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Common Core: High School - Number and Quantity : Vector & Matrix Quantities

Study concepts, example questions & explanations for Common Core: High School - Number and Quantity

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All Common Core: High School - Number and Quantity Resources

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Example Questions

Example Question #141 : Vector & Matrix Quantities

True or False:

The following matrix multiplication is possible

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

Possible Answers:

False

True

Correct answer:

True

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 $1\times 5$ .

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Example Question #5 : Understanding The Multiplication Concept In Matrices As The Associative And Distributive Properties: Ccss.Math.Content.Hsn Vm.C.9

True or False:

The following matrix multiplication is possible.

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

Possible Answers:

True

False

Correct answer:

False

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.

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Example Question #142 : Vector & Matrix Quantities

True or False:

The following matrix multiplication is possible.

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

Possible Answers:

True

False

Correct answer:

True

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.

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Example Question #143 : Vector & Matrix Quantities

True or False:

The following matrix multiplication is possible.

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

Possible Answers:

False

True

Correct answer:

False

Explanation:

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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Example Question #1 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

What does a $4\times4$ Identity matrix look like?

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Identity matrices have 's along the main diagonal (the diagonal that goes from the top left hand corner to the bottom right hand corner), and has 's in all the other entries.

Since we want to have a $4\times4$ Identity matrix, then this is the result that we want.

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

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Example Question #2 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Find the inverse of the following matrix.

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

Possible Answers:

$\begin{bmatrix} -\frac{1}{3}&\frac{1}{3} \\ \frac{1}{6}& -\frac{2}{3} \end{bmatrix}$

$\begin{bmatrix} -\frac{1}{3}&-\frac{1}{3} \\ -\frac{1}{6}& -\frac{2}{3} \end{bmatrix}$

$\begin{bmatrix} \frac{1}{3}&-\frac{1}{3} \\ -\frac{1}{6}& \frac{2}{3} \end{bmatrix}$

$\begin{bmatrix} \frac{1}{3}&\frac{1}{3} \\ \frac{1}{6}& \frac{2}{3} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{1}{3}&-\frac{1}{3} \\ -\frac{1}{6}& \frac{2}{3} \end{bmatrix}$

Explanation:

In order to find the inverse of a matrix, we need to recall the formula for finding an inverse of a 2x2 matrix.

$A^{-1}=\frac{1}{a\cdot d-b\cdot c}\begin{bmatrix} d&-b \\ -c& a \end{bmatrix}$ , where a,b,c,d refer to position within the general 2x2 matrix $\begin{bmatrix} a& b\\ c&d \end{bmatrix}$ .

The first step is to figure out what the fraction is.

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

In this case a=4 , b=2 , c=1 , and d=2 .

$\frac{1}{a\cdot d-b\cdot c}=\frac{1}{4\cdot2-2\cdot1}=\frac{1}{8-2}=\frac{1}{6}$

The next step is to swap the off diagonal entries, and the multiply by negative 1 on the off diagonal entries.

$\begin{bmatrix} 4&2 \\ 1& 2 \end{bmatrix}\rightarrow$ $\begin{bmatrix} 2&-2 \\ -1&4 \end{bmatrix}$

The last step is to multiply them together.

$\frac{1}{6}$ $\begin{bmatrix} 2&-2 \\ -1&4 \end{bmatrix}$

$=\begin{bmatrix} 2\cdot\frac{1}{6}&-2\cdot\frac{1}{6} \\ -1\cdot\frac{1}{6}& 4\cdot\frac{1}{6} \end{bmatrix}$

$=\begin{bmatrix} \frac{1}{3}&-\frac{1}{3} \\ -\frac{1}{6}& \frac{2}{3} \end{bmatrix}$

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Example Question #2 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Find the inverse of the following matrix.

$\begin{bmatrix} 10&8 \\ 3& 9 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} \frac{3}{22}&-\frac{4}{33} \\ -\frac{1}{22}& -\frac{5}{33} \end{bmatrix}$

$\begin{bmatrix} \frac{3}{22}&-\frac{4}{33} \\ \frac{1}{22}& \frac{5}{33} \end{bmatrix}$

$\begin{bmatrix} -\frac{3}{22}&-\frac{4}{33} \\ -\frac{1}{22}& -\frac{5}{33} \end{bmatrix}$

$\begin{bmatrix} \frac{3}{22}&-\frac{4}{33} \\ -\frac{1}{22}& \frac{5}{33} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{3}{22}&-\frac{4}{33} \\ -\frac{1}{22}& \frac{5}{33} \end{bmatrix}$

Explanation:

In order to find the inverse of a matrix, we need to recall the formula for finding an inverse of a 2x2 matrix.

