if `[{:(3,2),(7,5):}]A{:[(-1,1),(-2,1):}]={:[(2,-1),(0,4):}]` then trace of A is equal to

Options

A. `-11`
B. `-21`
C. `-25`
D. `-15`

Explanation

To solve the problem, we need to find the trace of matrix A given the equation:

\[ YAX = B \]

where \( Y = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix} \), \( X = \begin{pmatrix} -1 & 1 \\ -2 & 1 \end{pmatrix} \), and \( B = \begin{pmatrix} 2 & -1 \\ 0 & 4 \end{pmatrix} \).

### Step 1: Find the inverse of matrix Y

First, we calculate the determinant of matrix Y:

\[ \text{det}(Y) = (3)(5) - (2)(7) = 15 - 14 = 1 \]

Since the determinant is not zero, Y is invertible. The inverse of Y can be calculated as:

\[ Y^{-1} = \frac{1}{\text{det}(Y)} \text{adj}(Y) \]

The adjugate of Y is given by swapping the elements on the main diagonal and changing the signs of the off-diagonal elements:

\[ \text{adj}(Y) = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} \]

Thus, we have:

\[ Y^{-1} = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} \]

### Step 2: Multiply both sides of the equation by \( Y^{-1} \)

Now we will multiply both sides of the equation \( YAX = B \) by \( Y^{-1} \):

\[ Y^{-1}YAX = Y^{-1}B \]

Since \( Y^{-1}Y = I \) (the identity matrix), we simplify to:

\[ AX = Y^{-1}B \]

### Step 3: Calculate \( Y^{-1}B \)

Next, we compute \( Y^{-1}B \):

\[ Y^{-1}B = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ 0 & 4 \end{pmatrix} \]

Calculating this matrix multiplication:

1. First row, first column: \( (5)(2) + (-2)(0) = 10 \)
2. First row, second column: \( (5)(-1) + (-2)(4) = -5 - 8 = -13 \)
3. Second row, first column: \( (-7)(2) + (3)(0) = -14 \)
4. Second row, second column: \( (-7)(-1) + (3)(4) = 7 + 12 = 19 \)

So,

\[ Y^{-1}B = \begin{pmatrix} 10 & -13 \\ -14 & 19 \end{pmatrix} \]

### Step 4: Find matrix A

Now we have:

\[ AX = \begin{pmatrix} 10 & -13 \\ -14 & 19 \end{pmatrix} \]

To find A, we need to multiply both sides by \( X^{-1} \). First, we calculate \( X^{-1} \):

The determinant of X is:

\[ \text{det}(X) = (-1)(1) - (1)(-2) = -1 + 2 = 1 \]

The adjugate of X is:

\[ \text{adj}(X) = \begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix} \]

Thus,

\[ X^{-1} = \begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix} \]

Now we can find A:

\[ A = \begin{pmatrix} 10 & -13 \\ -14 & 19 \end{pmatrix} X^{-1} \]

Calculating this multiplication:

1. First row, first column: \( (10)(1) + (-13)(2) = 10 - 26 = -16 \)
2. First row, second column: \( (10)(-1) + (-13)(-1) = -10 + 13 = 3 \)
3. Second row, first column: \( (-14)(1) + (19)(2) = -14 + 38 = 24 \)
4. Second row, second column: \( (-14)(-1) + (19)(-1) = 14 - 19 = -5 \)

So, we have:

\[ A = \begin{pmatrix} -16 & 3 \\ 24 & -5 \end{pmatrix} \]

### Step 5: Calculate the trace of A

The trace of a matrix is the sum of its diagonal elements:

\[ \text{trace}(A) = -16 + (-5) = -21 \]

Thus, the trace of matrix A is:

\[ \text{trace}(A) = -21 \]

### Final Answer

The trace of A is equal to \(-21\).

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