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

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

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Example Questions

Example Question #11 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

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

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

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

Example Question #12 : Zero And Identity Matrices: Ccss.Math.Content.Hsn Vm.C.10

All Common Core: High School - Number and Quantity Resources

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