If the matrix A is such that `A[{:(-1,2),(3,1):}]=[{:(-4,1),(7,7):}]`,then A is equal to

Options

A. (a) `[{:(1,1),(2,-3):}]`
B. (b) `[{:(1,1),(-2,3):}]`
C. (c) `[{:(1,-1),(2,3):}]`
D. (d) `[{:(-1,1),(2,3):}]`

Answer

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

\[ A \begin{pmatrix} -1 & 2 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} -4 & 1 \\ 7 & 7 \end{pmatrix} \]

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

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

Now, we will perform the matrix multiplication:

\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -1 & 2 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} a(-1) + b(3) & a(2) + b(1) \\ c(-1) + d(3) & c(2) + d(1) \end{pmatrix} \]

This gives us:

\[ \begin{pmatrix} -a + 3b & 2a + b \\ -c + 3d & 2c + d \end{pmatrix} = \begin{pmatrix} -4 & 1 \\ 7 & 7 \end{pmatrix} \]

From this, we can set up the following equations by equating the corresponding entries:

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

### Step 1: Solve for \( a \) and \( b \)

From Equation 1:

\[ -a + 3b = -4 \implies a = 3b + 4 \]

Now substitute \( a \) in Equation 2:

\[ 2(3b + 4) + b = 1 \]
\[ 6b + 8 + b = 1 \]
\[ 7b + 8 = 1 \implies 7b = 1 - 8 \implies 7b = -7 \implies b = -1 \]

Now substitute \( b = -1 \) back into the expression for \( a \):

\[ a = 3(-1) + 4 = -3 + 4 = 1 \]

### Step 2: Solve for \( c \) and \( d \)

From Equation 3:

\[ -c + 3d = 7 \implies c = 3d - 7 \]

Now substitute \( c \) in Equation 4:

\[ 2(3d - 7) + d = 7 \]
\[ 6d - 14 + d = 7 \]
\[ 7d - 14 = 7 \implies 7d = 7 + 14 \implies 7d = 21 \implies d = 3 \]

Now substitute \( d = 3 \) back into the expression for \( c \):

\[ c = 3(3) - 7 = 9 - 7 = 2 \]

### Final Matrix

Now we have all the values:

\[ a = 1, \quad b = -1, \quad c = 2, \quad d = 3 \]

Thus, the matrix \( A \) is:

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

### Conclusion

The final answer is:

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