Precalculus Quiz: Properties Of Matrix Operations

What this quiz covers

This quiz focuses on Properties Of Matrix Operations, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Yes; because $A$ and $B$ are $2\times 2$, all $2\times 2$ matrices commute under multiplication.
2. No; matrix addition is not commutative, so $AB\ne BA$.
3. Yes; because $AI=IA$ for the identity matrix $I$, all matrices commute with each other.
4. No; matrix multiplication is not commutative in general, so $AB\ne BA$ is possible even for square matrices. (correct answer)

Explanation: This question tests understanding of the non-commutativity of matrix multiplication. Unlike multiplication of real numbers, matrix multiplication is not commutative: for most matrices A and B, AB ≠ BA, and even when both products are defined (square matrices), they typically give different results—only special cases like multiplication by the identity matrix commute. For example, with A = [[1,2],[0,1]] and B = [[3,0],[4,3]], computing AB gives [[11,6],[4,3]] while BA gives [[3,6],[4,11]], demonstrating that the order of multiplication matters—reversing the order changes the outcome. Choice D is correct because it accurately states that matrix multiplication is not commutative in general, so AB ≠ BA is possible even for square matrices. Choice A incorrectly claims that all 2×2 matrices commute under multiplication, but this is false—commutativity only holds in special cases, not for all square matrices of the same size. Key difference from numbers: matrix multiplication is NOT commutative (AB ≠ BA in general), so always pay attention to the order when multiplying matrices, but addition IS commutative (A + B = B + A), so order doesn't matter when adding. To remember the properties: addition and multiplication are both associative (can regroup) and both operations distribute over addition, but only addition is commutative (can reorder)—multiplication order matters.

Question 2

1. The distributive property fails because $MN \neq NM$, so $(M + N)^2 \neq M^2 + 2MN + N^2$
2. The distributive property works for addition but fails for subtraction when $MN \neq NM$
3. The distributive property is limited: it only applies when $M$ and $N$ commute
4. The distributive property works: $P(M + N) = PM + PN$ for any matrix $P$ (correct answer)

Explanation: When you encounter matrix operations questions, focus on which properties transfer from regular algebra and which don't. The key insight here is distinguishing between different types of distributive properties. The distributive property for matrices states that $P(M + N) = PM + PN$ and $(M + N)Q = MQ + NQ$ for any compatible matrices. This property always holds regardless of whether the matrices commute, because matrix addition and multiplication follow consistent rules about how entries combine. You can verify this by expanding the matrix products entry by entry. Choice A confuses the distributive property with the expansion of $(M + N)^2$. While it's true that $(M + N)^2 = M^2 + MN + NM + N^2$ (which doesn't simplify to $M^2 + 2MN + N^2$ when $MN \neq NM$), this isn't about the distributive property failing—it's about commutativity affecting algebraic expansions. Choice B incorrectly suggests the distributive property behaves differently for subtraction. The distributive property $P(M - N) = PM - PN$ works just as well as for addition. Choice C wrongly claims distributivity only works for commuting matrices. Commutativity affects whether $MN = NM$, but it doesn't impact whether $P(M + N) = PM + PN$. Remember this distinction: commutativity (whether $AB = BA$) is separate from distributivity (whether $A(B + C) = AB + AC$). Matrix multiplication isn't generally commutative, but distributivity over addition always holds.

Question 3

1. Yes; matrix multiplication is associative whenever the products are defined. (correct answer)
2. No; matrix multiplication is never associative.
3. Yes, but only when $AB=BA$.
4. No; $(AB)C=A(BC)$ holds only for $2\times2$ matrices.

Explanation: This question tests understanding of the associativity of matrix multiplication. Matrix multiplication has the associative property, meaning (AB)C = A(BC), so we can group operations differently without changing the result, but it lacks commutativity, so we cannot change the order of multiplication (AB ≠ BA in general). The associative property allows us to compute (AB)C or A(BC) and get the same result, so we can choose which multiplication to perform first based on computational convenience without worrying about different answers. Choice A is correct because it properly describes associativity, stating that matrix multiplication is associative whenever the products are defined. Choice B incorrectly claims that matrix multiplication is not associative, but (AB)C = A(BC) always holds (when dimensions are compatible). To remember the properties: addition and multiplication are both associative (can regroup) and both operations distribute over addition, but only addition is commutative (can reorder)—multiplication order matters. When simplifying matrix expressions, you can freely use associativity to regroup operations ((AB)C = A(BC)) and distributivity to expand or factor (A(B + C) = AB + AC), but you cannot change the order of multiplication without potentially changing the result.

