Practice cards

Front / question

In a grid P, which address does p_42 name?

Back / answer

Row 4, column 2 of P.

P names the whole grid; lowercase p with subscripts names one entry. The first subscript 4 chooses the row. The second subscript 2 chooses the column.

Front / question

Let A be 2 × 3, B be 3 × 2, and C = AB. Its entry rule is c_ij = Σ_(z=1)^3 a_iz b_zj. While finding c_21, which symbols stay fixed and which one runs?

Back / answer

i = 2 and j = 1 stay fixed; z runs through 1, 2, and 3.

The destination c_21 fixes row i at 2 and column j at 1. The bounds under and over Σ tell z to take the values 1, 2, and 3.

Front / question

Expand and evaluate Σ_(k=1)^4 (k + 1).

Back / answer

(1 + 1) + (2 + 1) + (3 + 1) + (4 + 1) = 2 + 3 + 4 + 5 = 14.

The bounds include both endpoints, so there are four terms. Evaluate (k + 1) at each value and add the four results.

Front / question

With A = [[2, 1, 3], [4, 2, 1]] and B = [[1, 2], [3, 1], [2, 4]], expand and calculate c_12 in C = AB.

Back / answer

c_12 = 2 × 2 + 1 × 1 + 3 × 4 = 4 + 1 + 12 = 17.

For c_12, fix i = 1 and j = 2; z still runs through 1, 2, and 3. A's first row is (2, 1, 3). B's second column is (2, 1, 4). Multiply corresponding positions and add: 4 + 1 + 12 = 17.

Front / question

Expand and evaluate Σ_(r=2)^4 3r. Does changing the running letter from r to t change the value?

Back / answer

3(2) + 3(3) + 3(4) = 6 + 9 + 12 = 27. Renaming r to t does not change the value.

The lower bound starts the sum at 2; the upper bound ends it at 4, giving three terms. The running letter is a placeholder. The same three values and rule give the same result with another letter.

Front / question

A rule says d_ij = Σ_(q=1)^5 x_iq y_qj. How many products make one entry d_ij, and does the 5 say how many entries D has?

Back / answer

Five products make each chosen entry. The 5 does not say how many entries D has.

The running index q takes 1, 2, 3, 4, and 5, so the sum has five product terms. The size of D comes from its available row and column addresses, which this sum bound alone does not specify.

Front / question

In A = [[2, 1, 3], [4, 2, 1]] and B = [[1, 2], [3, 1], [2, 4]], find a_23 and b_32. Say each address as row, column before reading its value.

Back / answer

a_23 = 1 and b_32 = 4.

A's second row is (4, 2, 1); its third entry is 1. B's third row is (2, 4); its second entry is 4. An address names one entry, not the whole matrix.

Front / question

Let A = [[2, 1, 3], [4, 2, 1]], B = [[1, 2], [3, 1], [2, 4]], and C = AB. For c_12, which row of A and column of B must meet? How many paired products make that one entry?

Back / answer

Use A row 1, (2, 1, 3), with B column 2, (2, 1, 4). That entry has three paired products.

The first subscript 1 selects A row 1; the second subscript 2 selects B column 2. Their positions match three times: 2 with 2, 1 with 1, and 3 with 4. The shape's shared 3 sets the count.

Front / question

Let A = [[2, 1, 3], [4, 2, 1]], B = [[1, 2], [3, 1], [2, 4]], and C = AB. Calculate c_22. Write the chosen row and column before the arithmetic.

Back / answer

c_22 = 4×2 + 2×1 + 1×4 = 8 + 2 + 4 = 14.

Address 22 fixes row 2 of A and column 2 of B. The corresponding pairs are 4 with 2, 2 with 1, and 1 with 4. Their products add to 14. Place 14 at row 2, column 2 of C.

Front / question

Let A = [[2, 1, 3], [4, 2, 1]] and B = [[1, 2], [3, 1], [2, 4]]. Is BA defined? If so, give its shape and top-left entry.

Back / answer

BA has shape 3 × 3. Its top-left entry is 1×2 + 2×4 = 10.

The adjacent dimensions in (3 × 2)(2 × 3) match at 2; the outside dimensions are 3 and 3. B row 1 is (1, 2), and A column 1 is (2, 4). Pair their positions and add 2 + 8 = 10.

Front / question

Let P = [[1, 0], [2, 3]] and Q = [[4], [5]]. Find PQ and its shape.

Back / answer

PQ = [[4], [23]], a 2 × 1 matrix.

