Question

part 3: inverses of matrices – make sure to show all work to receive full credit!

1. does the following matrix have an inverse? explain why or why not. \\(\

$$\begin{bmatrix} 1 & 2 \\\\ 6 & 12 \\end{bmatrix}$$

\\)
find the inverse of the following matrices. show all work
10)\\(\

$$\begin{bmatrix} 3 & 2 \\\\ 11 & 10 \\end{bmatrix}$$

\\)
11)\\(\

$$\begin{bmatrix} -6 & -7 \\\\ 3 & 6 \\end{bmatrix}$$

\\)

Explanation:

Problem 9:

Step1: Recall Invertibility Condition

A \( 2\times2 \) matrix \(

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

\) is invertible (has an inverse) if its determinant \( \det = ad - bc
eq 0 \).

Step2: Calculate Determinant

For matrix \(

$$\begin{bmatrix}1&2\\6&12\end{bmatrix}$$

\), \( a = 1 \), \( b = 2 \), \( c = 6 \), \( d = 12 \).
Determinant: \( (1)(12) - (2)(6) = 12 - 12 = 0 \).

Step3: Conclude Invertibility

Since determinant is \( 0 \), the matrix does not have an inverse (invertible matrices require non - zero determinant).

Step1: Recall Inverse Formula

For \( 2\times2 \) matrix \( A=

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

\), inverse \( A^{-1}=\frac{1}{\det(A)}

$$\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$

\), where \( \det(A)=ad - bc \).

Step2: Calculate Determinant

For \( A =

$$\begin{bmatrix}3&2\\11&10\end{bmatrix}$$

\), \( \det(A)=(3)(10)-(2)(11)=30 - 22 = 8 \).

Step3: Apply Inverse Formula

\( A^{-1}=\frac{1}{8}

$$\begin{bmatrix}10&-2\\-11&3\end{bmatrix}$$

=

$$\begin{bmatrix}\frac{10}{8}&\frac{-2}{8}\\\frac{-11}{8}&\frac{3}{8}\end{bmatrix}$$

=

$$\begin{bmatrix}\frac{5}{4}&-\frac{1}{4}\\-\frac{11}{8}&\frac{3}{8}\end{bmatrix}$$

\).

Step1: Recall Inverse Formula

For \( 2\times2 \) matrix \( A=

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

\), inverse \( A^{-1}=\frac{1}{\det(A)}

$$\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$

\), \( \det(A)=ad - bc \).

Step2: Calculate Determinant

For \( A=

$$\begin{bmatrix}-6&-7\\3&6\end{bmatrix}$$

\), \( \det(A)=(-6)(6)-(-7)(3)=- 36 + 21=-15 \).

Step3: Apply Inverse Formula

\( A^{-1}=\frac{1}{-15}

$$\begin{bmatrix}6&7\\-3&-6\end{bmatrix}$$

=

$$\begin{bmatrix}\frac{6}{-15}&\frac{7}{-15}\\\frac{-3}{-15}&\frac{-6}{-15}\end{bmatrix}$$

=

$$\begin{bmatrix}-\frac{2}{5}&-\frac{7}{15}\\\frac{1}{5}&\frac{2}{5}\end{bmatrix}$$

\).

Answer:

The matrix \(

$$\begin{bmatrix}1&2\\6&12\end{bmatrix}$$

\) does not have an inverse because its determinant \( (1\times12 - 2\times6)=0 \), and a \( 2\times2 \) matrix is invertible if and only if its determinant is non - zero.

Problem 10: