Precalculus Quiz: Properties Of Matrix Operations

20 questions · exam conditions

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Properties Of Matrix OperationsQuestion 1 of 20

Regarding matrix multiplication, the property being examined is associativity. Does $(AB)C=A(BC)$ for all matrices $A,B,C$ (with compatible dimensions)?

Yes; matrix multiplication is associative whenever the products are defined.

No; matrix multiplication is never associative.

Yes, but only when

AB=BA

No;

(AB)C=A(BC)

holds only for

2\times2

matrices.

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Precalculus Quiz

Precalculus Quiz: Properties Of Matrix Operations

Practice Properties Of Matrix Operations in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

All questions

Question 1

Regarding matrix multiplication, the property being examined is associativity. Does $(AB)C=A(BC)$ for all matrices $A,B,C$ (with compatible dimensions)?

Yes; matrix multiplication is associative whenever the products are defined. (correct answer)
No; matrix multiplication is never associative.
Yes, but only when $AB=BA$ .
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 2

For the matrices $A=\begin{pmatrix}2&0\\1&3\end{pmatrix},\quad B=\begin{pmatrix}1&4\\0&-1\end{pmatrix},$ what is the result of the calculation $AB$ ?

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

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 3

Let $R = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ and $S = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ . Compute $RS$ and $SR$ , then determine which property explains why $(RS)^2 \neq R^2S^2$ .

The associative property fails for these particular matrices
The distributive property doesn't apply to powers of matrix products
Matrix multiplication is not commutative, so $(RS)^2 = RSRS \neq R^2S^2 = RRSS$ (correct answer)
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 4

Consider matrices $A$ , $B$ , and $C$ where $A$ is $2 \times 3$ , $B$ is $3 \times 4$ , and $C$ is $4 \times 2$ . If we compute both $(AB)C$ and $A(BC)$ , what are the dimensions of each result?

$(AB)C$ is $2 \times 2$ and $A(BC)$ is $3 \times 3$
$(AB)C$ is $2 \times 2$ and $A(BC)$ is $2 \times 2$ (correct answer)
$(AB)C$ is $3 \times 4$ and $A(BC)$ is $2 \times 2$
$(AB)C$ 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 5

Simplify $(A+B)C$ using matrix properties (assume all matrices have compatible dimensions).

$(A+B)C=AC+BC$ (correct answer)
$(A+B)C=CA+CB$
$(A+B)C=A+BC$
$(A+B)C=AB+CC$

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 6

Does $(AB)C=A(BC)$ for all matrices $A,B,C$ with compatible dimensions (property being examined: associativity of matrix multiplication)?

No; matrix multiplication is not associative, so $(AB)C\ne A(BC)$ in general.
Yes; matrix multiplication is associative, so $(AB)C=A(BC)$ whenever both sides are defined. (correct answer)
Yes, but only when $AB=BA$ (commutativity is required for associativity).
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 7

For the matrices $A=\begin{pmatrix}1&2\\3&4\end{pmatrix} \text{ and } B=\begin{pmatrix}0&1\\-1&2\end{pmatrix},$ is $AB=BA$ in general? Why or why not?

Yes; matrix multiplication is commutative for all square matrices, so $AB=BA$ .
No; matrix multiplication is not commutative in general, so typically $AB\ne BA$ even when both products are defined. (correct answer)
Yes; because matrix addition is commutative, $AB=BA$ must also hold.
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 8

For $A=\begin{pmatrix}1&-2\\3&0\end{pmatrix}$ and the zero matrix $O=\begin{pmatrix}0&0\\0&0\end{pmatrix}$ , what is the result of $AO$ (property being examined: zero matrix absorption under multiplication)?

$AO=A$
$AO=O$ (correct answer)
$AO=I$
$AO=OA$ 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 9

For the matrices $A=\begin{pmatrix}1&0\\2&-1\end{pmatrix}$ , $B=\begin{pmatrix}3&1\\0&2\end{pmatrix}$ , and $C=\begin{pmatrix}-1&4\\5&0\end{pmatrix}$ , what is the expanded form of $(A+B)C$ using distributivity?

