Find a matrix B such that 
AB = I,
 where
A = 

4	1	

−4	4
  and  I = 

1	0	

0	1
.
Hint: Let 
B = 

a	b	

c	d
.

To find the matrix \( B \) such that \( AB = I \), we are looking for the inverse of matrix \( A \), where:

\[
A = \begin{pmatrix} 4 & 1 \\ -4 & 4 \end{pmatrix}
\quad \text{and} \quad I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]
We need to find a matrix \( B = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) such that:

\[
A \cdot B = I
\]
This implies:

\[
\begin{pmatrix} 4 & 1 \\ -4 & 4 \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]

Multiplying the two matrices on the left-hand side:

\[
\begin{pmatrix} 4a + 1c & 4b + 1d \\ -4a + 4c & -4b + 4d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]

This gives us a system of four equations:

1. \( 4a + c = 1 \)
2. \( 4b + d = 0 \)
3. \( -4a + 4c = 0 \)
4. \( -4b + 4d = 1 \)

### Solving the system of equations:

From equation (3):
\[
-4a + 4c = 0 \quad \Rightarrow \quad c = a
\]

Substitute \( c = a \) into equation (1):
\[
4a + a = 1 \quad \Rightarrow \quad 5a = 1 \quad \Rightarrow \quad a = \frac{1}{5}
\]

Thus, \( c = \frac{1}{5} \).

From equation (4):
\[
-4b + 4d = 1 \quad \Rightarrow \quad -b + d = \frac{1}{4} \quad \Rightarrow \quad d = b + \frac{1}{4}
\]

Substitute \( d = b + \frac{1}{4} \) into equation (2):
\[
4b + \left( b + \frac{1}{4} \right) = 0 \quad \Rightarrow \quad 5b + \frac{1}{4} = 0 \quad \Rightarrow \quad 5b = -\frac{1}{4} \quad \Rightarrow \quad b = -\frac{1}{20}
\]

Thus, \( d = b + \frac{1}{4} = -\frac{1}{20} + \frac{1}{4} = \frac{1}{5} \).

### Solution:

Therefore, the matrix \( B \) is:

\[
B = \begin{pmatrix} \frac{1}{5} & -\frac{1}{20} \\ \frac{1}{5} & \frac{1}{5} \end{pmatrix}
\]

### Conclusion:
The matrix \( B \) is the inverse of \( A \), and multiplying \( A \) by \( B \) will give the identity matrix \( I \).

Would you like further details on how matrix multiplication works? 

Here are 5 related questions to consider:
1. What is the determinant of a 2x2 matrix, and how is it used to find inverses?
2. How can you verify if a matrix is invertible?
3. What properties does the inverse matrix have?
4. Can every square matrix be inverted? Why or why not?
5. How would the process change for larger matrices?

**Tip**: When solving matrix equations, matrix multiplication is not commutative, meaning \( AB \neq BA \) in general.

Find a matrix B such that AB = I, where A = [[4, 1], [-4, 4]] and I = [[1, 0], [0, 1]]. Hint: Let B = [[a, b], [c, d]].

Explore how to find the inverse of matrix A to satisfy the equation AB = I, where A = [[4, 1], [-4, 4]] and B = [[a, b], [c, d]]. Learn the step-by-step process of solving the system of linear equations involved in this matrix inversion. Suitable for advanced high school or undergraduate students.

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