The shapes (2 × 2)(2 × 1) fit, and the outside dimensions give 2 × 1. The first entry is 1×4 + 0×5 = 4. The second is 2×4 + 3×5 = 8 + 15 = 23. Put the two entries in one column: [[4], [23]].

Front / question

A is 2 × 3, B is 3 × 2, and D is any 2 × 3 matrix. Which of AB, BA, AD, and DA are defined? Give each valid output shape.

Back / answer

AB is defined and is 2 × 2. BA is defined and is 3 × 3. AD and DA are undefined.

AB has shapes (2 × 3)(3 × 2): the adjacent 3s match, giving 2 × 2. BA has shapes (3 × 2)(2 × 3): the adjacent 2s match, giving 3 × 3. AD and DA both have shapes (2 × 3)(2 × 3). Their adjacent dimensions are 3 and 2, so neither product is defined.

Front / question

Without calculating entries, describe (4 × 9)(9 × 6): output shape, total output entries, and paired products per entry.

Back / answer

The output is 4 × 6, with 24 entries and 9 paired products in each entry.

The adjacent 9s match, so the product is defined. The outside dimensions 4 and 6 give its shape. A 4 × 6 matrix has 4×6 = 24 entries. The shared 9 means each entry adds nine corresponding products.

Front / question

Let A = [[2, 1, 3], [4, 2, 1]], B = [[1, 2], [3, 1], [2, 4]], and C = AB. Someone tries nine every-to-every products for c_21. Which row and column should they choose, how should they pair positions instead, and what is c_21?

Back / answer

Make three corresponding products: 4×1 + 2×3 + 1×2 = 12.

For c_21, A row 2 is (4, 2, 1) and B column 1 is (1, 3, 2). Pair first with first, second with second, and third with third, giving 4, 6, and 2. Add those three products to get 12. The shared dimension is 3, so nine every-to-every products answer a different question.

Front / question

P and Q are matrices. Which expression asks for scalar multiplication: 5P or PQ? What clue settles it?

Back / answer

5P asks for scalar multiplication; PQ asks for matrix multiplication.

In 5P, 5 is one number, so it multiplies every entry of P. In PQ, both named objects are matrices. Its entries require the matrix-product rule, assuming their shapes are compatible.

Front / question

M has 2 rows and 3 columns, and s is a scalar. What is the shape of sM? If the entry of M at row 2, column 3 is m₂₃, what is the entry at that address in sM?

Back / answer

sM is 2 × 3, and its row 2, column 3 entry is s × m₂₃.

M has six addresses arranged in 2 rows and 3 columns; scaling leaves those addresses in place. At each address, multiply the existing entry by s. At (2, 3), that gives s × m₂₃.

Front / question

Calculate −2 × [[3, −1], [0, 4]].

Back / answer

[[-6, 2], [0, -8]]

Keep the two rows: [[−2 × 3, −2 × (−1)], [−2 × 0, −2 × 4]]. The products are −6, 2, 0, and −8 in their original positions, giving [[−6, 2], [0, −8]].

Front / question

Calculate (1/2) × [[6, 0], [−4, 10]].

Back / answer

[[3, 0], [−2, 5]]

At each address, multiply by 1/2: [[6/2, 0/2], [−4/2, 10/2]]. Simplify without moving entries: [[3, 0], [−2, 5]].

Front / question

Let k(t) = 2t − 1 and P = [[1, 0], [2, −3]]. Find k(2)P. What would k(t)P mean before a value of t is chosen?

Back / answer

k(2)P = [[3, 0], [6, −9]]. Before choosing t, k(t)P = [[k(t), 0], [2k(t), −3k(t)]].

k(2) = 2 × 2 − 1 = 3, so the instruction becomes 3P. Multiply each entry of P by 3: [[3 × 1, 3 × 0], [3 × 2, 3 × (−3)]] = [[3, 0], [6, −9]]. With t unspecified, use the same factor k(t) at every address: [[k(t), 0], [2k(t), −3k(t)]].

Front / question

In complex-number work, i² = −1. Compare iP with pᵢⱼ, where P = [pᵢⱼ]. What job does i do in each expression? If P = [[1, 2]], what is iP?

Back / answer

In iP, i is the imaginary-unit scalar, so iP = [[i, 2i]]. In pᵢⱼ, i is the row index, not a multiplier.

The statement i² = −1 declares i to be the imaginary unit in this context. Placed before P, it scales each entry; this gives [[i × 1, i × 2]] = [[i, 2i]]. The output has complex entries. In pᵢⱼ, subscripts identify an entry at row i and column j. That i names a position. The same printed letter can have a different role in a different context.

Front / question

A square matrix gives Av = v for a nonzero v. Is v an eigenvector? What is λ?

