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

Options

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

Answer

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

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

### Step 1: Determine the order of matrix \( A \)

Since we are multiplying matrix \( A \) (which we denote as \( m \times n \)) with a \( 2 \times 2 \) matrix, the resulting matrix must also be \( 2 \times 2 \). Therefore, the order of matrix \( A \) must be \( 2 \times 2 \).

**Hint:** The order of the resulting matrix from the multiplication of two matrices is determined by the outer dimensions of the matrices involved.

### Step 2: Set up the equation

Let the matrix \( A \) be represented as:

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

Now we can set up the equation based on the multiplication:

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

### Step 3: Perform the matrix multiplication

Calculating the left-hand side:

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

This gives us the following equations:

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

**Hint:** Each entry in the resulting matrix corresponds to a linear equation formed by the elements of the matrices being multiplied.

### Step 4: Solve the equations for \( a \) and \( b \)

From Equation 2, we can express \( b \) in terms of \( a \):

\[ b = 1 - 2a \]

Substituting \( b \) into Equation 1:

\[ -a + 3(1 - 2a) = -4 \]

Expanding and simplifying:

\[ -a + 3 - 6a = -4 \\ -7a + 3 = -4 \\ -7a = -7 \\ a = 1 \]

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

\[ b = 1 - 2(1) = 1 - 2 = -1 \]

### Step 5: Solve the equations for \( c \) and \( d \)

From Equation 4, we can express \( d \) in terms of \( c \):

\[ d = 7 - 2c \]

Substituting \( d \) into Equation 3:

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

Expanding and simplifying:

\[ -c + 21 - 6c = 7 \\ -7c + 21 = 7 \\ -7c = -14 \\ c = 2 \]

Now substituting \( c = 2 \) back into the expression for \( d \):

\[ d = 7 - 2(2) = 7 - 4 = 3 \]

### Step 6: Write the final matrix \( A \)

Now that we have found \( a, b, c, \) and \( d \):

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

### Final Answer

Thus, the matrix \( A \) is:

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

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