$(A+B)C=AC+BC$ (correct answer)
$(A+B)C=CA+CB$
$(A+B)C=AB+CC$
$(A+B)C=AC+B$

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 10

For matrices $A,B,C$ of compatible dimensions, what is the expanded form of $A(B+C)$ (distributivity)?

$A(B+C)=AB+AC$ (correct answer)
$A(B+C)=AB+C$
$A(B+C)=A+BC$
$A(B+C)=BA+CA$

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).

Question 11

For the matrices $A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$ and $B=\begin{pmatrix}0&1\\-1&2\end{pmatrix}$ , the property being examined is commutativity of matrix multiplication. Is $AB=BA$ in general? Why or why not?

$AB=BA$ for all square matrices because multiplication is commutative.
$AB\ne BA$ in general; matrix multiplication is not commutative. (correct answer)
$AB=BA$ whenever $A$ and $B$ have the same dimensions.
$AB\ne BA$ because matrix addition is not commutative.

Explanation: This question tests understanding of the commutativity of matrix multiplication for matrices. 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 identifies the non-commutativity of matrix multiplication, stating AB ≠ BA in general. 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.

Question 12

Consider the matrix equation $(A + B)^2 = A^2 + 2AB + B^2$ , where $A$ and $B$ are $2 \times 2$ matrices. This equation holds if and only if:

$A$ and $B$ are both diagonal matrices with equal diagonal entries
$A$ and $B$ commute under multiplication (i.e., $AB = BA$ ) (correct answer)
$A$ and $B$ are both symmetric matrices with the same eigenvalues
$A$ and $B$ are both invertible matrices with determinant equal to $1$

Explanation: Expanding

(A + B)^2 = (A + B)(A + B) = A^2 + AB + BA + B^2

. For this to equal

A^2 + 2AB + B^2

, we need

AB + BA = 2AB

, which means

BA = AB

. This is precisely the condition that

A

and

B

commute. Choice A is too restrictive - many non-diagonal matrices commute. Choice C involves properties not directly related to commutativity. Choice D concerns invertibility, which is irrelevant to the equation holding.

Question 13

Let $M = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $N = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ . If we define $S = MN + NM$ and $T = MN - NM$ , what is the relationship between $\det(S)$ and $\det(T)$ ?

$\det(S) = \det(T)$
$\det(S) = -\det(T)$
$\det(S) + \det(T) = 0$
$\det(S) \cdot \det(T) = 0$ (correct answer)

Explanation: Calculate

MN = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}

and

NM = \begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix}

. Then

S = \begin{pmatrix} 5 & 5 \\ 5 & 5 \end{pmatrix}

and

T = \begin{pmatrix} -1 & -3 \\ 3 & 1 \end{pmatrix}

. Since

S

has identical rows,

\det(S) = 0

. For

T

\det(T) = (-1)(1) - (-3)(3) = -1 + 9 = 8 \neq 0

. Therefore

\det(S) \cdot \det(T) = 0 \cdot 8 = 0

. Choices A and B assume specific relationships that don't hold. Choice C would require

\det(T) = 0

as well.

Question 14

For the matrices $A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$ and $B=\begin{pmatrix}0&1\\-1&2\end{pmatrix}$ , is $AB=BA$ in general? Why or why not (based on this example)?

Yes; matrix multiplication is commutative for all square matrices, so $AB=BA$ .
No; matrix multiplication is not commutative in general, and for these matrices $AB\ne BA$ . (correct answer)
Yes; because $A+B=B+A$ , it follows that $AB=BA$ .
No; matrix multiplication is not associative, so $AB\ne BA$ .

Explanation: This question tests understanding of the properties of matrix operations, specifically the commutativity of matrix 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 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, and for these matrices AB ≠ BA. 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 15

For the matrices $A=\begin{pmatrix}2&-1\\0&3\end{pmatrix}$ , $B=\begin{pmatrix}1&4\\-2&0\end{pmatrix}$ , and $C=\begin{pmatrix}0&5\\1&-1\end{pmatrix}$ , what is the expanded form of $(A+B)C$ using the distributive property?