Back / answer

Yes. λ = 1.

Because v is nonzero and A is square, the definition can apply. Av = v = 1v, so one common scalar works for every coordinate. The input is unchanged, not merely kept on the same line.

Front / question

For A = [[3, 0], [0, 2]], someone says (1, 1) is an eigenvector because A(1, 1) = (3, 2) still points generally up and right. Check the claim.

Back / answer

No. (1, 1) is not an eigenvector of this A.

If (3, 2) = λ(1, 1), the first coordinate demands λ = 3 while the second demands λ = 2. One scalar cannot be both. A nearby-looking direction does not satisfy Av = λv.

Front / question

Keep A = [[3, 0], [0, 2]]. Test v = (0, −3). Give Av and the eigenvalue if v qualifies.

Back / answer

Av = (0, −6) = 2v, so v is an eigenvector with λ = 2.

The first output coordinate is 3×0 + 0×(−3) = 0. The second is 0×0 + 2×(−3) = −6. Because (0, −6) = 2(0, −3), one common scalar works.

Front / question

Let D = [[−2, 0], [0, 1]] and v = (3, 0). Find Dv, decide whether v is an eigenvector, and describe what the sign of λ does.

Back / answer

Dv = (−6, 0) = −2v. Yes: λ = −2, so the arrow reverses along the same line and doubles in length.

The first coordinate is (−2)×3 + 0×0 = −6; the second is 0×3 + 1×0 = 0. One scalar, −2, scales both coordinates. Its negative sign reverses orientation; its absolute value 2 doubles length.

Front / question

Let P = [[0, 0], [0, 1]] and v = (2, 0). P sends v to the origin. Does v count as an eigenvector? Would the zero vector itself count?

Back / answer

Yes: Pv = (0, 0) = 0v, so v is an eigenvector with λ = 0. The zero vector is not an eigenvector.

The input (2, 0) is nonzero. Multiplication gives (0×2 + 0×0, 0×2 + 1×0) = (0, 0). The scalar 0 works: 0(2, 0) = (0, 0). The output has no direction, but the eigenvector definition still holds. The zero input is excluded because it would satisfy A0 = λ0 for every λ.

Front / question

R = [[0, −1], [1, 0]] rotates every plane arrow 90° counterclockwise. Test (1, 0), then decide whether any nonzero real vector can be an eigenvector of R.

Back / answer

R(1, 0) = (0, 1), not a multiple of (1, 0). No nonzero real vector is an eigenvector of this 90° rotation.

R(1, 0) = (0×1 + (−1)×0, 1×1 + 0×0) = (0, 1). The input lies on the horizontal line and the output on the vertical line. The same quarter-turn moves every nonzero real arrow off its line, while preserving nonzero length. Real scaling cannot make that turn, and λ = 0 cannot describe a nonzero output. Thus no real λ and nonzero real v satisfy Rv = λv.

Front / question

Suppose Av = 3v for a nonzero vector v. Read A^4v aloud, then state its value as a multiple of v.

Back / answer

A^4v = 3^4v = 81v.

The equation Av = 3v verifies that v is an eigenvector with eigenvalue 3. Each application multiplies this same vector direction by 3; four applications use four factors of 3. Read A^4v as 'apply A four times to v,' so the result is 81v.

Front / question

A nonzero eigenvector v has λ = -2. List its positions at steps k = 0, 1, 2, 3 as multiples of v. Does its length grow or shrink with each application, and does it point to the same side of the origin?

Back / answer

Its length doubles each step and its direction alternates: v, -2v, 4v, -8v.

The multipliers at steps 0 through 3 are 1, -2, 4, and -8. Their absolute values are 1, 2, 4, and 8, so the arrow grows. The signs alternate, so the arrow switches sides of the origin every step.

Front / question

A nonzero eigenvector v has λ = -1/2. Write its positions at steps k = 0, 1, 2, 3, and 4 as multiples of v. Does any of these steps land on zero?

Back / answer

v, -v/2, v/4, -v/8, v/16. None is zero.

Begin with coefficient 1 at k = 0. Successive multiplication by -1/2 gives -1/2, 1/4, -1/8, and 1/16. Every listed coefficient is nonzero. Indeed, (-1/2)^k remains nonzero for every finite whole-number k.

Front / question

Let A = [[3, 0], [0, -1]] and v = (0, 2). Verify Av = λv, identify λ, and find A^2v.

Back / answer

Av = (0, -2) = -v, so λ = -1 and A^2v = v = (0, 2).

Matrix-vector multiplication gives Av = (3·0 + 0·2, 0·0 + (-1)·2) = (0, -2). Since v = (0, 2), the output is -1 times v. The second application uses (-1)^2 = 1, so A^2v = (0, 2).

