Matrix Multiplication Flashcards: Dimensions, Entries & Order
Practise 40 original matrix multiplication prompts: check dimensions, calculate row-by-column entries, compare AB and BA, and repair common errors.
Apie šį rinkinį
Check the dimensions before doing the arithmetic. These 40 matrix multiplication flashcards give English-reading learners original short problems on compatible shapes, row-by-column entries, small complete products, and the order of the factors.
You’ll decide whether a stated product exists and find its shape; calculate individual entries and short products with signed integers and zeros; and compare AB with BA. Selected reverse problems recover one missing dimension or entry from enough information to determine it uniquely. Local repairs target elementwise multiplication, a wrong row, a missed term, output dimensions, reversed factors, and a sign error. Every front supplies its own matrices or dimensions for mixed review.
The sequence begins with shape notation, compatibility, and the entry rule. It then introduces short calculations before order comparisons, constrained reverse problems, and error repair. Later cards revisit these decisions with rectangular matrices, cancellation, a zero column, a zero product from nonzero factors, and particular pairs that commute, including an identity matrix. Related calculation variants are separated by at least three other cards. All cards use the 'matrix-multiplication' tag; the review scheduler handles long-term spacing.
Know rows, columns, signed arithmetic, and simple equations before starting. Matrix indices start at 1, dimensions are positive integers, and a 1 × 1 product stays in brackets. This evergreen Mathematics deck has no exam or curriculum alignment. It excludes determinants, matrix inverses, elimination, eigenvalues, proofs, transformation composition, programming, and long 3 × 3 product drills. It does not mechanically reverse every calculation: arbitrary output matrices do not uniquely determine their inputs.
The compatibility and row-by-column rules were checked against Georgia Tech’s Interactive Linear Algebra. Only common mathematical facts are used. Every numerical example, prompt, answer, and arrangement is independently authored; no source exercises, textbook wording, competitor cards, or diagrams were copied or adapted.
AI tools assisted with drafting and the original decorative cover. Every final front and back was read in author review, and numerical answers were independently checked with exact integer arithmetic or substitution. This does not claim learner testing or guaranteed results. The CC0 dedication covers original wording, invented examples, organization, metadata, and generated cover to the extent applicable rights exist. Mathematical facts and cited materials retain their own status. This is independent study practice, with no affiliation to or endorsement by the cited organizations.
Šio rinkinio kortelės
1 kortelė
Klausimas
What are the dimensions (rows × columns) of this matrix?
[ 6 -1 0 ] [ 2 3 5 ]Atsakymas
2 × 3. There are two rows and three columns.
2 kortelė
Klausimas
A is m × n and B is n × p. What are the dimensions of AB?
Atsakymas
m × p. The shared inner dimension n makes the product defined; the outer dimensions give its shape. All dimensions here are positive integers.
3 kortelė
Klausimas
For a defined matrix product C = AB, which parts of A and B determine entry C(i, j)? Indices start at 1.
Atsakymas
Row i of A and column j of B. Multiply corresponding entries, then add every pairwise product.
4 kortelė
Klausimas
A is 4 × 2 and B is 2 × 3. Is AB defined, and what is its shape?
Atsakymas
Yes: 4 × 3. The inner dimensions are both 2.
5 kortelė
Klausimas
Find entry (2, 1) of AB. Indices start at 1.
A = [ -1 4 ] [ 3 2 ] B = [ 5 -2 ] [ 1 6 ]Atsakymas
17. Use row 2 of A and column 1 of B: 3 × 5 + 2 × 1 = 17.
6 kortelė
Klausimas
A is 2 × 3 and B is 2 × 4. Is AB defined?
Atsakymas
No. A has 3 columns, but B has 2 rows. The inner dimensions must match.
7 kortelė
Klausimas
Calculate AB, keeping the result in matrix brackets.
A = [ 2 -1 3 ] B = [ 4 ] [ 5 ] [ -2 ]Atsakymas
[ -3 ]The 1 × 1 entry is 2 × 4 + (−1) × 5 + 3 × (−2) = −3.
8 kortelė
Klausimas
Calculate AB. Is the product undefined or a valid matrix?
A = [ 0 2 ] B = [ 3 ] [ 0 ]Atsakymas
A valid 1 × 1 matrix:
[ 0 ]The dimensions match, and 0 × 3 + 2 × 0 = 0.
9 kortelė
Klausimas
Find entry (1, 2) of AB. Indices start at 1.
A = [ 1 -2 4 ] [ 3 0 -1 ] B = [ 2 5 ] [ -1 3 ] [ 4 -2 ]Atsakymas
−9. Row 1 meets column 2: 1 × 5 + (−2) × 3 + 4 × (−2) = 5 − 6 − 8 = −9.
10 kortelė
Klausimas
A is 3 × k and B is 5 × 2. AB is defined. What is k?
