Given `A=[{:(,-1,0),(,2,-4):}] and B=[{:(,3,-3),(,-2,0):}]` find the matrix X in each of the following
(i) A+X=B
(ii) A-X=B
(iii) X-B=A

Answer

To solve the given problem, we need to find the matrix \( X \) in three different scenarios involving matrices \( A \) and \( B \).

Given:
\[ A = \begin{pmatrix} -1 & 0 \\ 2 & -4 \end{pmatrix} \]
\[ B = \begin{pmatrix} 3 & -3 \\ -2 & 0 \end{pmatrix} \]

Let \( X = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \).

### (i) Solve for \( A + X = B \)

1. Start with the equation:
\[ A + X = B \]
Substituting the values of \( A \) and \( B \):
\[ \begin{pmatrix} -1 & 0 \\ 2 & -4 \end{pmatrix} + \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 3 & -3 \\ -2 & 0 \end{pmatrix} \]

2. Combine the matrices:
\[ \begin{pmatrix} -1 + a & 0 + b \\ 2 + c & -4 + d \end{pmatrix} = \begin{pmatrix} 3 & -3 \\ -2 & 0 \end{pmatrix} \]

3. Set up equations by equating corresponding elements:
- From the first element: \( -1 + a = 3 \)
- From the second element: \( 0 + b = -3 \)
- From the third element: \( 2 + c = -2 \)
- From the fourth element: \( -4 + d = 0 \)

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

5. Therefore, the matrix \( X \) is:
\[ X = \begin{pmatrix} 4 & -3 \\ -4 & 4 \end{pmatrix} \]

### (ii) Solve for \( A - X = B \)

1. Start with the equation:
\[ A - X = B \]
Substituting the values:
\[ \begin{pmatrix} -1 & 0 \\ 2 & -4 \end{pmatrix} - \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 3 & -3 \\ -2 & 0 \end{pmatrix} \]

2. Rearranging gives:
\[ \begin{pmatrix} -1 - a & 0 - b \\ 2 - c & -4 - d \end{pmatrix} = \begin{pmatrix} 3 & -3 \\ -2 & 0 \end{pmatrix} \]

3. Set up equations:
- From the first element: \( -1 - a = 3 \)
- From the second element: \( 0 - b = -3 \)
- From the third element: \( 2 - c = -2 \)
- From the fourth element: \( -4 - d = 0 \)

4. Solve each equation:
- \( -a = 3 + 1 \) → \( a = -4 \)
- \( -b = -3 \) → \( b = 3 \)
- \( -c = -2 - 2 \) → \( c = 4 \)
- \( -d = 0 + 4 \) → \( d = -4 \)

5. Therefore, the matrix \( X \) is:
\[ X = \begin{pmatrix} -4 & 3 \\ 4 & -4 \end{pmatrix} \]

### (iii) Solve for \( X - B = A \)

1. Start with the equation:
\[ X - B = A \]
Substituting the values:
\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} - \begin{pmatrix} 3 & -3 \\ -2 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 2 & -4 \end{pmatrix} \]

2. Rearranging gives:
\[ \begin{pmatrix} a - 3 & b + 3 \\ c + 2 & d - 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 2 & -4 \end{pmatrix} \]

3. Set up equations:
- From the first element: \( a - 3 = -1 \)
- From the second element: \( b + 3 = 0 \)
- From the third element: \( c + 2 = 2 \)
- From the fourth element: \( d = -4 \)

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

5. Therefore, the matrix \( X \) is:
\[ X = \begin{pmatrix} 2 & -3 \\ 0 & -4 \end{pmatrix} \]

### Summary of Solutions:
1. \( X = \begin{pmatrix} 4 & -3 \\ -4 & 4 \end{pmatrix} \)
2. \( X = \begin{pmatrix} -4 & 3 \\ 4 & -4 \end{pmatrix} \)
3. \( X = \begin{pmatrix} 2 & -3 \\ 0 & -4 \end{pmatrix} \)