If `A=[(2,-3),(0,1)]`, find B, such that `4A^(-1)+B=A^(2)`

Answer

To solve the problem, we need to find the matrix \( B \) such that:

\[ 4A^{-1} + B = A^2 \]

where \( A = \begin{pmatrix} 2 & -3 \\ 0 & 1 \end{pmatrix} \).

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we multiply matrix \( A \) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} 2 & -3 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 2 & -3 \\ 0 & 1 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \( 2 \cdot 2 + (-3) \cdot 0 = 4 \)
- First row, second column: \( 2 \cdot (-3) + (-3) \cdot 1 = -6 - 3 = -9 \)
- Second row, first column: \( 0 \cdot 2 + 1 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot (-3) + 1 \cdot 1 = 1 \)

Thus,

\[ A^2 = \begin{pmatrix} 4 & -9 \\ 0 & 1 \end{pmatrix} \]

### Step 2: Calculate \( A^{-1} \)

To find \( A^{-1} \), we first need to calculate the determinant of \( A \):

\[ \text{det}(A) = 2 \cdot 1 - 0 \cdot (-3) = 2 \]

Next, we find the adjoint of \( A \):

\[ \text{adj}(A) = \begin{pmatrix} 1 & 0 \\ 3 & 2 \end{pmatrix} \]

Now, we can find \( A^{-1} \):

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 0 \\ \frac{3}{2} & 1 \end{pmatrix} \]

### Step 3: Calculate \( 4A^{-1} \)

Now, we calculate \( 4A^{-1} \):

\[ 4A^{-1} = 4 \cdot \begin{pmatrix} \frac{1}{2} & 0 \\ \frac{3}{2} & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 6 & 4 \end{pmatrix} \]

### Step 4: Solve for \( B \)

Now we substitute \( A^2 \) and \( 4A^{-1} \) into the equation \( 4A^{-1} + B = A^2 \):

\[ B = A^2 - 4A^{-1} = \begin{pmatrix} 4 & -9 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 2 & 0 \\ 6 & 4 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \( 4 - 2 = 2 \)
- First row, second column: \( -9 - 0 = -9 \)
- Second row, first column: \( 0 - 6 = -6 \)
- Second row, second column: \( 1 - 4 = -3 \)

Thus,

\[ B = \begin{pmatrix} 2 & -9 \\ -6 & -3 \end{pmatrix} \]

### Final Answer

The required matrix \( B \) is:

\[ B = \begin{pmatrix} 2 & -9 \\ -6 & -3 \end{pmatrix} \]