Atsakymas
k = 5. A must have as many columns as B has rows. AB then has shape 3 × 2.
11 kortelė
Klausimas
Calculate AB.
A = [ 3 ] [ -2 ] B = [ 1 4 ]Atsakymas
[ 3 12 ] [ -2 -8 ]A 2 × 1 matrix times a 1 × 2 matrix gives a 2 × 2 matrix. Each entry has one pairwise product.
12 kortelė
Klausimas
A learner multiplies matching positions and reports entry (1, 1) of AB as 8. What should that entry be?
A = [ 2 1 ] [ 0 3 ] B = [ 4 -1 ] [ 5 2 ]Atsakymas
13. Ordinary matrix multiplication uses row 1 against column 1: 2 × 4 + 1 × 5 = 13. Multiplying matching positions omits the second term.
13 kortelė
Klausimas
A is 2 × 3 and B is 3 × 2. Give the shape of AB and the shape of BA, or say undefined.
Atsakymas
AB: 2 × 2. BA: 3 × 3. Both inner-dimension checks pass, but reversing the order changes the output shape.
14 kortelė
Klausimas
For AB below, a learner stops at 8 − 6 = 2. What term is missing, and what is the correct 1 × 1 result?
A = [ 2 -3 1 ] B = [ 4 ] [ 2 ] [ -5 ]Atsakymas
The missing term is 1 × (−5) = −5; the result is [−3]. All three pairs contribute: 8 − 6 − 5 = −3.
15 kortelė
Klausimas
Calculate AB.
A = [ 1 0 2 ] [ -1 3 1 ] B = [ 2 -1 ] [ 0 4 ] [ 3 2 ]Atsakymas
[ 8 3 ] [ 1 15 ]The entries are 2 + 0 + 6, −1 + 0 + 4, −2 + 0 + 3, and 1 + 12 + 2.
16 kortelė
Klausimas
For these matrices, compare entry (1, 2) of AB with entry (1, 2) of BA. Are the products equal?
A = [ 1 1 ] [ 0 1 ] B = [ 2 0 ] [ 0 3 ]Atsakymas
AB has entry 3; BA has entry 2. The products are unequal. AB uses 1 × 0 + 1 × 3; BA uses 2 × 1 + 0 × 1. Both products are 2 × 2, so equal shapes do not imply equal matrices.
17 kortelė
Klausimas
AB = [13]. Find x.
A = [ x 2 ] B = [ 3 ] [ -1 ]Atsakymas
x = 5. The entry equation is 3x − 2 = 13. Substitution gives 3 × 5 − 2 = 13.
18 kortelė
Klausimas
A learner uses row 1 of A and gets 4 for entry (2, 1) of AB. Correct the entry.
A = [ 2 -1 ] [ 4 3 ] B = [ 1 5 ] [ -2 0 ]Atsakymas
−2. Entry (2, 1) needs row 2 of A: 4 × 1 + 3 × (−2) = −2.
19 kortelė
Klausimas
A is 2 × 3 and B is 4 × 2. Give the shape of AB and BA, or say undefined.
Atsakymas
AB: undefined. BA: 4 × 3. For AB, 3 ≠ 4. For BA, the inner dimensions are both 2.
20 kortelė
Klausimas
Calculate AB.
A = [ -2 1 ] [ 3 0 ] B = [ 4 -1 ] [ 2 5 ]Atsakymas
[ -6 7 ] [ 12 -3 ]Row 1 gives −8 + 2 and 2 + 5; row 2 gives 12 + 0 and −3 + 0.
21 kortelė
Klausimas
A is m × 3, B is 3 × 4, and AB is 2 × 4. What is m?
Atsakymas
m = 2. AB keeps the number of rows of the left matrix A.
22 kortelė
Klausimas
Do AB and BA agree for this particular pair?
A = [ 2 0 ] [ 0 -1 ] B = [ 3 0 ] [ 0 4 ]Atsakymas
Yes. Both products are:
[ 6 0 ] [ 0 -4 ]Some matrix pairs commute. Matrix multiplication is not commutative in general.
23 kortelė
Klausimas
Find entry (3, 2) of AB. Indices start at 1.
A = [ 1 2 ] [ 0 -3 ] [ -2 4 ] B = [ 3 1 -1 ] [ 2 -2 5 ]Atsakymas
−10. Use row 3 and column 2: (−2) × 1 + 4 × (−2) = −10.
24 kortelė
Klausimas
A is 2 × 3 and B is 4 × 5. Is either AB or BA defined?
Atsakymas
Neither is defined. AB fails because 3 ≠ 4; BA fails because 5 ≠ 2.
25 kortelė
Klausimas
Entry (2, 1) of AB is 22. Find x.
A = [ 2 -1 ] [ 1 3 ] B = [ x 4 ] [ 5 -2 ]Atsakymas
x = 7. Entry (2, 1) gives x + 3 × 5 = 22. Substitution gives 7 + 15 = 22.