$A^{-1}=\frac{1}{a\cdot d-b\cdot c}\begin{bmatrix} d&-b \\ -c& a \end{bmatrix}$ , where a,b,c,d refer to position within the general 2x2 matrix $\begin{bmatrix} a& b\\ c&d \end{bmatrix}$ .

The first step is to figure out what the fraction is.

$\begin{bmatrix} 10&8 \\ 3& 9 \end{bmatrix}$

In this case a=10 , b=8 , c=3 , and d=9 .

$\frac{1}{a\cdot d-b\cdot c}=\frac{1}{10\cdot9-8\cdot3}=\frac{1}{90-24}=\frac{1}{66}$

The next step is to swap the off diagonal entries, and the multiply by negative 1 on the off diagonal entries.

$\begin{bmatrix} 10&8 \\ 3& 9 \end{bmatrix}\rightarrow$ $\begin{bmatrix} 9&-8 \\ -3&10 \end{bmatrix}$

The last step is to multiply them together.

$\frac{1}{66}$ $\begin{bmatrix} 9&-8 \\ -3&10 \end{bmatrix}$

$=\begin{bmatrix} 9\cdot\frac{1}{66}&-8\cdot\frac{1}{66} \\ -3\cdot\frac{1}{66}& 10\cdot\frac{1}{66} \end{bmatrix}$

$=\begin{bmatrix} \frac{3}{22}&-\frac{4}{33} \\ -\frac{1}{22}& \frac{5}{33} \end{bmatrix}$

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Example Question #3 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Find the inverse of the following matrix.

$\begin{bmatrix} 4&5 \\ -2& 2 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} \frac{1}{9}&-\frac{5}{18} \\ \frac{1}{9}& -\frac{2}{9} \end{bmatrix}$

$\begin{bmatrix} -\frac{1}{9}&-\frac{5}{18} \\ -\frac{1}{9}& -\frac{2}{9} \end{bmatrix}$

$\begin{bmatrix} \frac{1}{9}&-\frac{5}{18} \\ \frac{1}{9}& \frac{2}{9} \end{bmatrix}$

$\begin{bmatrix} \frac{1}{9}&-\frac{5}{18} \\ -\frac{1}{9}& \frac{2}{9} \end{bmatrix}$

$\begin{bmatrix} -\frac{1}{9}&-\frac{5}{18} \\ \frac{1}{9}& \frac{2}{9} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{1}{9}&-\frac{5}{18} \\ \frac{1}{9}& \frac{2}{9} \end{bmatrix}$

Explanation:

In order to find the inverse of a matrix, we need to recall the formula for finding an inverse of a 2x2 matrix.

$A^{-1}=\frac{1}{a\cdot d-b\cdot c}\begin{bmatrix} d&-b \\ -c& a \end{bmatrix}$ , where a,b,c,d refer to position within the general 2x2 matrix $\begin{bmatrix} a& b\\ c&d \end{bmatrix}$ .

The first step is to figure out what the fraction is.

$\begin{bmatrix} 4&5 \\ -2& 2 \end{bmatrix}$

In this case a=4 , b=5 , c=-2 , and d=2 .

$\frac{1}{a\cdot d-b\cdot c}=\frac{1}{4\cdot2-5\cdot(-2)}=\frac{1}{8+10}=\frac{1}{18}$

The next step is to swap the off diagonal entries, and the multiply by negative 1 on the off diagonal entries.

$\begin{bmatrix} 4&5 \\ -2& 2 \end{bmatrix}\rightarrow$ $\begin{bmatrix} 2&-5\\ 2&4 \end{bmatrix}$

The last step is to multiply them together.

$\frac{1}{18}$ $\begin{bmatrix} 2&-5\\ 2&4 \end{bmatrix}$

$=\begin{bmatrix} 2\cdot\frac{1}{18}&-5\cdot\frac{1}{18} \\ 2\cdot\frac{1}{18}& 4\cdot\frac{1}{18} \end{bmatrix}$

$=\begin{bmatrix} \frac{1}{9}&-\frac{5}{18} \\ \frac{1}{9}& \frac{2}{9} \end{bmatrix}$

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Example Question #1 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Find the inverse of the following matrix.

$\begin{bmatrix} 5&2 \\ -5& 3 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} \frac{3}{25}&\frac{2}{25} \\ \frac{1}{5}& \frac{1}{5} \end{bmatrix}$

$\begin{bmatrix} \frac{3}{25}&-\frac{2}{25} \\ -\frac{1}{5}& \frac{1}{5} \end{bmatrix}$

$\begin{bmatrix} -\frac{3}{25}&-\frac{2}{25} \\ -\frac{1}{5}& -\frac{1}{5} \end{bmatrix}$

$\begin{bmatrix} \frac{3}{25}&-\frac{2}{25} \\ \frac{1}{5}& \frac{1}{5} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{3}{25}&-\frac{2}{25} \\ \frac{1}{5}& \frac{1}{5} \end{bmatrix}$

Explanation:

In order to find the inverse of a matrix, we need to recall the formula for finding an inverse of a 2x2 matrix.

