Consider matrix `A=[a_(ij)]_(nxxn)`. Form the matrix `A-lamdal, lamda` being a number, real of complex.
`A-lamdal=[{:(a_11-lamda,a_12,...,a_(1n)),(s_21,a_22-lamda,...,a_(2n)),(...,...,...,...),(a_(n1),a_(n2),...,a_(n n)-lamda):}]`
Then det `(A-lamdaI)=(-1)^n[lamda^n+b_1lamda^(n-1)+b_2lamda^(n-2)+...+b_(n)]`.
An important rheorem tells us that the matrix A satisfies the equation `X^n+b_1x^(n-1)+b_2x^(n-2)+...+b_2=0.` This equation is called hte characteristic equation of A. For all the questions on theis passeage, take `A=[{:(1,4),(2,3):}]`
Which of the follwing is inverse fo A ?

Options

A. (a)`1/5(A-4l)`
B. (b)`1/5(A+4l)`
C. (c)`1/4(A-5l)`
D. (d)`1/4(A+5l)`

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the determinant of matrix \( A \)
The determinant of a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:
\[ \text{det}(A) = ad - bc \]
For our matrix \( A \):
- \( a = 1 \)
- \( b = 4 \)
- \( c = 2 \)
- \( d = 3 \)

Calculating the determinant:
\[ \text{det}(A) = (1)(3) - (4)(2) = 3 - 8 = -5 \]

### Step 2: Find the adjugate of matrix \( A \)
The adjugate of a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:
\[ \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
For our matrix \( A \):
\[ \text{adj}(A) = \begin{pmatrix} 3 & -4 \\ -2 & 1 \end{pmatrix} \]

### Step 3: Calculate the inverse of matrix \( A \)
The inverse of matrix \( A \) can be calculated using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Substituting the values we found:
\[ A^{-1} = \frac{1}{-5} \cdot \begin{pmatrix} 3 & -4 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} -\frac{3}{5} & \frac{4}{5} \\ \frac{2}{5} & -\frac{1}{5} \end{pmatrix} \]

### Step 4: Verify the inverse
To verify that this is indeed the inverse, we can multiply \( A \) and \( A^{-1} \) and check if we get the identity matrix:
\[ A \cdot A^{-1} = \begin{pmatrix} 1 & 4 \\ 2 & 3 \end{pmatrix} \cdot \begin{pmatrix} -\frac{3}{5} & \frac{4}{5} \\ \frac{2}{5} & -\frac{1}{5} \end{pmatrix} \]
Calculating the product:
- First row, first column: \( 1 \cdot -\frac{3}{5} + 4 \cdot \frac{2}{5} = -\frac{3}{5} + \frac{8}{5} = 1 \)
- First row, second column: \( 1 \cdot \frac{4}{5} + 4 \cdot -\frac{1}{5} = \frac{4}{5} - \frac{4}{5} = 0 \)
- Second row, first column: \( 2 \cdot -\frac{3}{5} + 3 \cdot \frac{2}{5} = -\frac{6}{5} + \frac{6}{5} = 0 \)
- Second row, second column: \( 2 \cdot \frac{4}{5} + 3 \cdot -\frac{1}{5} = \frac{8}{5} - \frac{3}{5} = 1 \)

Thus, \( A \cdot A^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \), confirming that \( A^{-1} \) is correct.

### Final Answer:
The inverse of matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} -\frac{3}{5} & \frac{4}{5} \\ \frac{2}{5} & -\frac{1}{5} \end{pmatrix} \]