26 kortelė
Klausimas
Calculate both AB and BA.
A = [ 1 -2 ] B = [ 3 ] [ 4 ]Atsakymas
AB:
[ -5 ]BA:
[ 3 -6 ] [ 4 -8 ]AB sums 3 − 8. BA is a 2 × 2 outer product.
27 kortelė
Klausimas
Does AB equal the zero matrix for this pair, even though neither factor is the zero matrix?
A = [ 1 -1 ] [ 2 -2 ] B = [ 3 0 ] [ 3 0 ]Atsakymas
Yes.
AB = [ 0 0 ] [ 0 0 ]The first column gives 3 − 3 and 6 − 6; the second gives zeros. A zero product does not force either factor to be the zero matrix.
28 kortelė
Klausimas
A is 3 × 2 and B is 2 × 5. A learner calls AB a 2 × 2 matrix because the inner dimensions are 2. Correct the shape.
Atsakymas
3 × 5. The matching inner dimensions allow multiplication; the outer dimensions determine the result.
29 kortelė
Klausimas
Calculate AB.
A = [ 2 0 -1 ] B = [ 3 1 ] [ 4 -2 ] [ -2 5 ]Atsakymas
[ 8 -3 ]The entries are 6 + 0 + 2 = 8 and 2 + 0 − 5 = −3. The result has shape 1 × 2.
30 kortelė
Klausimas
A is 2 × 4 and B is 4 × 3. How many pairwise products must be added to calculate entry (2, 3) of AB?
Atsakymas
Four. Row 2 of A and column 3 of B each contain four entries, so every output entry sums four pairwise products.
31 kortelė
Klausimas
AB = [10]. Find x.
A = [ 2 x -1 ] B = [ 3 ] [ -2 ] [ 4 ]Atsakymas
x = −4. The equation is 6 − 2x − 4 = 10, so −2x = 8. Substitution gives 6 + 8 − 4 = 10.
32 kortelė
Klausimas
What is column 2 of AB?
A = [ 1 -2 4 ] [ 0 3 5 ] B = [ 2 0 ] [ -1 0 ] [ 6 0 ]Atsakymas
[ 0 ] [ 0 ]Each entry in column 2 uses the all-zero column 2 of B, so every term is zero.
33 kortelė
Klausimas
A learner computes BA while asked for AB and reports entry (1, 1) as 3. What is entry (1, 1) of AB?
A = [ 1 2 ] [ 0 -1 ] B = [ 3 0 ] [ 4 2 ]Atsakymas
11. AB uses row 1 of A and column 1 of B: 1 × 3 + 2 × 4 = 11. The value 3 comes from reversing the factors.
34 kortelė
Klausimas
A and B are both 2 × 3. Does having the same shape make AB defined?
Atsakymas
No. The inner dimensions are 3 and 2, so AB is undefined. Equal shapes alone do not ensure compatibility.
35 kortelė
Klausimas
Find entry (1, 2) of AB. Indices start at 1.
A = [ 2 1 ] [ -1 3 ] B = [ 4 -2 ] [ 2 4 ]Atsakymas
0. The entry is 2 × (−2) + 1 × 4 = −4 + 4 = 0. Cancellation is a valid result.
36 kortelė
Klausimas
A is 4 × 2 and B is k × 3. AB is defined and has shape 4 × 3. What is k?
Atsakymas
k = 2. B must have two rows to match A’s two columns.
37 kortelė
Klausimas
For AB below, a learner writes −2 + 12 = 10. Correct the sign error and give the 1 × 1 result.
A = [ 1 -3 ] B = [ -2 ] [ 4 ]Atsakymas
[−14]. The second term is (−3) × 4 = −12, so the entry is −2 − 12 = −14.
38 kortelė
Klausimas
Calculate AI and IA for these matrices.
A = [ 2 -1 ] [ 4 3 ] I = [ 1 0 ] [ 0 1 ]Atsakymas
Both equal A:
[ 2 -1 ] [ 4 3 ]Multiplication by this 2 × 2 identity matrix preserves A in either order.
39 kortelė
Klausimas
Find entry (3, 1) of AB. Indices start at 1.
A = [ 2 0 ] [ -1 4 ] [ 5 -2 ] B = [ 3 ] [ 7 ]Atsakymas
1. Row 3 of A meets column 1 of B: 5 × 3 + (−2) × 7 = 15 − 14 = 1.
40 kortelė
Klausimas
Calculate AB.
A = [ 3 -1 ] [ 2 4 ] B = [ 2 ] [ -3 ]Atsakymas
[ 9 ] [ -8 ]The rows give 6 + 3 = 9 and 4 − 12 = −8. The result has shape 2 × 1.
40 kortelių
Matrix Multiplication Flashcards: Dimensions, Entries & Order
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