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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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Example Questions

Example Question #4 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Find the inverse of the following matrix.

$\begin{bmatrix} -2&-3 \\ 4& 5 \end{bmatrix}$

Possible Answers:

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

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

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

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

Correct answer:

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

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.

$\begin{bmatrix} -2&-3 \\ 4& 5 \end{bmatrix}$

In this case a=-2 , b=-3 , c=4 , and d=5 .

$\frac{1}{a\cdot d-b\cdot c}=\frac{1}{-2\cdot5-(-3)\cdot4}=\frac{1}{-10+12}=\frac{1}{2}$

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

$\begin{bmatrix} -2&-3 \\ 4& 5 \end{bmatrix}\rightarrow$ $\begin{bmatrix} 5&3 \\ -4&-2 \end{bmatrix}$

The last step is to multiply them together.

$\frac{1}{2}$ $\begin{bmatrix} 5&3 \\ -4&-2 \end{bmatrix}$

$=\begin{bmatrix} 5\cdot\frac{1}{2}&3\cdot\frac{1}{2} \\ -4\cdot\frac{1}{2}& -2\cdot\frac{1}{2} \end{bmatrix}$

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

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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.

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

Possible Answers:

$\begin{bmatrix} \frac{4}{19}&-\frac{5}{19} \\ -\frac{3}{19}& \frac{1}{19} \end{bmatrix}$

$\begin{bmatrix} -\frac{4}{19}&-\frac{5}{19} \\ -\frac{3}{19}& \frac{1}{19} \end{bmatrix}$

$\begin{bmatrix} \frac{4}{19}&\frac{5}{19} \\ -\frac{3}{19}& \frac{1}{19} \end{bmatrix}$

$\begin{bmatrix} \frac{4}{19}&\frac{5}{19} \\ \frac{3}{19}& \frac{1}{19} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{4}{19}&\frac{5}{19} \\ -\frac{3}{19}& \frac{1}{19} \end{bmatrix}$

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.

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

In this case a=1 , b=-5 , c=3 , and d=4 .

$\frac{1}{a\cdot d-b\cdot c}=\frac{1}{1\cdot4-(-5)\cdot3}=\frac{1}{4+15}=\frac{1}{19}$

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

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

The last step is to multiply them together.

$\frac{1}{19}$ $\begin{bmatrix} 4&5\\ -3&1 \end{bmatrix}$

$=\begin{bmatrix} 4\cdot\frac{1}{19}&5\cdot\frac{1}{19} \\ -3\cdot\frac{1}{19}& 1\cdot\frac{1}{19} \end{bmatrix}$

$=\begin{bmatrix} \frac{4}{19}&\frac{5}{19} \\ -\frac{3}{19}& \frac{1}{19} \end{bmatrix}$

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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.

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

Possible Answers:

$\begin{bmatrix} \frac{5}{14}&-\frac{1}{14} \\ -\frac{1}{14}& -\frac{3}{14} \end{bmatrix}$

$\begin{bmatrix} \frac{5}{14}&-\frac{1}{14} \\ \frac{1}{14}& \frac{3}{14} \end{bmatrix}$

$\begin{bmatrix} \frac{5}{14}&\frac{1}{14} \\ -\frac{1}{14}& \frac{3}{14} \end{bmatrix}$

$\begin{bmatrix} \frac{5}{14}&-\frac{1}{14} \\ -\frac{1}{14}& \frac{3}{14} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{5}{14}&-\frac{1}{14} \\ -\frac{1}{14}& \frac{3}{14} \end{bmatrix}$

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.

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

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

$\frac{1}{a\cdot d-b\cdot c}=\frac{1}{3\cdot5-1\cdot1}=\frac{1}{15-1}=\frac{1}{14}$

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

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

The last step is to multiply them together.

$\frac{1}{14}$ $\begin{bmatrix} 5&-1 \\ -1&3 \end{bmatrix}$

$=\begin{bmatrix} 5\cdot\frac{1}{14}&-1\cdot\frac{1}{14} \\ -1\cdot\frac{1}{14}& 3\cdot\frac{1}{14} \end{bmatrix}$

$=\begin{bmatrix} \frac{5}{14}&-\frac{1}{14} \\ -\frac{1}{14}& \frac{3}{14} \end{bmatrix}$

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Example Question #4 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Find the inverse of the following matrix.

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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.

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

In this case a=-3 , b=2 , c=-4 , and d=3 .

$\frac{1}{a\cdot d-b\cdot c}=\frac{1}{(-3)\cdot3-2\cdot(-4)}=\frac{1}{-9+8}=\frac{1}{-1}=-1$

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

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

The last step is to multiply them together.

