Exercises: Understand Matrix Multiplication Properties

Work through each section in order. Show your work where indicated.

Grade 9·18 problems·~30 min·Common Core Math - HS Number and Quantity·standard·hsn-vm-c-9

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Recall / Warm-Up

For real numbers, $ab = ba$ for all $a, b$ . Is this true for matrices?

Yes — multiplication of matrices is always commutative.

No — matrix multiplication is generally NOT commutative; $AB \neq BA$ in most cases.

Yes — but only when the matrices are square.

No — but $AB = BA$ is true whenever both products are defined.

Matrix multiplication is associative. Which equation correctly expresses this?

$(AB)C = A(CB)$

$(AB)C = A(BC)$

$A(BC) = B(AC)$

$(AB)C = (BA)C$

Which equation correctly expresses the left distributive law for matrix multiplication?

$A(B + C) = AB + AC$

$A(B + C) = BA + CA$

$(B + C)A = AB + AC$

$A(B + C) = (AB)(AC)$

Let $A = \begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $B = \begin{bmatrix}1&0\\1&1\end{bmatrix}$ . Compute $AB$ and $BA$ . Which statement is true?

$AB = BA = \begin{bmatrix}3&2\\1&1\end{bmatrix}$

$AB = \begin{bmatrix}3&2\\1&1\end{bmatrix}$ and $BA = \begin{bmatrix}1&2\\1&3\end{bmatrix}$ ; they are different.

$AB = BA = \begin{bmatrix}1&2\\1&3\end{bmatrix}$

$AB$ is defined but $BA$ is undefined for these matrices.

Matrix $A$ is $2 \times 3$ and $B$ is $3 \times 4$ . Is $BA$ defined?

Yes — matrix multiplication is always defined for any two matrices.

Yes — because $AB$ is defined, $BA$ is too.

No — $B$ is $3 \times 4$ and $A$ is $2 \times 3$ . The inner dimensions are $4$ and $2$ , which don't match.

Yes — both $AB$ and $BA$ are always defined for the same pair of matrices.

The associative property says $(AB)C = A(BC)$ . Which manipulation does this allow?

You can swap $A$ and $C$ in the product.

You can move the parentheses to group the factors differently, as long as the left-to-right order of $A$ , $B$ , $C$ is preserved.

You can multiply in any order you choose.

You can move any one matrix to the front of the product.

Which correctly expands $(A + B)C$ using the right distributive law?

$(A + B)C = CA + CB$

$(A + B)C = AC + BC$

$(A + B)C = A + BC$

$(A + B)C = AC + BA$

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