Question 4

1. $\begin{pmatrix}2&8\\1&1\end{pmatrix}$ (correct answer)
2. $\begin{pmatrix}2&4\\1&11\end{pmatrix}$
3. $\begin{pmatrix}2&-8\\1&-1\end{pmatrix}$
4. $\begin{pmatrix}1&4\\0&-1\end{pmatrix}$

Explanation: This question tests understanding of matrix multiplication computation. Matrix multiplication is performed by taking the dot product of rows from the first matrix with columns from the second matrix, where each element (i,j) in the product is the sum of products of corresponding elements from row i of the first matrix and column j of the second matrix. For A = [[2,0],[1,3]] and B = [[1,4],[0,-1]], to find AB we compute: element (1,1) = 2(1) + 0(0) = 2, element (1,2) = 2(4) + 0(-1) = 8, element (2,1) = 1(1) + 3(0) = 1, and element (2,2) = 1(4) + 3(-1) = 1, giving AB = [[2,8],[1,1]]. Choice A is correct because it shows the accurate result of the matrix multiplication AB = [[2,8],[1,1]]. Choice B shows [[2,4],[1,11]], which contains computational errors in the calculation of the matrix product. Unlike multiplication of real numbers, matrix multiplication is not commutative: for most matrices A and B, AB ≠ BA, and even when both products are defined (square matrices), they typically give different results—only special cases like multiplication by the identity matrix commute. Key difference from numbers: matrix multiplication is NOT commutative (AB ≠ BA in general), so always pay attention to the order when multiplying matrices, but addition IS commutative (A + B = B + A), so order doesn't matter when adding.

Question 5

1. The associative property fails for these particular matrices
2. The distributive property doesn't apply to powers of matrix products
3. Matrix multiplication is not commutative, so $(RS)^2 = RSRS \neq R^2S^2 = RRSS$ (correct answer)
4. These matrices are not invertible, which prevents the usual power rules from applying

Explanation: When you encounter matrix multiplication problems involving powers and products, the key insight is that matrix multiplication is fundamentally different from regular number multiplication—it's not commutative. Let's compute the required products. First, $RS = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$. Next, $SR = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$. Notice that $RS \neq SR$—this confirms these matrices don't commute. Now for the key insight: $(RS)^2 = (RS)(RS) = RSRS$, while $R^2S^2 = (RR)(SS) = RRSS$. These are only equal if we can rearrange the middle terms, which requires $SR = RS$. Since we just showed $RS \neq SR$, we have $(RS)^2 \neq R^2S^2$. Choice C correctly identifies that matrix multiplication's non-commutativity prevents the usual power rule $(ab)^n = a^n b^n$ from working. Choice A is wrong because the associative property does hold for matrices—we can group multiplications differently. Choice B misses the point; the issue isn't about distributive properties but commutativity. Choice D is incorrect because both matrices are actually invertible (their determinants are non-zero), and invertibility doesn't affect this power rule anyway. Remember: when working with matrix powers and products, always check whether the matrices commute before applying familiar algebraic rules.

Question 6

1. Yes; matrix multiplication is commutative for all square matrices, so $AB=BA$.
2. No; matrix multiplication is not commutative in general, so typically $AB\ne BA$ even when both products are defined. (correct answer)
3. No; matrix addition is not commutative, so $AB\ne BA$.
4. Yes; because $A$ and $B$ are $2\times 2$, order does not matter in multiplication.

