Understanding the Multiplication Concept in Matrices as the Associative and Distributive Properties: CCSS.Math.Content.HSN-VM.C.9 - Common Core: High School - Number and Quantity

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Question

Which of the following properties does not apply to matrices?

Answer

Commutative does not apply to matrices because if we have matrices , and . It is not necessarily true that $A\times B=B\times A$ , even though in some cases it's true.

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Question

Which is an example of two matrices satisfying the associative and distributive properties? Let a be a scalar, and A, B, and C be three unique matrices.

Answer

is the correct answer because it is the only answer that involves both the associative and distributive properties.

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Question

Which matrix when multiplied with

$\begin{bmatrix} 1&3 \ 5& -4 \end{bmatrix}$

will yield the same result regardless of the order in which they're multiplied?

Answer

The only matrix that works is $\begin{bmatrix} 0&0 \ 0& 0 \end{bmatrix}$ , because regardless of the order of matrix multiplication, the result will always be $\begin{bmatrix} 0&0 \ 0& 0 \end{bmatrix}$ .

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Question

Why doesn't the commutative property hold for matrix multiplication?

Answer

The reason that the commutative property doesn't apply to matrix multiplication is because order of multiplication matters. We multiply by the entry in the row of the first matrix by the entry in the column of the second matrix.

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Question

Which is an example of two matrices satisfying the distributive properties? Let be a scalar, and ,, and be three unique matrices.

Answer

The only answer that satisfies the distributive property is .

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Question

True or False: If and are square matrices, is ?

Answer

Let

$A=\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}$ , $B=\begin{bmatrix} b_1& b_2\ b_3&b_4 \end{bmatrix}$ , $C=\begin{bmatrix} c_1& c_2\ c_3&c_4 \end{bmatrix}$ .

$(AB)C=\left(\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}\begin{bmatrix} b_1& b_2\ b_3&b_4 \end{bmatrix} \right )\begin{bmatrix} c_1&c_2 \ c_3& c_4 \end{bmatrix}$

Now do matrix multiplication inside the parenthesis.

$\left(\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}\begin{bmatrix} b_1& b_2\ b_3&b_4 \end{bmatrix} \right )\begin{bmatrix} c_1&c_2 \ c_3& c_4 \end{bmatrix} =\begin{bmatrix} a_1b_1+a_2b_3 &a_1b_2+a_2b_4 \ a_3b_1+a_4b_3&a_3b_2+a_4b_4 \end{bmatrix} \begin{bmatrix} c_1&c_2 \ c_3&c_4 \end{bmatrix}$

Now multiply the result by the other matrix to get

$=\begin{bmatrix} c_1(a_1b_1+a_2b_3)+c_3(a_1b_2+a_2b_4)& c_2(a_1b_1+a_2b_3)+c_4(a_1b_2+a_2b_4) \ c_1(a_3b_1+a_4b_3)+c_3(a_3b_2+a_4b_4)& c_2(a_3b_1+a_4b_3)+c_4(a_3b_2+a_4b_4) \end{bmatrix}$

Now lets do it from the other side

$A(BC)=\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}\left(\begin{bmatrix} b_1& b_2\ b_3&b_4 \end{bmatrix} \begin{bmatrix} c_1&c_2 \ c_3& c_4 \end{bmatrix}\right )$

Do the matrix multiplication inside the parenthesis first

$\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}\left(\begin{bmatrix} b_1& b_2\ b_3&b_4 \end{bmatrix} \begin{bmatrix} c_1&c_2 \ c_3& c_4 \end{bmatrix}\right )=\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}\begin{bmatrix} b_1c_1+b_2c_3&b_1c_2+b_2c_4 \ b_3c_1+b_4c_3&b_3c_2+b_4c_4 \end{bmatrix}$

Now multiply the result by the other matrix to get

$\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}\begin{bmatrix} b_1c_1+b_2c_3&b_1c_2+b_2c_4 \ b_3c_1+b_4c_3&b_3c_2+b_4c_4 \end{bmatrix}=\begin{bmatrix} a_1(b_1c_1+b_2c_3)+a_2(b_3c_1+b_4c_3)&a_1(b_1c_2+b_2c_4)+a_2(b_3c_2+b_4c_4) \ a_3(b_1c_1+b_2c_3)+a_4(b_3c_1+b_4c_3)& a_3(b_1c_2+b_2c_4)+a_4(b_3c_2+b_4c_4) \end{bmatrix}$

