math-2017:N.VM.9math-2017:N.VM.10
math-2017:N.VM.9 Understand that, unlike multiplication of numbers, matrix multiplication for square matrices is not a commutative operation, but still satisfies the associative and distributive properties.
math-2017:N.VM.10 Understand that the zero and identity matrices play a role in matrix addition and multiplication analogous to the role of 0 and 1 in the real numbers. The determinant of a square matrix is nonzero if and only if the matrix has a multiplicative inverse.
Learning goal: Understand that, unlike multiplication of numbers, matrix multiplication for square matrices is not a commutative operation but still satisfies the associative and distributive properties; understand that the zero and identity matrices play a role in matrix addition and multiplication analogous to the role of 0 and 1 in the real numbers; and understand that the determinant of a square matrix is nonzero if and only if the matrix has a multiplicative inverse.
Scope note. N.VM.9 and N.VM.10 are marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate standards "for students who are planning on taking advanced courses". They are taught here in full, as extension content for students on that path.
Square matrices and the question of order. For real numbers, 3 × 5 and 5 × 3 are the same, and we rarely think about it. For two square matrices of the same size, both AB and BA are defined and have the same dimensions, so the question "are they equal?" makes sense, and the answer is usually no. Take A with rows [1 2] and [3 4], and B with rows [0 1] and [1 0]. Compute AB entry by entry: (1,1) = 1(0) + 2(1) = 2; (1,2) = 1(1) + 2(0) = 1; (2,1) = 3(0) + 4(1) = 4; (2,2) = 3(1) + 4(0) = 3. So AB has rows [2 1] and [4 3]. Now BA: (1,1) = 0(1) + 1(3) = 3; (1,2) = 0(2) + 1(4) = 4; (2,1) = 1(1) + 0(3) = 1; (2,2) = 1(2) + 0(4) = 2. BA has rows [3 4] and [1 2]. AB ≠ BA. Multiplying by B on the right swapped the columns of A; multiplying on the left swapped the rows. Different operations, different answers. Matrix multiplication is not commutative, and a single example like this one is enough to prove it, because a property that claims to hold always is refuted by one failure.
What still works: associativity. Although the order of the factors matters, the grouping does not: (AB)C = A(BC) whenever the products are defined. With C having rows [1 1] and [0 1]: (AB)C = [2 1; 4 3][1 1; 0 1] has entries (1,1) = 2 + 0 = 2, (1,2) = 2 + 1 = 3, (2,1) = 4 + 0 = 4, (2,2) = 4 + 3 = 7, so rows [2 3] and [4 7]. For A(BC), first BC = [0 1; 1 0][1 1; 0 1] = rows [0 1] and [1 1]; then A(BC) = [1 2; 3 4][0 1; 1 1] has entries (1,1) = 0 + 2 = 2, (1,2) = 1 + 2 = 3, (2,1) = 0 + 4 = 4, (2,2) = 3 + 4 = 7, again rows [2 3] and [4 7]. The same matrix, so we may write ABC without brackets, as long as we keep the factors in order.
What still works: distributivity. A(B + C) = AB + AC, and (B + C)A = BA + CA, with the multiplier kept on the same side throughout. Check the first with our matrices. B + C has rows [1 2] and [1 1]. A(B + C) = [1 2; 3 4][1 2; 1 1] has entries (1,1) = 1 + 2 = 3, (1,2) = 2 + 2 = 4, (2,1) = 3 + 4 = 7, (2,2) = 6 + 4 = 10: rows [3 4] and [7 10]. Separately, AC = [1 2; 3 4][1 1; 0 1] has rows [1 3] and [3 7], and AB + AC = [2 1; 4 3] + [1 3; 3 7] = rows [3 4] and [7 10]. Equal. Distributivity is what lets you expand brackets, but because of non-commutativity you must not reorder: (A + B)² = (A + B)(A + B) = A² + AB + BA + B², and AB + BA does not collapse to 2AB.
The zero matrix. The zero matrix O of a given size has every entry 0. For matrices of matching dimensions, A + O = A and A − A = O, so O plays the role that 0 plays in addition. For multiplication, AO = O and OA = O whenever the products are defined, just as multiplying a number by 0 gives 0. One difference from numbers deserves care: for real numbers, ab = 0 forces a = 0 or b = 0, but for matrices a product can be O with neither factor zero. The matrix with rows [1 0], [0 0] times the matrix with rows [0 0], [0 1] is O, entry by entry: (1,1) = 0, (1,2) = 0, (2,1) = 0, (2,2) = 0.
The identity matrix. The identity matrix I of size n is the square matrix with 1s on the main diagonal, top-left to bottom-right, and 0s everywhere else. For 2 × 2 it has rows [1 0] and [0 1]. For any matrix A of compatible size, AI = A and IA = A; check with A: AI = [1 2; 3 4][1 0; 0 1] gives (1,1) = 1 + 0 = 1, (1,2) = 0 + 2 = 2, and so on, reproducing A. I plays for multiplication the role that 1 plays for numbers, and it is one of the few matrices that commutes with everything of its size.