Explanation: This question tests understanding of the properties of matrix operations, specifically commutativity for multiplication. Unlike multiplication of real numbers, matrix multiplication is not commutative: for most matrices A and B, AB ≠ BA, and even when both products are defined (square matrices), they typically give different results—only special cases like multiplication by the identity matrix commute. For example, with A = [[1,2],[3,4]] and B = [[0,1],[2,-1]], computing AB gives [[4,-1],[8,-1]] while BA gives [[3,4],[-1,0]], demonstrating that the order of multiplication matters—reversing the order changes the outcome. Choice B is correct because it accurately states which properties hold and correctly identifies non-commutativity of multiplication. Choice A incorrectly claims that matrix multiplication is commutative (AB = BA), but this is false in general—only special matrices like the identity commute with all matrices. Key difference from numbers: matrix multiplication is NOT commutative (AB ≠ BA in general), so always pay attention to the order when multiplying matrices, but addition IS commutative (A + B = B + A), so order doesn't matter when adding. To remember the properties: addition and multiplication are both associative (can regroup) and both operations distribute over addition, but only addition is commutative (can reorder)—multiplication order matters.

Question 7

1. $(AB)C$ is $2 \times 2$ and $A(BC)$ is $3 \times 3$
2. $(AB)C$ is $2 \times 2$ and $A(BC)$ is $2 \times 2$ (correct answer)
3. $(AB)C$ is $3 \times 4$ and $A(BC)$ is $2 \times 2$
4. $(AB)C$ is $2 \times 4$ and $A(BC)$ is $2 \times 2$

Explanation: For $(AB)C$: First $AB$ gives a $2 \times 4$ matrix, then $(AB)C$ gives $2 \times 2$. For $A(BC)$: First $BC$ gives a $3 \times 2$ matrix, then $A(BC)$ gives $2 \times 2$. Both results have the same dimensions, which is consistent with the associative property. Choice A incorrectly determines dimensions of $A(BC)$. Choice C confuses intermediate dimensions with final dimensions. Choice D incorrectly gives dimensions of the intermediate product $AB$.

Question 8

1. $(A+B)C=AC+BC$ (correct answer)
2. $(A+B)C=CA+CB$
3. $(A+B)C=A+BC$
4. $(A+B)C=AB+CC$

Explanation: This question tests understanding of the properties of matrix operations, specifically distributivity. The distributive property states that matrix multiplication distributes over addition: A(B + C) = AB + AC and (A + B)C = AC + BC, allowing us to expand or factor matrix expressions just like with numbers. To expand (A + B)C, we apply the distributive property: (A + B)C = AC + BC, meaning we multiply A by C and B by C separately, then add the results, just as we would distribute a factor across a sum in algebra. Choice A is correct because it properly applies distributive property. Choice B reverses the order in the expanded form, writing CA + CB instead of AC + BC, but order matters in matrix multiplication. The distributive property A(B + C) = AB + AC is crucial for expanding matrix expressions, and it works just like distributing with numbers, but remember to keep the multiplication order the same (A on the left in both terms). When simplifying matrix expressions, you can freely use associativity to regroup operations ((AB)C = A(BC)) and distributivity to expand or factor (A(B + C) = AB + AC), but you cannot change the order of multiplication without potentially changing the result.

Question 9

1. commutativity of addition
2. distributivity of multiplication over addition
3. associativity of addition (correct answer)
4. commutativity of multiplication

Explanation: This question tests understanding of the properties of matrix operations, specifically the associativity of addition. Matrix addition is commutative (A + B = B + A) and associative ((A + B) + C = A + (B + C)), but matrix multiplication is generally NOT commutative (AB ≠ BA in general) even though it is associative ((AB)C = A(BC)). The associative property allows us to compute (A + B) + C or A + (B + C) and get the same result, so we can choose which addition to perform first without changing the outcome. Choice C is correct because it correctly identifies the associativity of addition. Choice A confuses the properties, stating that it is commutativity of addition when actually it is associativity—commutativity is about reordering, not regrouping. To remember the properties: addition and multiplication are both associative (can regroup) and both operations distribute over addition, but only addition is commutative (can reorder)—multiplication order matters. When simplifying matrix expressions, you can freely use associativity to regroup operations ((AB)C = A(BC)) and distributivity to expand or factor (A(B + C) = AB + AC), but you cannot change the order of multiplication without potentially changing the result.

Question 10

1. No; matrix multiplication is not associative, so $(AB)C\ne A(BC)$ in general.
2. Yes; matrix multiplication is associative, so $(AB)C=A(BC)$ whenever both sides are defined. (correct answer)
3. Yes, but only when $AB=BA$ (commutativity is required for associativity).
4. No; $(AB)C=A(BC)$ holds only when $A$ is the identity matrix $I$.

