-2\\2&-1\end{vmatrix}\\ &=1+4=5\\ A_{32}&=(-1)^{3+2}\begin{vmatrix}3&-2\\0&-1\end{vmatrix}\\ &=-(-3-0)=3\\ A_{33}&=(-1)^{3+3}\begin{vmatrix}3&-1\\0&2\end{vmatrix}\\ &=6-0=6 \end{aligned}

    Check: row 1 times its cofactors should also give ∣A∣|A|⁠:

    3(−5)+(−1)(−3)+(−2)(−6)=−15+3+12=0=∣A∣. ✓\begin{aligned} &3(-5)+(-1)(-3)+(-2)(-6)\\ &=-15+3+12=0=|A|.\ \checkmark \end{aligned}
    Each minor deletes the row and column of its entry; for instance A23A_{23} deletes row 2 and column 3, leaving columns 1 and 2 of rows 1 and 3. Expanding along a different row and getting 0 again is a good check that the cofactors, and ∣A∣=0|A| = 0⁠, are right.
    Definitions 1 and 2, Section 4.4. Lesson: minors and cofactors

  4. adj⁡A\operatorname{adj}A

    Transpose the matrix of cofactors: the cofactors of row 1 (−5,−3,−6-5, -3, -6⁠) form column 1, and so on.

    adj⁡A=[−5105−363−6126]\operatorname{adj}A=\begin{bmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{bmatrix}
    The adjoint exists for every square matrix, singular or not; only A−1=1∣A∣adj⁡AA^{-1} = \frac{1}{|A|}\operatorname{adj}A needs ∣A∣≠0|A| \ne 0⁠.
    Definition 3, Section 4.5.1. Lesson: the adjoint

  5. Compute (adj⁡A)B(\operatorname{adj}A)B and conclude

    A 3×33\times3 times a 3×13\times1 gives a 3×13\times1 column:

    (adj⁡A)B=[−5(2)+10(−1)+5(3)−3(2)+6(−1)+3(3)−6(2)+12(−1)+6(3)]=[−10−10+15−6−6+9−12−12+18]=[−5−3−6]≠O\begin{aligned} (\operatorname{adj}A)B&=\begin{bmatrix}-5(2)+10(-1)+5(3)\\-3(2)+6(-1)+3(3)\\-6(2)+12(-1)+6(3)\end{bmatrix}\\ &=\begin{bmatrix}-10-10+15\\-6-6+9\\-12-12+18\end{bmatrix}\\ &=\begin{bmatrix}-5\\-3\\-6\end{bmatrix}\ne O \end{aligned}

    Since ∣A∣=0|A| = 0 and (adj⁡A)B≠O(\operatorname{adj}A)B \ne O⁠, the system is inconsistent.

    Why the rule holds: if AX0=BAX_0 = B for some X0X_0⁠, then (adj⁡A)B=(adj⁡A)AX0(\operatorname{adj}A)B = (\operatorname{adj}A)AX_0⁠, which equals ∣A∣X0=0⋅X0=O|A|X_0 = 0\cdot X_0 = O⁠, using Theorem 1, (adj⁡A)A=∣A∣I(\operatorname{adj}A)A = |A|I⁠. Our (adj⁡A)B(\operatorname{adj}A)B is not OO⁠, so no such X0X_0 exists.
    Section 4.6.1, Case II; Theorem 1, Section 4.5.1. Lesson: solving systems AX = B

  6. See it directly: the equations contradict each other

    Take the first equation and subtract twice the second:

    (3x−y−2z)−2(2y−z)=2−2(−1)3x−5y=4\begin{aligned} (3x-y-2z)-2(2y-z)&=2-2(-1)\\ 3x-5y&=4 \end{aligned}

    The zz terms cancel (−2z+2z=0-2z + 2z = 0⁠). Any solution of the first two equations therefore has 3x−5y=43x - 5y = 4⁠, but the third equation demands 3x−5y=33x - 5y = 3⁠. Both cannot hold, so there is no solution.

    This is the relation hiding behind ∣A∣=0|A| = 0⁠: row 3 of AA⁠, (3,−5,0)(3, -5, 0)⁠, equals row 1 minus twice row 2. The left side of equation 3 is that same combination of the first two left sides, but the right side 3 is not the matching combination of 2 and −1-1⁠, which is 4.
    Adding multiples of equations keeps every solution.
Result

∣A∣=0|A| = 0 and (adj⁡A)B(\operatorname{adj}A)B has entries −5,−3,−6-5, -3, -6⁠, so (adj⁡A)B≠O(\operatorname{adj}A)B \ne O⁠. The system is inconsistent. (Textbook answer: Inconsistent.)

Check it with numbers

Satisfy the first two equations and test the third. Take y=1y = 1⁠. The second equation gives z=2y+1=3z = 2y + 1 = 3⁠. The first then gives 3x=2+y+2z3x = 2 + y + 2z⁠, which is 2+1+6=92 + 1 + 6 = 9⁠, so x=3x = 3⁠. Now the third: 3x−5y=9−5=43x - 5y = 9 - 5 = 4⁠, not 3. ✗ Trying y=0y = 0⁠: z=1z = 1⁠, 3x=2+0+2=43x = 2 + 0 + 2 = 4⁠, and 3x−5y=43x - 5y = 4 again. Every solution of the first two equations gives 4, as Step 6 predicts.

Common mistakes
  1. Dropping the zero coefficients. Writing row 2 of AA as (2,−1)(2, -1) or row 3 as (3,−5)(3, -5) gives a matrix that is not square.
  2. Getting the sign of a row 2 term wrong. Row 2 starts with −-⁠: the signs are − + −-\,+\,-⁠, not + − ++\,-\,+⁠.
  3. Concluding "inconsistent" from ∣A∣=0|A| = 0 alone. It is (adj⁡A)B≠O(\operatorname{adj}A)B \ne O that proves it.
  4. Forgetting to transpose the cofactors before multiplying by BB⁠. The matrix of cofactors times BB gives the column (−25,50,25)(-25, 50, 25) instead: non-zero here too, but the working is wrong and in another question it could change the verdict.
What to take away
  1. Write every missing term as a zero coefficient, and use zeros to choose the expansion row.
  2. ∣A∣=0|A| = 0 means one equation's left side is a combination of the others. Whether the right sides match decides between "none" and "infinitely many".
  3. (adj⁡A)B≠O(\operatorname{adj}A)B \ne O is a quick, mechanical proof that the right sides do not match.

Every fact used, and where it comes from

FactUsed inSource
Consistent and inconsistent systemsSteps 5, 6Section 4.6 (Lesson: solving systems AX = B)
Expansion of a 3×33\times3 determinant along any rowSteps 2, 3Section 4.2.3
Aij=(−1)i+jMijA_{ij} = (-1)^{i+j}M_{ij}⁠; ∣A∣|A| = row times its cofactorsStep 3Definitions 1, 2, Section 4.4
adj⁡A\operatorname{adj}A = transpose of the matrix of cofactorsStep 4Definition 3, Section 4.5.1
(adj⁡A)A=∣A∣I(\operatorname{adj}A)A = |A|IStep 5Theorem 1, Section 4.5.1
∣A∣=0|A| = 0⁠, (adj⁡A)B≠O(\operatorname{adj}A)B \ne O ⇒\Rightarrow no solutionStep 5Section 4.6.1, Case II
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