If we rearrange the terms in this matrix we get

$=\begin{bmatrix} c_1(a_1b_1+a_2b_3)+c_3(a_1b_2+a_2b_4)& c_2(a_1b_1+a_2b_3)+c_4(a_1b_2+a_2b_4) \ c_1(a_3b_1+a_4b_3)+c_3(a_3b_2+a_4b_4)& c_2(a_3b_1+a_4b_3)+c_4(a_3b_2+a_4b_4) \end{bmatrix}$

Since these are the same matrix, we have evidence that the statement is true, .

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Question

True or False: The following matrix product is possible.

$\begin{bmatrix} 3& 2 \end{bmatrix}\begin{bmatrix} 3& 4 \end{bmatrix}$

Answer

The answer is false because the dimensions are $1\times 2$ for each matrix.

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Question

True or False:

The following matrix multiplication is possible.

$\begin{bmatrix} 4&3 &2 \end{bmatrix} \begin{bmatrix} 6\ 7\ 8 \end{bmatrix}$

Answer

The matrix multiplication is possible since the dimensions will work out. The result will be a $3\times 3$ since the dimensions are $3\times1$ , and $1\times 3$ .

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Question

True or False:

The following matrix multiplication is possible

$\begin{bmatrix} 3 \end{bmatrix} \begin{bmatrix} 4&5 &6 &7 &8 \end{bmatrix}$

Answer

The matrix multiplication is possible because the dimensions work out. The resulting matrix will be $1\times 5$ , because the matrices are $1\times1$ , and $1\times 5$ .

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Question

True or False:

The following matrix multiplication is possible.

$\begin{bmatrix} 4& 5& 6& 7\ 9& 0& 5&4 \end{bmatrix} \begin{bmatrix} 3& 4& 6& 7\ 34& 76& 9& 90\ 1& 3& 4& 67 \end{bmatrix}$

Answer

The matrix multiplication is not possible because the dimensions do not work out. You can't multiply a $2\times 4$ and a $3\times4$ matrix together.

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Question

True or False:

The following matrix multiplication is possible.

$\begin{bmatrix} 4& 5& 6\ 2& 3& 4 \end{bmatrix}\begin{bmatrix} 1&2 &3 \ 8&6 & 4\ 2& 4& 10 \end{bmatrix}$

Answer

The matrix multiplication is possible because the dimensions work out. Since we have a $2\times3$ , and a $3\times3$ matrix, we will have a $2\times3$ as our result.

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Question

True or False:

The following matrix multiplication is possible.

$\begin{bmatrix} 2\ 3 \end{bmatrix} \begin{bmatrix} 1&0 \ 0& 1 \end{bmatrix}$

Answer

The matrix multiplication is not possible because the dimensions do not work. You can't multiply a $2\times1$ with a $2\times2$ matrix.

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Question

Which of the following properties does not apply to matrices?

Answer

Commutative does not apply to matrices because if we have matrices , and . It is not necessarily true that $A\times B=B\times A$ , even though in some cases it's true.

Compare your answer with the correct one above

Question

Which is an example of two matrices satisfying the associative and distributive properties? Let a be a scalar, and A, B, and C be three unique matrices.

Answer

is the correct answer because it is the only answer that involves both the associative and distributive properties.

Compare your answer with the correct one above

Question

Which matrix when multiplied with

$\begin{bmatrix} 1&3 \ 5& -4 \end{bmatrix}$

will yield the same result regardless of the order in which they're multiplied?

Answer

The only matrix that works is $\begin{bmatrix} 0&0 \ 0& 0 \end{bmatrix}$ , because regardless of the order of matrix multiplication, the result will always be $\begin{bmatrix} 0&0 \ 0& 0 \end{bmatrix}$ .

Compare your answer with the correct one above

Question

Why doesn't the commutative property hold for matrix multiplication?

Answer

The reason that the commutative property doesn't apply to matrix multiplication is because order of multiplication matters. We multiply by the entry in the row of the first matrix by the entry in the column of the second matrix.

Compare your answer with the correct one above

Question

Which is an example of two matrices satisfying the distributive properties? Let be a scalar, and ,, and be three unique matrices.

Answer

The only answer that satisfies the distributive property is .