Explanation: This question tests understanding of the properties of matrix operations, specifically the associativity of matrix multiplication. Matrix multiplication has the associative property, meaning (AB)C = A(BC), so we can group operations differently without changing the result, but it lacks commutativity, so we cannot change the order of multiplication (AB ≠ BA in general). The associative property allows us to compute (AB)C or A(BC) and get the same result, so we can choose which multiplication to perform first based on computational convenience without worrying about different answers. Choice B is correct because it correctly describes associativity. Choice A incorrectly claims that matrix multiplication is not associative, but (AB)C = A(BC) always holds (when dimensions are compatible). When simplifying matrix expressions, you can freely use associativity to regroup operations ((AB)C = A(BC)) and distributivity to expand or factor (A(B + C) = AB + AC), but you cannot change the order of multiplication without potentially changing the result. To remember the properties: addition and multiplication are both associative (can regroup) and both operations distribute over addition, but only addition is commutative (can reorder)—multiplication order matters.

Question 11

1. commutativity of matrix multiplication
2. associativity of matrix multiplication
3. distributivity of matrix multiplication over addition (correct answer)
4. commutativity of matrix addition implies commutativity of multiplication

Explanation: This question tests understanding of identifying which property is illustrated by a given equation. The distributive property states that matrix multiplication distributes over addition: A(B + C) = AB + AC and (A + B)C = AC + BC, allowing us to expand or factor matrix expressions just like with numbers. The equation A(B + C) = AB + AC specifically shows how matrix multiplication distributes over addition from the left, meaning we can multiply A by the sum (B + C) or multiply A by each term separately and add the results. Choice C is correct because it correctly identifies this as the distributive property of matrix multiplication over addition. Choice A incorrectly identifies this as commutativity of multiplication, but commutativity would be AB = BA, not the distribution shown. To remember the properties: the distributive property A(B + C) = AB + AC shows how multiplication interacts with addition, while commutativity (AB = BA) and associativity ((AB)C = A(BC)) describe how operations behave with themselves.

Question 12

1. Yes; matrix multiplication is commutative for all square matrices, so $AB=BA$.
2. No; matrix multiplication is not commutative in general, so typically $AB\ne BA$ even when both products are defined. (correct answer)
3. Yes; because matrix addition is commutative, $AB=BA$ must also hold.
4. No; matrix multiplication is not associative, so $AB\ne BA$.

Explanation: This question tests understanding of the commutativity property for matrix multiplication. Unlike multiplication of real numbers, matrix multiplication is not commutative: for most matrices A and B, AB ≠ BA, and even when both products are defined (square matrices), they typically give different results—only special cases like multiplication by the identity matrix commute. For example, with A = [[1,2],[3,4]] and B = [[0,1],[-1,2]], computing AB gives [[−2,5],[−4,11]] while BA gives [[3,4],[5,6]], demonstrating that the order of multiplication matters—reversing the order changes the outcome. Choice B is correct because it accurately states that matrix multiplication is not commutative in general, so typically AB ≠ BA even when both products are defined. Choice A incorrectly claims that matrix multiplication is commutative for all square matrices, but this is false—commutativity only holds in special cases, not generally. Key difference from numbers: matrix multiplication is NOT commutative (AB ≠ BA in general), so always pay attention to the order when multiplying matrices, but addition IS commutative (A + B = B + A), so order doesn't matter when adding.

Question 13

1. $A(B+C)=AB+C$.
2. $A(B+C)=AB+AC$. (correct answer)
3. $A(B+C)=A+B+C$.
4. $A(B+C)=BA+CA$.

Explanation: This question tests understanding of the distributive property for matrix operations. The distributive property states that matrix multiplication distributes over addition: A(B + C) = AB + AC and (A + B)C = AC + BC, allowing us to expand or factor matrix expressions just like with numbers. To expand A(B + C), we apply the distributive property: A(B + C) = AB + AC, meaning we multiply A by B and A by C separately, then add the results, just as we would distribute a factor across a sum in algebra. Choice B is correct because it properly applies the distributive property to get A(B + C) = AB + AC. Choice D reverses the order in the expanded form, writing BA + CA instead of AB + AC, but order matters in matrix multiplication. The distributive property A(B + C) = AB + AC is crucial for expanding matrix expressions, and it works just like distributing with numbers, but remember to keep the multiplication order the same (A on the left in both terms). When simplifying matrix expressions, you can freely use associativity to regroup operations ((AB)C = A(BC)) and distributivity to expand or factor (A(B + C) = AB + AC), but you cannot change the order of multiplication without potentially changing the result.

