Sure! Let's solve for the inverses of the given matrices step-by-step.
### Part (a)
We are given the matrix:
[tex]\[ A = \begin{pmatrix} 4 & 5 \\ 3 & 4 \end{pmatrix} \][/tex]
To find the inverse of a 2x2 matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex], use the formula:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For matrix [tex]\( A \)[/tex]:
[tex]\[ a = 4, \quad b = 5, \quad c = 3, \quad d = 4 \][/tex]
Calculate the determinant:
[tex]\[ \text{det}(A) = ad - bc = (4 \cdot 4) - (5 \cdot 3) = 16 - 15 = 1 \][/tex]
Now apply the formula for the inverse:
[tex]\[ A^{-1} = \frac{1}{1} \begin{pmatrix} 4 & -5 \\ -3 & 4 \end{pmatrix} = \begin{pmatrix} 4 & -5 \\ -3 & 4 \end{pmatrix} \][/tex]
Hence, the inverse of matrix [tex]\( A \)[/tex] is:
[tex]\[ A^{-1} = \begin{pmatrix} 4 & -5 \\ -3 & 4 \end{pmatrix} \][/tex]
### Part (b)
We are given the matrix:
[tex]\[ B = \begin{pmatrix} 5 & 6 \\ 2 & 4 \end{pmatrix} \][/tex]
Using the same formula for the inverse of a 2x2 matrix:
[tex]\[ B^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For matrix [tex]\( B \)[/tex]:
[tex]\[ a = 5, \quad b = 6, \quad c = 2, \quad d = 4 \][/tex]
Calculate the determinant:
[tex]\[ \text{det}(B) = ad - bc = (5 \cdot 4) - (6 \cdot 2) = 20 - 12 = 8 \][/tex]
Now apply the formula for the inverse:
[tex]\[ B^{-1} = \frac{1}{8} \begin{pmatrix} 4 & -6 \\ -2 & 5 \end{pmatrix} = \begin{pmatrix} \frac{4}{8} & \frac{-6}{8} \\ \frac{-2}{8} & \frac{5}{8} \end{pmatrix} = \begin{pmatrix} 0.5 & -0.75 \\ -0.25 & 0.625 \end{pmatrix} \][/tex]
Hence, the inverse of matrix [tex]\( B \)[/tex] is:
[tex]\[ B^{-1} = \begin{pmatrix} 0.5 & -0.75 \\ -0.25 & 0.625 \end{pmatrix} \][/tex]
So the inverses of the given matrices are:
- For matrix [tex]\( A \)[/tex]:
[tex]\[ \begin{pmatrix} 4 & -5 \\ -3 & 4 \end{pmatrix} \][/tex]
- For matrix [tex]\( B \)[/tex]:
[tex]\[ \begin{pmatrix} 0.5 & -0.75 \\ -0.25 & 0.625 \end{pmatrix} \][/tex]
