If I is the unit matrix of order `2 xx 2` find the matrix M, such that
(i) `M-2I=3 [{:(,-1,0),(,4,1):}]`
(ii) `5M+3I=4[{:(,2,-5),(,0,-3):}]`

Answer

To solve the problem, we need to find the matrix \( M \) for the two given equations involving the identity matrix \( I \) of order \( 2 \times 2 \).

### Step 1: Solve the first equation \( M - 2I = 3 \begin{pmatrix} -1 & 0 \\ 4 & 1 \end{pmatrix} \)

1. **Write down the identity matrix \( I \)**:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

2. **Multiply the matrix by 3**:
\[ 3 \begin{pmatrix} -1 & 0 \\ 4 & 1 \end{pmatrix} = \begin{pmatrix} -3 & 0 \\ 12 & 3 \end{pmatrix} \]

3. **Substitute this into the equation**:
\[ M - 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -3 & 0 \\ 12 & 3 \end{pmatrix} \]

4. **Calculate \( 2I \)**:
\[ 2I = 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \]

5. **Rearranging the equation to solve for \( M \)**:
\[ M = \begin{pmatrix} -3 & 0 \\ 12 & 3 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \]

6. **Add the matrices**:
\[ M = \begin{pmatrix} -3 + 2 & 0 + 0 \\ 12 + 0 & 3 + 2 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 12 & 5 \end{pmatrix} \]

### Step 2: Solve the second equation \( 5M + 3I = 4 \begin{pmatrix} 2 & -5 \\ 0 & -3 \end{pmatrix} \)

1. **Multiply the matrix by 4**:
\[ 4 \begin{pmatrix} 2 & -5 \\ 0 & -3 \end{pmatrix} = \begin{pmatrix} 8 & -20 \\ 0 & -12 \end{pmatrix} \]

2. **Substitute this into the equation**:
\[ 5M + 3 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 8 & -20 \\ 0 & -12 \end{pmatrix} \]

3. **Calculate \( 3I \)**:
\[ 3I = 3 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]

4. **Rearranging the equation to solve for \( 5M \)**:
\[ 5M = \begin{pmatrix} 8 & -20 \\ 0 & -12 \end{pmatrix} - \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]

5. **Subtract the matrices**:
\[ 5M = \begin{pmatrix} 8 - 3 & -20 - 0 \\ 0 - 0 & -12 - 3 \end{pmatrix} = \begin{pmatrix} 5 & -20 \\ 0 & -15 \end{pmatrix} \]

6. **Divide by 5 to find \( M \)**:
\[ M = \frac{1}{5} \begin{pmatrix} 5 & -20 \\ 0 & -15 \end{pmatrix} = \begin{pmatrix} 1 & -4 \\ 0 & -3 \end{pmatrix} \]

### Final Result:
The matrix \( M \) is:
\[ M = \begin{pmatrix} -1 & 0 \\ 12 & 5 \end{pmatrix} \text{ from the first equation and } M = \begin{pmatrix} 1 & -4 \\ 0 & -3 \end{pmatrix} \text{ from the second equation.} \]