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Common Core: High School - Number and Quantity : Zero and Identity Matrices: CCSS.Math.Content.HSN-VM.C.10

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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:

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

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

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

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

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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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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Common Core: High School - Number and Quantity : Zero and Identity Matrices: CCSS.Math.Content.HSN-VM.C.10

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