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.
Σχετικά με αυτήν τη δέσμη
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.
Κάρτες σε αυτήν τη δέσμη
Κάρτα 1
Ερώτηση
What are the dimensions (rows × columns) of this matrix?
[ 6 -1 0 ] [ 2 3 5 ]Απάντηση
2 × 3. There are two rows and three columns.
Κάρτα 2
Ερώτηση
A is m × n and B is n × p. What are the dimensions of AB?
Απάντηση
m × p. The shared inner dimension n makes the product defined; the outer dimensions give its shape. All dimensions here are positive integers.
Κάρτα 3
Ερώτηση
For a defined matrix product C = AB, which parts of A and B determine entry C(i, j)? Indices start at 1.
Απάντηση
Row i of A and column j of B. Multiply corresponding entries, then add every pairwise product.
Κάρτα 4
Ερώτηση
A is 4 × 2 and B is 2 × 3. Is AB defined, and what is its shape?
Απάντηση
Yes: 4 × 3. The inner dimensions are both 2.
Κάρτα 5
Ερώτηση
Find entry (2, 1) of AB. Indices start at 1.
A = [ -1 4 ] [ 3 2 ] B = [ 5 -2 ] [ 1 6 ]Απάντηση
17. Use row 2 of A and column 1 of B: 3 × 5 + 2 × 1 = 17.
Κάρτα 6
Ερώτηση
A is 2 × 3 and B is 2 × 4. Is AB defined?
Απάντηση
No. A has 3 columns, but B has 2 rows. The inner dimensions must match.
Κάρτα 7
Ερώτηση
Calculate AB, keeping the result in matrix brackets.
A = [ 2 -1 3 ] B = [ 4 ] [ 5 ] [ -2 ]Απάντηση
[ -3 ]The 1 × 1 entry is 2 × 4 + (−1) × 5 + 3 × (−2) = −3.
Κάρτα 8
Ερώτηση
Calculate AB. Is the product undefined or a valid matrix?
A = [ 0 2 ] B = [ 3 ] [ 0 ]Απάντηση
A valid 1 × 1 matrix:
[ 0 ]The dimensions match, and 0 × 3 + 2 × 0 = 0.
Κάρτα 9
Ερώτηση
Find entry (1, 2) of AB. Indices start at 1.
A = [ 1 -2 4 ] [ 3 0 -1 ] B = [ 2 5 ] [ -1 3 ] [ 4 -2 ]Απάντηση
−9. Row 1 meets column 2: 1 × 5 + (−2) × 3 + 4 × (−2) = 5 − 6 − 8 = −9.
Κάρτα 10
Ερώτηση
A is 3 × k and B is 5 × 2. AB is defined. What is k?
Απάντηση
k = 5. A must have as many columns as B has rows. AB then has shape 3 × 2.
Κάρτα 11
Ερώτηση
Calculate AB.
A = [ 3 ] [ -2 ] B = [ 1 4 ]Απάντηση
[ 3 12 ] [ -2 -8 ]A 2 × 1 matrix times a 1 × 2 matrix gives a 2 × 2 matrix. Each entry has one pairwise product.
Κάρτα 12
Ερώτηση
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 ]Απάντηση
13. Ordinary matrix multiplication uses row 1 against column 1: 2 × 4 + 1 × 5 = 13. Multiplying matching positions omits the second term.
Κάρτα 13
Ερώτηση
A is 2 × 3 and B is 3 × 2. Give the shape of AB and the shape of BA, or say undefined.
Απάντηση
AB: 2 × 2. BA: 3 × 3. Both inner-dimension checks pass, but reversing the order changes the output shape.
Κάρτα 14
Ερώτηση
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 ]Απάντηση
The missing term is 1 × (−5) = −5; the result is [−3]. All three pairs contribute: 8 − 6 − 5 = −3.
Κάρτα 15
Ερώτηση
Calculate AB.
A = [ 1 0 2 ] [ -1 3 1 ] B = [ 2 -1 ] [ 0 4 ] [ 3 2 ]Απάντηση
[ 8 3 ] [ 1 15 ]The entries are 2 + 0 + 6, −1 + 0 + 4, −2 + 0 + 3, and 1 + 12 + 2.
Κάρτα 16
Ερώτηση
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 ]Απάντηση
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.
Κάρτα 17
Ερώτηση
AB = [13]. Find x.
A = [ x 2 ] B = [ 3 ] [ -1 ]Απάντηση
x = 5. The entry equation is 3x − 2 = 13. Substitution gives 3 × 5 − 2 = 13.
Κάρτα 18
Ερώτηση
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 ]Απάντηση
−2. Entry (2, 1) needs row 2 of A: 4 × 1 + 3 × (−2) = −2.
Κάρτα 19
Ερώτηση
A is 2 × 3 and B is 4 × 2. Give the shape of AB and BA, or say undefined.
Απάντηση
AB: undefined. BA: 4 × 3. For AB, 3 ≠ 4. For BA, the inner dimensions are both 2.
Κάρτα 20
Ερώτηση
Calculate AB.
A = [ -2 1 ] [ 3 0 ] B = [ 4 -1 ] [ 2 5 ]Απάντηση
[ -6 7 ] [ 12 -3 ]Row 1 gives −8 + 2 and 2 + 5; row 2 gives 12 + 0 and −3 + 0.
Κάρτα 21
Ερώτηση
A is m × 3, B is 3 × 4, and AB is 2 × 4. What is m?