Compare your answer with the correct one above

Question

True or False: If and are square matrices, is ?

Answer

Let

$A=\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}$ , $B=\begin{bmatrix} b_1& b_2\ b_3&b_4 \end{bmatrix}$ , $C=\begin{bmatrix} c_1& c_2\ c_3&c_4 \end{bmatrix}$ .

$(AB)C=\left(\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}\begin{bmatrix} b_1& b_2\ b_3&b_4 \end{bmatrix} \right )\begin{bmatrix} c_1&c_2 \ c_3& c_4 \end{bmatrix}$

Now do matrix multiplication inside the parenthesis.

$\left(\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}\begin{bmatrix} b_1& b_2\ b_3&b_4 \end{bmatrix} \right )\begin{bmatrix} c_1&c_2 \ c_3& c_4 \end{bmatrix} =\begin{bmatrix} a_1b_1+a_2b_3 &a_1b_2+a_2b_4 \ a_3b_1+a_4b_3&a_3b_2+a_4b_4 \end{bmatrix} \begin{bmatrix} c_1&c_2 \ c_3&c_4 \end{bmatrix}$

Now multiply the result by the other matrix to get

$=\begin{bmatrix} c_1(a_1b_1+a_2b_3)+c_3(a_1b_2+a_2b_4)& c_2(a_1b_1+a_2b_3)+c_4(a_1b_2+a_2b_4) \ c_1(a_3b_1+a_4b_3)+c_3(a_3b_2+a_4b_4)& c_2(a_3b_1+a_4b_3)+c_4(a_3b_2+a_4b_4) \end{bmatrix}$

Now lets do it from the other side

$A(BC)=\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}\left(\begin{bmatrix} b_1& b_2\ b_3&b_4 \end{bmatrix} \begin{bmatrix} c_1&c_2 \ c_3& c_4 \end{bmatrix}\right )$

Do the matrix multiplication inside the parenthesis first

$\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}\left(\begin{bmatrix} b_1& b_2\ b_3&b_4 \end{bmatrix} \begin{bmatrix} c_1&c_2 \ c_3& c_4 \end{bmatrix}\right )=\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}\begin{bmatrix} b_1c_1+b_2c_3&b_1c_2+b_2c_4 \ b_3c_1+b_4c_3&b_3c_2+b_4c_4 \end{bmatrix}$

Now multiply the result by the other matrix to get

$\begin{bmatrix} a_1& a_2\ a_3&a_4 \end{bmatrix}\begin{bmatrix} b_1c_1+b_2c_3&b_1c_2+b_2c_4 \ b_3c_1+b_4c_3&b_3c_2+b_4c_4 \end{bmatrix}=\begin{bmatrix} a_1(b_1c_1+b_2c_3)+a_2(b_3c_1+b_4c_3)&a_1(b_1c_2+b_2c_4)+a_2(b_3c_2+b_4c_4) \ a_3(b_1c_1+b_2c_3)+a_4(b_3c_1+b_4c_3)& a_3(b_1c_2+b_2c_4)+a_4(b_3c_2+b_4c_4) \end{bmatrix}$

If we rearrange the terms in this matrix we get

$=\begin{bmatrix} c_1(a_1b_1+a_2b_3)+c_3(a_1b_2+a_2b_4)& c_2(a_1b_1+a_2b_3)+c_4(a_1b_2+a_2b_4) \ c_1(a_3b_1+a_4b_3)+c_3(a_3b_2+a_4b_4)& c_2(a_3b_1+a_4b_3)+c_4(a_3b_2+a_4b_4) \end{bmatrix}$

Since these are the same matrix, we have evidence that the statement is true, .

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Question

True or False: The following matrix product is possible.

$\begin{bmatrix} 3& 2 \end{bmatrix}\begin{bmatrix} 3& 4 \end{bmatrix}$

Answer

The answer is false because the dimensions are $1\times 2$ for each matrix.

Compare your answer with the correct one above

Question

True or False:

The following matrix multiplication is possible.

$\begin{bmatrix} 4&3 &2 \end{bmatrix} \begin{bmatrix} 6\ 7\ 8 \end{bmatrix}$

Answer

The matrix multiplication is possible since the dimensions will work out. The result will be a $3\times 3$ since the dimensions are $3\times1$ , and $1\times 3$ .

Compare your answer with the correct one above

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