$(A+B)C=AC+BC$ (correct answer)
$(A+B)C=CA+CB$
$(A+B)C=AB+C$
$(A+B)C=A(C+B)$

Explanation: This question tests understanding of the properties of matrix operations, specifically the distributive property of 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 = 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 the 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). 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 16

How does $k(AB)$ relate to $(kA)B$ for a scalar $k$ and compatible matrices $A$ and $B$ (property being examined: scalar compatibility with matrix multiplication)?

$k(AB)=(kA)B=A(kB)$ (correct answer)
$k(AB)=kA+kB$
$k(AB)=(Ak)B$ only when $AB=BA$
$k(AB)=ABk$ is true only if $k=0$

Explanation: This question tests understanding of the properties of matrix operations, specifically scalar compatibility with matrix multiplication. 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. For scalars, the property extends such that k(AB) = (kA)B = A(kB), demonstrating how scalar multiplication associates with matrix multiplication. Choice A is correct because it correctly describes the scalar compatibility property. Choice B incorrectly expands k(AB) as kA + kB, but the distributive property requires multiplying the scalar by the product, not adding. 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 17

For matrices $A,B,C$ of compatible dimensions, what is the expanded form of $A(B+C)$ (property being examined: distributivity of multiplication over addition)?

$A(B+C)=AB+AC$ (correct answer)
$A(B+C)=AB+C$
$A(B+C)=A+B C$
$A(B+C)=BA+CA$

Explanation: This question tests understanding of the properties of matrix operations, specifically the distributivity of 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. 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). 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 18

For matrices $A$ and $B$ of the same dimensions, which statement is true about matrix addition (commutativity)?

$A+B=B+A$ (correct answer)
$A+B\ne B+A$ in general, even when dimensions match.
$A+B=AB$
$A+B=BA$

Explanation: This question tests understanding of the commutativity property for matrix 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)). Matrix addition is commutative, so A + B = B + A always holds when A and B have the same dimensions, meaning we can add matrices in any order just like adding numbers. Choice A is correct because it accurately states that A + B = B + A, which is the commutativity property for matrix addition. Choice B incorrectly states that matrix addition is not commutative, but A + B = B + A always holds for matrices of the same dimensions. 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 19

Let $I$ be the $2\times 2$ identity matrix and let $A$ be any $2\times 2$ matrix. When does $AI=IA$ hold for square matrices?

Only when $A$ is the zero matrix $O$ .
Only when $A$ is diagonal.
Always: $AI=IA=A$ for any square matrix $A$ of the same size as $I$ . (correct answer)
Never: matrix multiplication cannot commute with any matrix.

Explanation: This question tests understanding of special cases where matrix multiplication commutes. 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. Matrix multiplication commutes in special cases: AB = BA when B is the identity matrix I (since AI = IA = A), when both are diagonal matrices with specific patterns, or when one is a scalar multiple of the identity, but these are exceptions, not the general rule. Choice C is correct because it accurately states that AI = IA = A always holds for any square matrix A of the same size as I. Choice D incorrectly claims that matrix multiplication never commutes with any matrix, but the identity matrix always commutes with all matrices of the same size. Special cases where AB = BA: when B is the identity matrix (AI = IA = A), when both matrices are scalar multiples of I, or when both are certain diagonal matrices, but don't assume commutativity 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 20

When does $AB=BA$ for square matrices? (This question is about commutativity and its special cases.)

Always, for any square matrices $A$ and $B$ of the same size.
Never, for any nonzero square matrices $A$ and $B$ .
Sometimes; for example, $AB=BA$ when $B=I$ (the identity matrix) of the same size as $A$ . (correct answer)
Only when $A$ and $B$ have the same entries in the same positions.

Explanation: This question tests understanding of the properties of matrix operations, specifically commutativity of multiplication and its special cases. 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. Matrix multiplication commutes in special cases: AB = BA when B is the identity matrix I (since AI = IA = A), when both are diagonal matrices with specific patterns, or when one is a scalar multiple of the identity, but these are exceptions, not the general rule. Choice C is correct because it correctly identifies non-commutativity while noting a special case like the identity matrix. 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. Special cases where AB = BA: when B is the identity matrix (AI = IA = A), when both matrices are scalar multiples of I, or when both are certain diagonal matrices, but don't assume commutativity 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.