$A^{-1}=\frac{1}{a\cdot d-b\cdot c}\begin{bmatrix} d&-b \\ -c& a \end{bmatrix}$ , where a,b,c,d refer to position within the general 2x2 matrix $\begin{bmatrix} a& b\\ c&d \end{bmatrix}$ .

The first step is to figure out what the fraction is.

$\begin{bmatrix} 5&2 \\ -5& 3 \end{bmatrix}$

In this case a=5 , b=2 , c=-5 , and d=3 .

$\frac{1}{a\cdot d-b\cdot c}=\frac{1}{5\cdot3-2\cdot(-5)}=\frac{1}{15+10}=\frac{1}{25}$

The next step is to swap the off diagonal entries, and the multiply by negative 1 on the off diagonal entries.

$\begin{bmatrix} 5&2 \\ -5& 3 \end{bmatrix}\rightarrow$ $\begin{bmatrix} 3&-2 \\ 5&5 \end{bmatrix}$

The last step is to multiply them together.

$\frac{1}{25}$ $\begin{bmatrix} 3&-2 \\ 5&5 \end{bmatrix}$

$=\begin{bmatrix} 3\cdot\frac{1}{25}&-2\cdot\frac{1}{25} \\ 5\cdot\frac{1}{25}& 5\cdot\frac{1}{25} \end{bmatrix}$

$=\begin{bmatrix} \frac{3}{25}&-\frac{2}{25} \\ \frac{1}{5}& \frac{1}{5} \end{bmatrix}$

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Example Question #6 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Find the inverse of the following matrix.

$\begin{bmatrix} -3&1 \\ 5& 0 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 0&\frac{1}{5} \\ 1& -\frac{3}{5} \end{bmatrix}$

$\begin{bmatrix} 0&-\frac{1}{5} \\ -1& \frac{3}{5} \end{bmatrix}$

$\begin{bmatrix} 0&\frac{1}{5} \\ 1& -\frac{3}{5} \end{bmatrix}$

$\begin{bmatrix} 0&\frac{1}{5} \\ 1& \frac{3}{5} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 0&\frac{1}{5} \\ 1& \frac{3}{5} \end{bmatrix}$

Explanation:

n order to find the inverse of a matrix, we need to recall the formula for finding an inverse of a 2x2 matrix.

$A^{-1}=\frac{1}{a\cdot d-b\cdot c}\begin{bmatrix} d&-b \\ -c& a \end{bmatrix}$ , where a,b,c,d refer to position within the general 2x2 matrix $\begin{bmatrix} a& b\\ c&d \end{bmatrix}$ .

The first step is to figure out what the fraction is.

$\begin{bmatrix} -3&1 \\ 5& 0 \end{bmatrix}$

In this case a=-3 , b=1 , c=5 , and d=0 .

$\frac{1}{-3\cdot 0-1\cdot 5}=\frac{1}{0-5}=-\frac{1}{5}$

The next step is to swap the off diagonal entries, and the multiply by negative 1 on the off diagonal entries.

$\begin{bmatrix} -3&1 \\ 5& 0 \end{bmatrix}\rightarrow$ $\begin{bmatrix} 0&-1 \\ -5&-3 \end{bmatrix}$

The last step is to multiply them together.

$-\frac{1}{5}$ $\begin{bmatrix} 0&-1 \\ -5&-3 \end{bmatrix}$

$=\begin{bmatrix} 0\cdot-\frac{1}{5}&-1\cdot-\frac{1}{5} \\ -5\cdot-\frac{1}{5}& -3\cdot-\frac{1}{5} \end{bmatrix}$

$=\begin{bmatrix} 0&\frac{1}{5} \\ 1& \frac{3}{5} \end{bmatrix}$

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Common Core: High School - Number and Quantity : Vector & Matrix Quantities

Study concepts, example questions & explanations for Common Core: High School - Number and Quantity

All Common Core: High School - Number and Quantity Resources

Example Questions

Example Question #141 : Vector & Matrix Quantities

Example Question #5 : Understanding The Multiplication Concept In Matrices As The Associative And Distributive Properties: Ccss.Math.Content.Hsn Vm.C.9

Example Question #142 : Vector & Matrix Quantities

Example Question #143 : Vector & Matrix Quantities

Example Question #1 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Example Question #2 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Example Question #2 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Example Question #3 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Example Question #1 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Example Question #6 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

All Common Core: High School - Number and Quantity Resources

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