Inverses and the determinant. For a nonzero number a, the reciprocal 1/a satisfies a × (1/a) = 1. The analogous object for a square matrix A is a multiplicative inverse A⁻¹ with AA⁻¹ = A⁻¹A = I. Not every matrix has one, and the determinant decides. For a 2 × 2 matrix with rows [a b] and [c d], the determinant is det A = ad − bc. The determinant of a square matrix is nonzero if and only if the matrix has a multiplicative inverse, and for 2 × 2 the inverse is then A⁻¹ = (1/det A) × [d −b; −c a]: swap the diagonal entries, negate the other two, divide by the determinant.
A fully worked model. For A with rows [1 2] and [3 4], det A = 1(4) − 2(3) = 4 − 6 = −2, nonzero, so A⁻¹ exists: (1/−2)[4 −2; −3 1] = rows [−2 1] and [1.5 −0.5]. Verify AA⁻¹: (1,1) = 1(−2) + 2(1.5) = −2 + 3 = 1; (1,2) = 1(1) + 2(−0.5) = 1 − 1 = 0; (2,1) = 3(−2) + 4(1.5) = −6 + 6 = 0; (2,2) = 3(1) + 4(−0.5) = 3 − 2 = 1. That is I. Verifying A⁻¹A gives I as well: a matrix and its inverse always commute, although most pairs of matrices do not. Now take D with rows [2 4] and [1 2]: det D = 2(2) − 4(1) = 0, so no inverse exists. The formula would divide by zero, but the reason is deeper: row 1 of D is twice row 2, so in any product DX the first row is twice the second, and no such matrix can equal I, whose rows are not multiples of each other.
Why inverses matter. If A has an inverse, the equation Ax = b is solved by x = A⁻¹b, exactly as ax = b is solved by x = b/a. With A as above and b = [5; 11]: A⁻¹b = [−2(5) + 1(11); 1.5(5) − 0.5(11)] = [1; 2]. Check: 1(1) + 2(2) = 5 and 3(1) + 4(2) = 11. When the determinant is zero the system has either no solution or infinitely many, and no single inverse can produce it.
The misconception. Students carry over habits from numbers: they cancel matrices across an equation, reorder products, and write A/B. None of these is safe. To undo multiplication by A on the left you multiply by A⁻¹ on the left, and only if A⁻¹ exists; "dividing by a matrix" has no meaning, and A⁻¹B and BA⁻¹ are generally different. Every manipulation must respect the side on which a factor sits.
Worked, showing non-commutativity: A has rows [1 2], [3 4]; B has rows [0 1], [1 0]. AB: (1,1) = 0 + 2 = 2, (1,2) = 1 + 0 = 1, (2,1) = 0 + 4 = 4, (2,2) = 3 + 0 = 3, so rows [2 1], [4 3]. BA: (1,1) = 0 + 3 = 3, (1,2) = 0 + 4 = 4, (2,1) = 1 + 0 = 1, (2,2) = 2 + 0 = 2, so rows [3 4], [1 2]. Since AB ≠ BA, matrix multiplication is not commutative. Note the pattern: B on the right swaps the columns of A; B on the left swaps the rows.
Worked, associativity check: With C having rows [1 1], [0 1]: (AB)C = [2 1; 4 3][1 1; 0 1] = rows [2 3], [4 7]. BC = rows [0 1], [1 1], and A(BC) = [1 2; 3 4][0 1; 1 1] = rows [0 + 2, 1 + 2], [0 + 4, 3 + 4] = rows [2 3], [4 7]. The groupings agree, so ABC is unambiguous.
Worked, distributivity check: B + C = rows [1 2], [1 1]. A(B + C) = rows [1 + 2, 2 + 2], [3 + 4, 6 + 4] = rows [3 4], [7 10]. AB = rows [2 1], [4 3] and AC = rows [1 3], [3 7], so AB + AC = rows [3 4], [7 10]. Equal, as distributivity promises.
Worked, inverse from the determinant: det A = 1(4) − 2(3) = −2 ≠ 0, so A⁻¹ = (1/−2)[4 −2; −3 1] = rows [−2 1], [1.5 −0.5]. Check AA⁻¹: (1,1) = −2 + 3 = 1, (1,2) = 1 − 1 = 0, (2,1) = −6 + 6 = 0, (2,2) = 3 − 2 = 1: the identity. Solve Ax = [5; 11]: x = A⁻¹[5; 11] = [−10 + 11; 7.5 − 5.5] = [1; 2]; substituting back gives 5 and 11.
Worked, no inverse: D has rows [2 4], [1 2]. det D = 2(2) − 4(1) = 0. Suppose some X had DX = I. Row 1 of D is 2 × row 2, so row 1 of DX is 2 × row 2 of DX for any X; but row 1 of I, [1 0], is not 2 × [0 1]. Contradiction, so no inverse exists. The zero determinant flags exactly this collapse.
Worked, zero and identity: With O the 2 × 2 zero matrix, A + O = A and AO = OA = O. With I the 2 × 2 identity, AI = IA = A; for example row 2 of IA is 0[1 2] + 1[3 4] = [3 4]. And the product [1 0; 0 0][0 0; 0 1] = O although neither factor is O, a behavior real numbers never show.