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

$=\begin{bmatrix} 3\cdot(-1)&(-2)\cdot(-1) \\ 4\cdot(-1)& (-3)\cdot(-1)) \end{bmatrix}$

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

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Example Question #151 : Vector & Matrix Quantities

Find the inverse of the following matrix.

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

Possible Answers:

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

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

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

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

Correct answer:

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

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.

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

In this case a=-2 , b=-3 , c=2 , and d=0 .

$\frac{1}{a\cdot d-b\cdot c}=\frac{1}{(-2)\cdot0-(-3)\cdot2}=\frac{1}{0+6}=\frac{1}{6}$

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

$\begin{bmatrix} -2&-3 \\ 2& 0 \end{bmatrix}\rightarrow$ $\begin{bmatrix} 0&3 \\ -2&-2 \end{bmatrix}$

The last step is to multiply them together.

$\frac{1}{6}$ $\begin{bmatrix} 0&3 \\ -2&-2 \end{bmatrix}$

$=\begin{bmatrix} 0\cdot\frac{1}{6}&3\cdot\frac{1}{6} \\ -2\cdot\frac{1}{6}& -2\cdot\frac{1}{6} \end{bmatrix}$

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

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Example Question #152 : Vector & Matrix Quantities

Find the inverse of the following matrix.

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

Possible Answers:

$\begin{bmatrix} 0&-\frac{1}{3} \\ \frac{1}{3}& \frac{5}{9} \end{bmatrix}$

$\begin{bmatrix} 0&-\frac{1}{3} \\ -\frac{1}{3}& \frac{5}{9} \end{bmatrix}$

$\begin{bmatrix} 0&-\frac{1}{3} \\ -\frac{1}{3}& -\frac{5}{9} \end{bmatrix}$

$\begin{bmatrix} 0&\frac{1}{3} \\ \frac{1}{3}& \frac{5}{9} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 0&-\frac{1}{3} \\ -\frac{1}{3}& -\frac{5}{9} \end{bmatrix}$

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.

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

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

$\frac{1}{a\cdot d-b\cdot c}=\frac{1}{5\cdot0-(-3)\cdot(-3)}=\frac{1}{0-9}=-\frac{1}{9}$

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

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

The last step is to multiply them together.

$-\frac{1}{9}$ $\begin{bmatrix} 0&3 \\3&5 \end{bmatrix}$

$=\begin{bmatrix} 0\cdot-\frac{1}{9}&3\cdot-\frac{1}{9} \\ 3\cdot-\frac{1}{9}& 5\cdot-\frac{1}{9} \end{bmatrix}$

$=\begin{bmatrix} 0&-\frac{1}{3} \\ -\frac{1}{3}& -\frac{5}{9} \end{bmatrix}$

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Example Question #1 : Vector Multiplication (Matrices Being Transformations Of Vectors): Ccss.Math.Content.Hsn Vm.C.11

Calculate

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

Possible Answers:

$\begin{bmatrix} 5\\ 11 \end{bmatrix}$

Not possible

$\begin{bmatrix} 7\\ 2 \end{bmatrix}$

$\begin{bmatrix} 11\\ 5 \end{bmatrix}$

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

Correct answer:

$\begin{bmatrix} 11\\ 5 \end{bmatrix}$

Explanation:

In order to do matrix multiplication, we need to check if the dimensions check out. The matrix on the left is $2\times2$ , and the matrix on the right is $2\times1$ , so the dimensions check out. The resulting matrix will be $2\times1$ . To do matrix multiplication, we take the rows of the matrix on the left, and multiply by the columns of the right matrix. Then we sum the results together. This is what it looks like in general.

$\begin{bmatrix} a& b\\ c& d \end{bmatrix} \times \begin{bmatrix} e\\ f \end{bmatrix} = \begin{bmatrix} a\cdot e+b\cdot f\\ c\cdot e+d\cdot f \end{bmatrix}$

Now we apply the above to get the following solution.

$\begin{bmatrix} 3&4 \\ 1& 2 \end{bmatrix} \times \begin{bmatrix} 1\\ 2 \end{bmatrix}=\begin{bmatrix} 3\cdot1+4\cdot2\\ 1\cdot 1+2\cdot2 \end{bmatrix} =\begin{bmatrix} 3+8\\ 1+4 \end{bmatrix} =\begin{bmatrix} 11\\ 5 \end{bmatrix}$

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Example Question #2 : Vector Multiplication (Matrices Being Transformations Of Vectors): Ccss.Math.Content.Hsn Vm.C.11

Calculate

$\begin{bmatrix} 17&14 \\ 4& -19 \end{bmatrix} \times \begin{bmatrix} 10\\ -16 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} -54\\ -344 \end{bmatrix}$

$\begin{bmatrix} 54\\ 344 \end{bmatrix}$

$\begin{bmatrix} -344\\ 54 \end{bmatrix}$

$\begin{bmatrix} 344\\ 54 \end{bmatrix}$

$\begin{bmatrix} -54\\ 344 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -54\\ 344 \end{bmatrix}$

Explanation:

$\begin{bmatrix} a& b\\ c& d \end{bmatrix} \times \begin{bmatrix} e\\ f \end{bmatrix} = \begin{bmatrix} a\cdot e+b\cdot f\\ c\cdot e+d\cdot f \end{bmatrix}$

Now we apply the above to get the following solution.

