Understanding the Multiplication Concept in Matrices as the Associative and Distributive Properties: CCSS.Math.Content.HSN-VM.C.9

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

Questions 1 - 10

True or False:

The following matrix multiplication is possible.

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

True

False

Explanation

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 $1\times 3$ .

True or False:

The following matrix multiplication is possible.

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

True

False

Explanation

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.

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

Order of multiplication matters

Commutative property doesn't work for regular multiplication

Explanation

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.

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?

$\begin{bmatrix} 0&0 \ 0& 0 \end{bmatrix}$

$\begin{bmatrix} 1&1 \ 1& 1 \end{bmatrix}$

$\begin{bmatrix} 1&4 \ 4& 1 \end{bmatrix}$

$\begin{bmatrix} 1&2 \ 3& 4 \end{bmatrix}$

$\begin{bmatrix} 10&-10 \ -2& 3 \end{bmatrix}$

Explanation

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 $\begin{bmatrix} 0&0 \ 0& 0 \end{bmatrix}$ .

True or False: The following matrix product is possible.

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

False

True

Explanation

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

True or False:

The following matrix multiplication is possible.

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

False

True

Explanation

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.

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

$A\times B=B\times A$

Explanation

The only answer that satisfies the distributive property is .

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

True

False

Explanation

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, .

True or False:

The following matrix multiplication is possible

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

True

False

Explanation

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 $1\times 5$ .

Which of the following properties does not apply to matrices?

Commutative

Associative

Distributive

None of the answers

Explanation

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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