Recording task: Using A = [1 2; 3 4], B = [0 1; 1 0] and C = [1 1; 0 1], compute and record in a table: AB, BA, (AB)C, A(BC), A(B + C) and AB + AC, showing the row-times-column sum for every entry of AB and BA. Under the table write three sentences: one stating which pair of results differs and what property fails, one stating which pairs agree and which properties they illustrate, and one computing det A = −2 and A⁻¹ = [−2 1; 1.5 −0.5] with the check that AA⁻¹ = I written entry by entry.
Independent performance: Choose two 2 × 2 matrices of your own with no zero entries and compute both products to show they differ. Then choose a third matrix and verify associativity and the distributive property with full entry-by-entry work. Next, compute the determinant of each of your matrices, find the inverse of one whose determinant is nonzero, verify the inverse in both orders, and use it to solve a system Ax = b with a b you choose, checking by substitution. Finally construct a 2 × 2 matrix with nonzero entries whose determinant is zero, and explain in one paragraph, without using the formula, why no inverse can exist for it. Success criteria: every product entry shows its sum; the non-commutative pair is explicitly compared; associativity and distributivity are each verified with the same matrices; the inverse checks to I in both orders; the solved system checks by substitution; and the zero-determinant explanation refers to the rows or columns being proportional.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Compute AB and BA for A = [1 2; 3 4] and B = [0 1; 1 0]. What does the comparison prove about matrix multiplication?
AB has rows [2 1] and [4 3]; BA has rows [3 4] and [1 2]. They differ, so matrix multiplication is not commutative. One counterexample is enough, because commutativity would have to hold for every pair. Unlike multiplication of numbers, the order of the factors is part of the product's definition.
Show that (AB)C = A(BC) for the matrices A = [1 2; 3 4], B = [0 1; 1 0], C = [1 1; 0 1], and explain why this lets us write ABC.
AB = [2 1; 4 3], so (AB)C = [2 3; 4 7]. BC = [0 1; 1 1], so A(BC) = [0 + 2, 1 + 2; 0 + 4, 3 + 4] = [2 3; 4 7]. Equal, which is the associative property. Because either grouping gives the same matrix, ABC is unambiguous as long as the factors stay in order.
Explain why (A + B)² is A² + AB + BA + B² and not A² + 2AB + B² for matrices.
Distributivity gives (A + B)(A + B) = A(A + B) + B(A + B) = A² + AB + BA + B². For numbers AB and BA are equal, so they combine into 2AB. For matrices AB and BA are generally different, as A = [1 2; 3 4] and B = [0 1; 1 0] show, so the two middle terms cannot be merged.
In what sense do the zero matrix and the identity matrix play the roles of 0 and 1? Give one way the zero matrix behaves differently from the number 0.
A + O = A, so O is the additive identity like 0; AO = OA = O, like multiplying by 0. AI = IA = A, so I is the multiplicative identity like 1. A difference: for numbers ab = 0 forces a = 0 or b = 0, but [1 0; 0 0][0 0; 0 1] = O with neither factor zero, so a zero product does not force a zero factor.
Find the determinant of A = [1 2; 3 4] and use it to find A⁻¹. Verify by multiplication.
det A = 1(4) − 2(3) = −2, nonzero, so an inverse exists: A⁻¹ = (1/−2)[4 −2; −3 1] = [−2 1; 1.5 −0.5]. Check AA⁻¹: (1,1) = −2 + 3 = 1, (1,2) = 1 − 1 = 0, (2,1) = −6 + 6 = 0, (2,2) = 3 − 2 = 1, the identity matrix. The formula swaps the diagonal, negates the off-diagonal and divides by the determinant.
D = [2 4; 1 2] has determinant 0. Without using the inverse formula, explain why D has no multiplicative inverse.
Row 1 of D is twice row 2. In any product DX, row 1 is computed from row 1 of D and row 2 from row 2 of D, so row 1 of DX is twice row 2 of DX. The identity matrix has rows [1 0] and [0 1], and [1 0] is not twice [0 1]. So DX = I is impossible for every X. A zero determinant signals exactly this proportionality of rows.
Use A⁻¹ = [−2 1; 1.5 −0.5] to solve the system x + 2y = 5, 3x + 4y = 11, and check.
The system is A[x; y] = [5; 11] with A = [1 2; 3 4]. Multiply both sides on the left by A⁻¹: [x; y] = A⁻¹[5; 11] = [−2(5) + 1(11); 1.5(5) − 0.5(11)] = [1; 2]. Check: 1 + 2(2) = 5 and 3(1) + 4(2) = 11. Multiplying on the left of both sides keeps the order consistent, which is required since the product is not commutative.
State the condition under which a square matrix has a multiplicative inverse, and give one 2 × 2 example with an inverse and one without, with the determinant of each.
A square matrix has an inverse if and only if its determinant is nonzero. A = [1 2; 3 4] has det −2, so it has an inverse. D = [2 4; 1 2] has det 2(2) − 4(1) = 0, so it has none. The determinant test works in both directions: nonzero guarantees an inverse, zero guarantees there is none.