12 multiple-choice questions, progressively harder.
For which value of kkk does the system with augmented matrix [12324k]\left[\begin{array}{cc|c}1 & 2 & 3 \\ 2 & 4 & k\end{array}\right][12243k] have infinitely many solutions?
Solution
Correct answer: A
Apply R2→R2−2R1R_2 \to R_2 - 2R_1R2→R2−2R1: the row becomes (0, 0∣k−6)(0,\ 0 \mid k - 6)(0, 0∣k−6). Infinitely many solutions need this to be 0=00 = 00=0.
k−6=0 ⇒ k=6k - 6 = 0 \;\Rightarrow\; k = 6k−6=0⇒k=6
For any other kkk the row says 0=(nonzero)0 = (\text{nonzero})0=(nonzero) and the system has no solution.
Solve the system with augmented matrix [111212312113]\left[\begin{array}{ccc|c}1 & 1 & 1 & 2 \\ 1 & 2 & 3 & 1 \\ 2 & 1 & 1 & 3\end{array}\right]112121131213.
Correct answer: D
Use R2→R2−R1R_2 \to R_2 - R_1R2→R2−R1 to get (0, 1, 2∣−1)(0,\ 1,\ 2 \mid -1)(0, 1, 2∣−1) and R3→R3−2R1R_3 \to R_3 - 2R_1R3→R3−2R1 to get (0, −1, −1∣−1)(0,\ -1,\ -1 \mid -1)(0, −1, −1∣−1). Then R3→R3+R2R_3 \to R_3 + R_2R3→R3+R2 gives (0, 0, 1∣−2)(0,\ 0,\ 1 \mid -2)(0, 0, 1∣−2), so z=−2z = -2z=−2. Back-substitute: y+2z=−1y + 2z = -1y+2z=−1 gives y=3y = 3y=3.
x+y+z=2 ⇒ x+3−2=2 ⇒ x=1x + y + z = 2 \;\Rightarrow\; x + 3 - 2 = 2 \;\Rightarrow\; x = 1x+y+z=2⇒x+3−2=2⇒x=1
So (x,y,z)=(1,3,−2)(x, y, z) = (1, 3, -2)(x,y,z)=(1,3,−2).
After clearing column 111 of [1214253936411]\left[\begin{array}{ccc|c}1 & 2 & 1 & 4 \\ 2 & 5 & 3 & 9 \\ 3 & 6 & 4 & 11\end{array}\right]1232561344911 using row 111, what does row 333 become?
Clear the leading 333 with R3→R3−3R1R_3 \to R_3 - 3R_1R3→R3−3R1.
(3−3, 6−6, 4−3, 11−12)=(0, 0, 1, −1)(3 - 3,\ 6 - 6,\ 4 - 3,\ 11 - 12) = (0,\ 0,\ 1,\ -1)(3−3, 6−6, 4−3, 11−12)=(0, 0, 1, −1)
So row 333 becomes [001−1]\left[\begin{array}{ccc|c}0 & 0 & 1 & -1\end{array}\right][001−1].
Given [336912140125]\left[\begin{array}{ccc|c}3 & 3 & 6 & 9 \\ 1 & 2 & 1 & 4 \\ 0 & 1 & 2 & 5\end{array}\right]310321612945, the cleanest first step is:
Row 111 has a common factor of 333, so scaling by 13\tfrac{1}{3}31 turns it into a leading 111.
13(3, 3, 6∣9)=(1, 1, 2∣3)\tfrac{1}{3}(3,\ 3,\ 6 \mid 9) = (1,\ 1,\ 2 \mid 3)31(3, 3, 6∣9)=(1, 1, 2∣3)
A leading 111 makes clearing the column below it easy. Multiplying by 333 makes entries larger, and multiplying a row by 000 is illegal.
Which reduced bottom row makes a system inconsistent (no solution)?
Correct answer: B
An inconsistent row has all-zero coefficients but a nonzero constant, standing for 0=(nonzero)0 = (\text{nonzero})0=(nonzero).