Απάντηση
m = 2. AB keeps the number of rows of the left matrix A.
Κάρτα 22
Ερώτηση
Do AB and BA agree for this particular pair?
A = [ 2 0 ] [ 0 -1 ] B = [ 3 0 ] [ 0 4 ]Απάντηση
Yes. Both products are:
[ 6 0 ] [ 0 -4 ]Some matrix pairs commute. Matrix multiplication is not commutative in general.
Κάρτα 23
Ερώτηση
Find entry (3, 2) of AB. Indices start at 1.
A = [ 1 2 ] [ 0 -3 ] [ -2 4 ] B = [ 3 1 -1 ] [ 2 -2 5 ]Απάντηση
−10. Use row 3 and column 2: (−2) × 1 + 4 × (−2) = −10.
Κάρτα 24
Ερώτηση
A is 2 × 3 and B is 4 × 5. Is either AB or BA defined?
Απάντηση
Neither is defined. AB fails because 3 ≠ 4; BA fails because 5 ≠ 2.
Κάρτα 25
Ερώτηση
Entry (2, 1) of AB is 22. Find x.
A = [ 2 -1 ] [ 1 3 ] B = [ x 4 ] [ 5 -2 ]Απάντηση
x = 7. Entry (2, 1) gives x + 3 × 5 = 22. Substitution gives 7 + 15 = 22.
Κάρτα 26
Ερώτηση
Calculate both AB and BA.
A = [ 1 -2 ] B = [ 3 ] [ 4 ]Απάντηση
AB:
[ -5 ]BA:
[ 3 -6 ] [ 4 -8 ]AB sums 3 − 8. BA is a 2 × 2 outer product.
Κάρτα 27
Ερώτηση
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 ]Απάντηση
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.
Κάρτα 28
Ερώτηση
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.
Απάντηση
3 × 5. The matching inner dimensions allow multiplication; the outer dimensions determine the result.
Κάρτα 29
Ερώτηση
Calculate AB.
A = [ 2 0 -1 ] B = [ 3 1 ] [ 4 -2 ] [ -2 5 ]Απάντηση
[ 8 -3 ]The entries are 6 + 0 + 2 = 8 and 2 + 0 − 5 = −3. The result has shape 1 × 2.
Κάρτα 30
Ερώτηση
A is 2 × 4 and B is 4 × 3. How many pairwise products must be added to calculate entry (2, 3) of AB?
Απάντηση
Four. Row 2 of A and column 3 of B each contain four entries, so every output entry sums four pairwise products.
Κάρτα 31
Ερώτηση
AB = [10]. Find x.
A = [ 2 x -1 ] B = [ 3 ] [ -2 ] [ 4 ]Απάντηση
x = −4. The equation is 6 − 2x − 4 = 10, so −2x = 8. Substitution gives 6 + 8 − 4 = 10.
Κάρτα 32
Ερώτηση
What is column 2 of AB?
A = [ 1 -2 4 ] [ 0 3 5 ] B = [ 2 0 ] [ -1 0 ] [ 6 0 ]Απάντηση
[ 0 ] [ 0 ]Each entry in column 2 uses the all-zero column 2 of B, so every term is zero.
Κάρτα 33
Ερώτηση
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 ]Απάντηση
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.
Κάρτα 34
Ερώτηση
A and B are both 2 × 3. Does having the same shape make AB defined?
Απάντηση
No. The inner dimensions are 3 and 2, so AB is undefined. Equal shapes alone do not ensure compatibility.
Κάρτα 35
Ερώτηση
Find entry (1, 2) of AB. Indices start at 1.
A = [ 2 1 ] [ -1 3 ] B = [ 4 -2 ] [ 2 4 ]Απάντηση
0. The entry is 2 × (−2) + 1 × 4 = −4 + 4 = 0. Cancellation is a valid result.
Κάρτα 36
Ερώτηση
A is 4 × 2 and B is k × 3. AB is defined and has shape 4 × 3. What is k?
Απάντηση
k = 2. B must have two rows to match A’s two columns.
Κάρτα 37
Ερώτηση
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 ]Απάντηση
[−14]. The second term is (−3) × 4 = −12, so the entry is −2 − 12 = −14.
Κάρτα 38
Ερώτηση
Calculate AI and IA for these matrices.
A = [ 2 -1 ] [ 4 3 ] I = [ 1 0 ] [ 0 1 ]Απάντηση
Both equal A:
[ 2 -1 ] [ 4 3 ]Multiplication by this 2 × 2 identity matrix preserves A in either order.
Κάρτα 39
Ερώτηση
Find entry (3, 1) of AB. Indices start at 1.
A = [ 2 0 ] [ -1 4 ] [ 5 -2 ] B = [ 3 ] [ 7 ]Απάντηση
1. Row 3 of A meets column 1 of B: 5 × 3 + (−2) × 7 = 15 − 14 = 1.
Κάρτα 40
Ερώτηση
Calculate AB.
A = [ 3 -1 ] [ 2 4 ] B = [ 2 ] [ -3 ]Απάντηση
[ 9 ] [ -8 ]The rows give 6 + 3 = 9 and 4 − 12 = −8. The result has shape 2 × 1.
40 κάρτες
Matrix Multiplication Flashcards: Dimensions, Entries & Order
Το Nibomo ανοίγει για να ξεκινήσετε τη μελέτη.