Question

2 - 01 use the inverse formula to find the inverse of each matrix or explain why it cannot be found. your answer must be simplified and exact, and check your answers with a calculator.
(a) $c=\begin{bmatrix}1/2&3/5\\5& - 4end{bmatrix}$
(b) $d=\begin{bmatrix}0.3&5.1\\2.4&0.7end{bmatrix}$
(c) $a=\begin{bmatrix}-10&5\\-6&3end{bmatrix}$
(d) $b=\begin{bmatrix}0.4& - 0.5\\2&3end{bmatrix}$
(e) $a=\begin{bmatrix}-7&5\\4& - 3end{bmatrix}$
(f) $b=\begin{bmatrix}4&5\\2&3end{bmatrix}$

Explanation:

Step1: Recall the inverse - formula for a 2x2 matrix

For a 2x2 matrix $M=

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

$, the inverse is $M^{-1}=\frac{1}{ad - bc}

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

$, provided $ad - bc
eq0$.

Step2: Calculate the inverse of matrix $C=

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

$
First, find the determinant $ad - bc$:
\[

$$\begin{align*} \det(C)&=\frac{1}{2}\times(-4)-\frac{3}{5}\times5\\ &=- 2-3\\ &=-5 \end{align*}$$

\]
Then, $C^{-1}=\frac{1}{-5}

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

=

$$\begin{bmatrix}\frac{4}{5}&\frac{3}{25}\\1&-\frac{1}{10}\end{bmatrix}$$

$

Step3: Calculate the inverse of matrix $D=

$$\begin{bmatrix}0.3&5.1\\2.4&0.7\end{bmatrix}$$

$
Find the determinant:
\[

$$\begin{align*} \det(D)&=0.3\times0.7 - 5.1\times2.4\\ &=0.21-12.24\\ &=-12.03 \end{align*}$$

\]
Then, $D^{-1}=\frac{1}{-12.03}

$$\begin{bmatrix}0.7&-5.1\\-2.4&0.3\end{bmatrix}$$

=

$$\begin{bmatrix}-\frac{0.7}{12.03}&\frac{5.1}{12.03}\\\frac{2.4}{12.03}&-\frac{0.3}{12.03}\end{bmatrix}$$

=

$$\begin{bmatrix}-\frac{70}{1203}&\frac{510}{1203}\\\frac{240}{1203}&-\frac{30}{1203}\end{bmatrix}$$

$

Step4: Analyze matrix $A=

$$\begin{bmatrix}-10&5\\-6&3\end{bmatrix}$$

$
Find the determinant:
\[

$$\begin{align*} \det(A)&=(-10)\times3-5\times(-6)\\ &=-30 + 30\\ &=0 \end{align*}$$

\]
Since the determinant is 0, the inverse of matrix $A$ does not exist.

Step5: Calculate the inverse of matrix $B=

$$\begin{bmatrix}0.4&-0.5\\2&3\end{bmatrix}$$

$
Find the determinant:
\[

$$\begin{align*} \det(B)&=0.4\times3-(-0.5)\times2\\ &=1.2 + 1\\ &=2.2 \end{align*}$$

\]
Then, $B^{-1}=\frac{1}{2.2}

$$\begin{bmatrix}3&0.5\\-2&0.4\end{bmatrix}$$

=

$$\begin{bmatrix}\frac{3}{2.2}&\frac{0.5}{2.2}\\-\frac{2}{2.2}&\frac{0.4}{2.2}\end{bmatrix}$$

=

$$\begin{bmatrix}\frac{30}{22}&\frac{5}{22}\\-\frac{20}{22}&\frac{4}{22}\end{bmatrix}$$

=

$$\begin{bmatrix}\frac{15}{11}&\frac{5}{22}\\-\frac{10}{11}&\frac{2}{11}\end{bmatrix}$$

$

Step6: Calculate the inverse of matrix $A=

$$\begin{bmatrix}-7&5\\4&-3\end{bmatrix}$$

$
Find the determinant:
\[

$$\begin{align*} \det(A)&=(-7)\times(-3)-5\times4\\ &=21 - 20\\ &=1 \end{align*}$$

\]
Then, $A^{-1}=

$$\begin{bmatrix}-3&-5\\-4&-7\end{bmatrix}$$

$

Step7: Calculate the inverse of matrix $B=

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

$
Find the determinant:
\[

$$\begin{align*} \det(B)&=4\times3-5\times2\\ &=12 - 10\\ &=2 \end{align*}$$

\]
Then, $B^{-1}=\frac{1}{2}

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

=

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

$

Answer:

(a) $C^{-1}=

$$\begin{bmatrix}\frac{4}{5}&\frac{3}{25}\\1&-\frac{1}{10}\end{bmatrix}$$

$
(b) $D^{-1}=

$$\begin{bmatrix}-\frac{70}{1203}&\frac{510}{1203}\\\frac{240}{1203}&-\frac{30}{1203}\end{bmatrix}$$

$
(c) Inverse does not exist since $\det(A) = 0$
(d) $B^{-1}=

$$\begin{bmatrix}\frac{15}{11}&\frac{5}{22}\\-\frac{10}{11}&\frac{2}{11}\end{bmatrix}$$

$
(e) $A^{-1}=

$$\begin{bmatrix}-3&-5\\-4&-7\end{bmatrix}$$

$
(f) $B^{-1}=

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

$