If B and C are two matrices such that `B=[{:(,1,3),(,-2,0):}] and C=[{:(,17,7),(,-4,-8):}]`, find the matrix M so that BM=C.

Answer

To solve the problem, we need to find the matrix \( M \) such that \( BM = C \), where:

\[ B = \begin{pmatrix} 1 & 3 \\ -2 & 0 \end{pmatrix}, \quad C = \begin{pmatrix} 17 & 7 \\ -4 & -8 \end{pmatrix} \]

Let's denote the matrix \( M \) as:

\[ M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

### Step 1: Set up the equation

We need to multiply matrix \( B \) by matrix \( M \):

\[ BM = \begin{pmatrix} 1 & 3 \\ -2 & 0 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

### Step 2: Perform the multiplication

Using matrix multiplication, we calculate \( BM \):

\[ BM = \begin{pmatrix} 1 \cdot a + 3 \cdot c & 1 \cdot b + 3 \cdot d \\ -2 \cdot a + 0 \cdot c & -2 \cdot b + 0 \cdot d \end{pmatrix} \]

This simplifies to:

\[ BM = \begin{pmatrix} a + 3c & b + 3d \\ -2a & -2b \end{pmatrix} \]

### Step 3: Set the equation equal to matrix \( C \)

Now, we set \( BM \) equal to \( C \):

\[ \begin{pmatrix} a + 3c & b + 3d \\ -2a & -2b \end{pmatrix} = \begin{pmatrix} 17 & 7 \\ -4 & -8 \end{pmatrix} \]

### Step 4: Create a system of equations

From the equality of the matrices, we can create the following equations:

1. \( a + 3c = 17 \) (Equation 1)
2. \( b + 3d = 7 \) (Equation 2)
3. \( -2a = -4 \) (Equation 3)
4. \( -2b = -8 \) (Equation 4)

### Step 5: Solve for \( a \) and \( b \)

From Equation 3:

\[ -2a = -4 \implies a = 2 \]

From Equation 4:

\[ -2b = -8 \implies b = 4 \]

### Step 6: Substitute \( a \) and \( b \) into the remaining equations

Substituting \( a = 2 \) into Equation 1:

\[ 2 + 3c = 17 \implies 3c = 15 \implies c = 5 \]

Substituting \( b = 4 \) into Equation 2:

\[ 4 + 3d = 7 \implies 3d = 3 \implies d = 1 \]

### Step 7: Write the matrix \( M \)

Now that we have the values of \( a \), \( b \), \( c \), and \( d \):

\[ M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 5 & 1 \end{pmatrix} \]

### Final Answer

Thus, the matrix \( M \) is:

\[ M = \begin{pmatrix} 2 & 4 \\ 5 & 1 \end{pmatrix} \]