Front / question

Let A = [[2, 0], [0, 1/2]] and w = (1, 1). Compute Aw and A^2w. Why can you not choose one λ and write A^2w = λ^2w here?

Back / answer

Aw = (2, 1/2) and A^2w = (4, 1/4). The first coordinate doubles while the second halves, so w is not an eigenvector and has no single λ for this rule.

One application gives Aw = (2·1, (1/2)·1) = (2, 1/2). A second application gives A(Aw) = (2·2, (1/2)·(1/2)) = (4, 1/4). If Aw were λw, both coordinates would equal the same λ. They are 2 and 1/2, so the eigenvector shortcut does not apply to w.

Front / question

For M = [[1, 1], [0, 1]], compare the true matrix square M^2 with the array obtained by squaring each entry. Are they equal?

Back / answer

M^2 = MM = [[1, 2], [0, 1]], while entry-by-entry squaring gives [[1, 1], [0, 1]]. They are different.

For MM, the top-right entry is 1·1 + 1·1 = 2; the other entries are 1, 0, and 1. Squaring the four entries separately gives 1, 1, 0, and 1. A matrix power represents composition through matrix multiplication, which combines rows and columns.

Front / question

For F(x) = 3x - 4 on real numbers, test whether x* = 2 is fixed. Is x = 1 fixed?

Back / answer

2 is fixed because F(2) = 2. 1 is not fixed because F(1) = -1.

F(2) = 3·2 - 4 = 2, so applying the rule leaves 2 unchanged. F(1) = 3·1 - 4 = -1, which differs from 1. The fixed-point test asks for equality after one application, not just a nearby result.

Front / question

Let F(x) = 2 - x. Starting at 5 gives 5 → -3 → 5. Is 5 fixed or on a genuine two-step cycle? Find one fixed state of this F.

Back / answer

5 is on a genuine two-step cycle, not fixed. The fixed state is x* = 1.

F(5) = 2 - 5 = -3, so 5 changes after one step. F(-3) = 2 - (-3) = 5, so it returns after exactly two steps. Solving F(x) = x gives 2 - x = x, hence x = 1; indeed F(1) = 1.

Front / question

Let x_(k+1) = A x_k: each new state is A times the current state. For A = [[1, 0], [0, -1/2]] and x_0 = (2, 4), find x_1, x_2, x_3 and the fixed state this path approaches. Does it ever reach that state in a finite number of steps?

Back / answer

x_1 = (2, -2), x_2 = (2, 1), x_3 = (2, -1/2). They approach (2, 0) but never reach it at any finite step.

The matrix leaves the first coordinate at 2 and multiplies the second by -1/2 each time. The second coordinates are 4, -2, 1, -1/2, and in general 4(-1/2)^k. Their magnitudes tend to zero, but 4(-1/2)^k is nonzero for every finite k. Also A(2, 0) = (2, 0), so (2, 0) is fixed.

Front / question

A nonzero vector v satisfies Av = -v. Classify the repeated path v → Av → A^2v as fixed, a genuine two-step cycle, or neither.

Back / answer

It is a genuine two-step cycle: v → -v → v.

The first step is Av = -v, which differs from v because v is nonzero. The second is A(-v) = -Av = -(-v) = v. Returning after two steps, with a change after one, is precisely a period-two cycle.

Front / question

A = [[1, 0], [0, 2]] fixes v = (1, 0). Starting instead at w = (1, 1), find Aw and A^2w. Does the fixed vector v prove that every state approaches a fixed state?

Back / answer

Aw = (1, 2), A^2w = (1, 4). No: the vertical coordinate grows, so this starting state does not approach the fixed horizontal axis.

Av = (1, 0) verifies v is fixed. But Aw = (1, 2) and A(Aw) = (1, 4); the second coordinate doubles each time. One fixed eigendirection describes what happens along that direction only. It cannot establish stability for all states.

Front / question

Compare two rules with the same fixed state 7. Rule P(x) = 7; rule Q(x) = 7 + (x - 7)/2. Starting at x_0 = 2, which rule reaches 7 in a finite step, and which merely approaches it?

Back / answer

P reaches 7 after one step and stays there. Q approaches 7 but never reaches it from 2 in a finite number of steps.

Both have fixed state 7 because P(7) = Q(7) = 7. P(2) = 7, so its path is 2 → 7 → 7 → ... . For Q, the initial error is 2 - 7 = -5 and each step halves it: x_k - 7 = -5/2^k. This is nonzero for finite k but tends to zero.