Zero and Identity Matrices: CCSS.Math.Content.HSN-VM.C.10

Help Questions

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

Questions 1 - 10

Find the inverse of the following matrix.

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

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

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

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

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

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 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 , , , and .

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

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.

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

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

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

Find the inverse of the following matrix.

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

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

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

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

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

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

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 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 , , , and .

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

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.

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

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

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

Find the inverse of the following matrix.

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

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

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

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

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

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 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 , , , and .

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

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.

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

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

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

Find the inverse of the following matrix.

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

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

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

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

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

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 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 , , , and .

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

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.

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

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

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

Find the inverse of the following matrix.

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

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

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

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

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

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 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 , , , and .

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

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.

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

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

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

Find the inverse of the following matrix.

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

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

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

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

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

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 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 , , , and .

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

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.

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

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

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

Find the inverse of the following matrix.

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

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

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

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

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

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 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 , , , and .

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

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.

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

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

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

Find the inverse of the following matrix.

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

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

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

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

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

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 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 , , , and .

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

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.

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

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

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

Find the inverse of the following matrix.

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

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

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

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

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

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

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 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 , , , and .

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

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.

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

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

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

Find the inverse of the following matrix.

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

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

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

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

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

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 refer to position within the general 2x2 matrix $\begin{bmatrix} a& b\ c&d \end{bmatrix}$ .