[0002]⇒0=2\left[\begin{array}{ccc|c}0 & 0 & 0 & 2\end{array}\right] \Rightarrow 0 = 2[0002]⇒0=2
The others give z=2z = 2z=2, 0=00 = 00=0, or x=2x = 2x=2, all of which are satisfiable.
Two rows of an augmented matrix are identical. After reduction this produces:
Subtracting one row from an identical copy zeros it out entirely.
R2−R1=[0000]R_2 - R_1 = \left[\begin{array}{ccc|c}0 & 0 & 0 & 0\end{array}\right]R2−R1=[0000]
The duplicate carried no new information, so reduction produces a row of all zeros; if the rest of the system is consistent, the system is dependent with infinitely many solutions.
For [123358]\left[\begin{array}{cc|c}1 & 2 & 3 \\ 3 & 5 & 8\end{array}\right][132538], apply R2→R2−3R1R_2 \to R_2 - 3R_1R2→R2−3R1 and then read the bottom row. What is yyy?
Correct answer: C
After R2→R2−3R1R_2 \to R_2 - 3R_1R2→R2−3R1 the bottom row is (0, −1∣−1)(0,\ -1 \mid -1)(0, −1∣−1), so it says −y=−1-y = -1−y=−1.
−y=−1 ⇒ y=1-y = -1 \;\Rightarrow\; y = 1−y=−1⇒y=1
Solve the system with augmented matrix [32121−1−1]\left[\begin{array}{cc|c}3 & 2 & 12 \\ 1 & -1 & -1\end{array}\right][312−112−1].
Swap to put the leading 111 on top: x−y=−1x - y = -1x−y=−1, 3x+2y=123x + 2y = 123x+2y=12. Then R2→R2−3R1R_2 \to R_2 - 3R_1R2→R2−3R1 gives (0, 5∣15)(0,\ 5 \mid 15)(0, 5∣15), so y=3y = 3y=3. Back-substitute into x−y=−1x - y = -1x−y=−1.
x−3=−1 ⇒ x=2x - 3 = -1 \;\Rightarrow\; x = 2x−3=−1⇒x=2
So (x,y)=(2,3)(x, y) = (2, 3)(x,y)=(2,3).
Which situation is impossible for a system of linear equations?
A linear system always has exactly one solution, no solution, or infinitely many, never a finite number greater than one.
one / none / infinitely many\text{one} \;/\; \text{none} \;/\; \text{infinitely many}one/none/infinitely many
So exactly two solutions is impossible.
For [12324c]\left[\begin{array}{cc|c}1 & 2 & 3 \\ 2 & 4 & c\end{array}\right][12243c], which statement is true for all values of the constant ccc?
The rows are proportional on the left (2:4=1:22 : 4 = 1 : 22:4=1:2). Reducing gives (0, 0∣c−6)(0,\ 0 \mid c - 6)(0, 0∣c−6).
c=6⇒infinitely many;c≠6⇒no solutionc = 6 \Rightarrow \text{infinitely many};\qquad c \ne 6 \Rightarrow \text{no solution}c=6⇒infinitely many;c=6⇒no solution
A unique solution is impossible, because the second column never earns its own pivot.
The reduced matrix [1−24000]\left[\begin{array}{cc|c}1 & -2 & 4 \\ 0 & 0 & 0\end{array}\right][10−2040] (columns xxx, yyy) describes:
The zero row gives no constraint, so yyy is free; set y=ty = ty=t. The first row gives x−2y=4x - 2y = 4x−2y=4.
x=4+2t,y=tx = 4 + 2t,\quad y = tx=4+2t,y=t
The solutions form a whole line, so there are infinitely many.
Row operations never change a system's solution set because each one is:
Every row operation can be undone (swap back, divide by the same constant, subtract the same multiple), so the reduced system carries exactly the same constraints as the original.
reversible⇒same solution set\text{reversible} \Rightarrow \text{same solution set}reversible⇒same solution set
That is why you may read the answer off the reduced matrix.
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