$\begin{bmatrix} 17&14 \\ 4& -19 \end{bmatrix} \times \begin{bmatrix} 10\\ -16 \end{bmatrix}=\begin{bmatrix} 17\cdot10+14\cdot-16\\ 4\cdot 10+(-19)\cdot(-16) \end{bmatrix} =\begin{bmatrix} 170-224\\ 40+304 \end{bmatrix} =\begin{bmatrix} -54\\ 344 \end{bmatrix}$

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Example Question #2 : Vector Multiplication (Matrices Being Transformations Of Vectors): Ccss.Math.Content.Hsn Vm.C.11

Calculate

$\begin{bmatrix} 7&-1 \\ -16& -1 \end{bmatrix} \times \begin{bmatrix} 17\\ 3 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} -116\\ -275 \end{bmatrix}$

$\begin{bmatrix} 116\\ 275 \end{bmatrix}$

$\begin{bmatrix} 116\\ -275 \end{bmatrix}$

$\begin{bmatrix} -275\\ 116 \end{bmatrix}$

$\begin{bmatrix} -275\\ -116 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 116\\ -275 \end{bmatrix}$

Explanation:

$\begin{bmatrix} a& b\\ c& d \end{bmatrix} \times \begin{bmatrix} e\\ f \end{bmatrix} = \begin{bmatrix} a\cdot e+b\cdot f\\ c\cdot e+d\cdot f \end{bmatrix}$

Now we apply the above to get the following solution.

$\begin{bmatrix} 7&-1 \\ -16& -1 \end{bmatrix} \times \begin{bmatrix} 17\\ 3 \end{bmatrix}=\begin{bmatrix} 7\cdot17+(-1)\cdot3\\ (-16)\cdot 17+(-1)\cdot3 \end{bmatrix} =\begin{bmatrix} 119-3\\ -272-3 \end{bmatrix} =\begin{bmatrix} 116\\ -275 \end{bmatrix}$

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Example Question #4 : Vector Multiplication (Matrices Being Transformations Of Vectors): Ccss.Math.Content.Hsn Vm.C.11

Calculate

$\begin{bmatrix} -2&-19 \\ 9& 5 \end{bmatrix} \times \begin{bmatrix} -12\\ 10 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} -166\\ 58 \end{bmatrix}$

$\begin{bmatrix} 166\\ 58 \end{bmatrix}$

$\begin{bmatrix} 166\\ -58 \end{bmatrix}$

$\begin{bmatrix} -58\\ -166 \end{bmatrix}$

$\begin{bmatrix} -166\\ -58 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -166\\ -58 \end{bmatrix}$

Explanation:

$\begin{bmatrix} a& b\\ c& d \end{bmatrix} \times \begin{bmatrix} e\\ f \end{bmatrix} = \begin{bmatrix} a\cdot e+b\cdot f\\ c\cdot e+d\cdot f \end{bmatrix}$

Now we apply the above to get the following solution.

$\\ \begin{bmatrix} -2&-19 \\ 9& 5 \end{bmatrix} \times \begin{bmatrix} -12\\ 10 \end{bmatrix}=\begin{bmatrix}(-2)\cdot(-12)+(-19)\cdot10\\ 9\cdot (-12)+5\cdot10 \end{bmatrix} =\begin{bmatrix} 24-190\\ -108+50 \end{bmatrix} =\begin{bmatrix} -166\\ -58 \end{bmatrix}$

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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 #4 : 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 #3 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Example Question #4 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

Example Question #151 : Vector & Matrix Quantities

Example Question #152 : Vector & Matrix Quantities

Example Question #1 : Vector Multiplication (Matrices Being Transformations Of Vectors): Ccss.Math.Content.Hsn Vm.C.11

Example Question #2 : Vector Multiplication (Matrices Being Transformations Of Vectors): Ccss.Math.Content.Hsn Vm.C.11

Example Question #2 : Vector Multiplication (Matrices Being Transformations Of Vectors): Ccss.Math.Content.Hsn Vm.C.11

Example Question #4 : Vector Multiplication (Matrices Being Transformations Of Vectors): Ccss.Math.Content.Hsn Vm.C.11

All Common Core: High School - Number and Quantity Resources

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