Question 14

1. $k(AB)=(kA)B=A(kB)$. (correct answer)
2. $k(AB)=AB+k$.
3. $k(AB)=kA+kB$.
4. $k(AB)=(Ak)B$ is undefined because scalars cannot multiply matrices.

Explanation: This question tests understanding of scalar multiplication properties with matrix operations. Matrix multiplication has the associative property, meaning (AB)C = A(BC), so we can group operations differently without changing the result, but it lacks commutativity, so we cannot change the order of multiplication (AB ≠ BA in general). For scalar multiplication with matrices, the scalar can be associated with either matrix in a product: k(AB) = (kA)B = A(kB), which follows from the associative property and the fact that scalar multiplication commutes with matrix multiplication in this specific sense. Choice A is correct because it accurately states that k(AB) = (kA)B = A(kB), showing that the scalar k can be associated with either matrix in the product. Choice C incorrectly expands k(AB) as kA + kB, but scalar multiplication doesn't distribute over matrix multiplication like this—it associates with individual matrices in the product. To remember the properties: addition and multiplication are both associative (can regroup) and both operations distribute over addition, but only addition is commutative (can reorder)—multiplication order matters. When simplifying matrix expressions, you can freely use associativity to regroup operations ((AB)C = A(BC)) and distributivity to expand or factor (A(B + C) = AB + AC), but you cannot change the order of multiplication without potentially changing the result.

Question 15

1. $(XY)Z = \begin{pmatrix} 1 & 3 & 0 \\ 0 & 1 & 2 \\ 2 & 4 & 1 \end{pmatrix}$
2. $(XY)Z = \begin{pmatrix} 1 & 0 & 2 \\ 3 & 1 & 4 \\ 0 & 2 & 1 \end{pmatrix}$ (correct answer)
3. $(XY)Z$ cannot be determined without knowing the individual matrices
4. $(XY)Z = \begin{pmatrix} 2 & 0 & 4 \\ 6 & 2 & 8 \\ 0 & 4 & 2 \end{pmatrix}$

Explanation: By the associative property of matrix multiplication, $X(YZ) = (XY)Z$ whenever both products are defined. Since we're given that $X(YZ) = \begin{pmatrix} 1 & 0 & 2 \\ 3 & 1 & 4 \\ 0 & 2 & 1 \end{pmatrix}$, we can conclude that $(XY)Z$ equals the same matrix. Choice A incorrectly transposes the given matrix. Choice C fails to recognize the associative property. Choice D incorrectly doubles all entries of the given matrix.

Question 16

1. $AO=A$
2. $AO=O$ (correct answer)
3. $AO=I$
4. $AO=OA$ only if $A$ is diagonal

Explanation: This question tests understanding of the properties of matrix operations, specifically the zero matrix in multiplication. Unlike matrix addition, which is commutative (A + B = B + A), matrix multiplication is generally NOT commutative (AB ≠ BA in general) even though it is associative ((AB)C = A(BC)). For the zero matrix O, multiplying any matrix A by O gives O, as each entry in AO is a sum of zeros, resulting in the zero matrix. Choice B is correct because it accurately states that AO = O. Choice A claims AO = A, but this confuses the zero matrix with the identity, as multiplication by zero absorbs to zero, not leaving A unchanged. Key difference from numbers: matrix multiplication is NOT commutative (AB ≠ BA in general), so always pay attention to the order when multiplying matrices, but addition IS commutative (A + B = B + A), so order doesn't matter when adding. When simplifying matrix expressions, you can freely use associativity to regroup operations ((AB)C = A(BC)) and distributivity to expand or factor (A(B + C) = AB + AC), but you cannot change the order of multiplication without potentially changing the result.

Question 17

1. Both commutativity and associativity
2. Commutativity only
3. Associativity only (correct answer)
4. Neither commutativity nor associativity

Explanation: This question tests understanding of the properties of matrix operations, specifically commutativity and associativity for multiplication. Matrix addition is commutative (A + B = B + A) and associative ((A + B) + C = A + (B + C)), but matrix multiplication is generally NOT commutative (AB ≠ BA in general) even though it is associative ((AB)C = A(BC)). The associative property allows us to compute (AB)C or A(BC) and get the same result, so we can choose which multiplication to perform first based on computational convenience without worrying about different answers. Choice C is correct because it accurately states which properties hold and correctly describes associativity. Choice A incorrectly claims that matrix multiplication is commutative (AB = BA), but this is false in general—only special matrices like the identity commute with all matrices. To remember the properties: addition and multiplication are both associative (can regroup) and both operations distribute over addition, but only addition is commutative (can reorder)—multiplication order matters. When simplifying matrix expressions, you can freely use associativity to regroup operations ((AB)C = A(BC)) and distributivity to expand or factor (A(B + C) = AB + AC), but you cannot change the order of multiplication without potentially changing the result.

Question 18

1. No; matrix multiplication is not associative in general.
2. Yes; matrix multiplication is associative whenever the products are defined. (correct answer)
3. Yes, but only when $AB=BA$.
4. No; because matrix addition is commutative, regrouping changes the product.

Explanation: This question tests understanding of the associative property of matrix multiplication. Matrix multiplication has the associative property, meaning (AB)C = A(BC), so we can group operations differently without changing the result, but it lacks commutativity, so we cannot change the order of multiplication (AB ≠ BA in general). The associative property allows us to compute (AB)C or A(BC) and get the same result, so we can choose which multiplication to perform first based on computational convenience without worrying about different answers. Choice B is correct because it accurately states that matrix multiplication is associative whenever the products are defined. Choice A incorrectly claims that matrix multiplication is not associative, but (AB)C = A(BC) always holds (when dimensions are compatible). When simplifying matrix expressions, you can freely use associativity to regroup operations ((AB)C = A(BC)) and distributivity to expand or factor (A(B + C) = AB + AC), but you cannot change the order of multiplication without potentially changing the result.

Question 19

1. $(A+B)C=AC+BC$ (correct answer)
2. $(A+B)C=CA+CB$
3. $(A+B)C=AB+CC$
4. $(A+B)C=AC+B$

Explanation: This question tests understanding of the distributive property when multiplication occurs on the right. The distributive property states that matrix multiplication distributes over addition: A(B + C) = AB + AC and (A + B)C = AC + BC, allowing us to expand or factor matrix expressions just like with numbers. To expand (A + B)C, we apply the right distributive property: (A + B)C = AC + BC, meaning we multiply A by C and B by C separately, then add the results, maintaining the order with C on the right in both terms. Choice A is correct because it properly applies the distributive property to get (A + B)C = AC + BC. Choice B reverses the order in the expanded form, writing CA + CB instead of AC + BC, but order matters in matrix multiplication since it's not commutative. The distributive property works from both sides: A(B + C) = AB + AC (left distribution) and (A + B)C = AC + BC (right distribution), but remember to keep the multiplication order consistent in the expanded form.

Question 20

1. $A(B+C)=AB+AC$ (correct answer)
2. $A(B+C)=AB+C$
3. $A(B+C)=A+BC$
4. $A(B+C)=BA+CA$

Explanation: This question tests understanding of the distributive property for matrix multiplication over addition. The distributive property states that matrix multiplication distributes over addition: A(B + C) = AB + AC and (A + B)C = AC + BC, allowing us to expand or factor matrix expressions just like with numbers. To expand A(B + C), we apply the distributive property: A(B + C) = AB + AC, meaning we multiply A by B and A by C separately, then add the results, just as we would distribute a factor across a sum in algebra. Choice A is correct because it properly applies the distributive property to get A(B + C) = AB + AC. Choice D reverses the order in the expanded form, writing BA + CA instead of AB + AC, but order matters in matrix multiplication. The distributive property A(B + C) = AB + AC is crucial for expanding matrix expressions, and it works just like distributing with numbers, but remember to keep the multiplication order the same (A on the left in both terms).