The number of `2xx2` matrices A with real entries, such that `A+A^(T)=3I` and `A A^(T)=5I`, is equal to

Options

A. 0
B. 1
C. 2
D. infinite

Answer

To solve the problem, we need to find the number of \(2 \times 2\) matrices \(A\) with real entries that satisfy the equations:

1. \(A + A^T = 3I\)
2. \(AA^T = 5I\)

Here, \(I\) is the identity matrix.

### Step 1: Define the Matrix \(A\)

Let \(A\) be defined as:

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

### Step 2: Calculate \(A^T\)

The transpose of \(A\) is given by:

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

### Step 3: Use the First Equation

From the first equation, we have:

\[ A + A^T = 3I \]

Substituting the matrices, we get:

\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} + \begin{pmatrix} a & c \\ b & d \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]

This simplifies to:

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

From this, we can equate the corresponding elements:

1. \(2a = 3 \Rightarrow a = \frac{3}{2}\)
2. \(2d = 3 \Rightarrow d = \frac{3}{2}\)
3. \(b + c = 0 \Rightarrow c = -b\)

### Step 4: Substitute Values into Matrix \(A\)

Now substituting the values of \(a\) and \(d\) back into the matrix \(A\):

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

### Step 5: Use the Second Equation

Now, we apply the second equation:

\[ AA^T = 5I \]

Calculating \(AA^T\):

\[ AA^T = \begin{pmatrix} \frac{3}{2} & b \\ -b & \frac{3}{2} \end{pmatrix} \begin{pmatrix} \frac{3}{2} & -b \\ b & \frac{3}{2} \end{pmatrix} \]

Calculating the elements:

1. First element: \(\left(\frac{3}{2}\right)^2 + b^2 = \frac{9}{4} + b^2\)
2. Second element: \(\frac{3}{2}(-b) + b\left(\frac{3}{2}\right) = 0\)
3. Third element: \(-b\left(\frac{3}{2}\right) + \frac{3}{2}b = 0\)
4. Fourth element: \((-b)^2 + \left(\frac{3}{2}\right)^2 = b^2 + \frac{9}{4}\)

Thus, we have:

\[ AA^T = \begin{pmatrix} \frac{9}{4} + b^2 & 0 \\ 0 & b^2 + \frac{9}{4} \end{pmatrix} \]

Setting this equal to \(5I\):

\[ \begin{pmatrix} \frac{9}{4} + b^2 & 0 \\ 0 & b^2 + \frac{9}{4} \end{pmatrix} = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]

This gives us two equations:

1. \(\frac{9}{4} + b^2 = 5\)
2. \(b^2 + \frac{9}{4} = 5\)

### Step 6: Solve for \(b^2\)

From the first equation:

\[ b^2 = 5 - \frac{9}{4} = \frac{20}{4} - \frac{9}{4} = \frac{11}{4} \]

Thus, we have:

\[ b = \pm \sqrt{\frac{11}{4}} = \pm \frac{\sqrt{11}}{2} \]

### Step 7: Determine the Number of Matrices

Since \(b\) can take two values (\(\frac{\sqrt{11}}{2}\) and \(-\frac{\sqrt{11}}{2}\)), we can construct two distinct matrices \(A\):

1. For \(b = \frac{\sqrt{11}}{2}\):
\[ A_1 = \begin{pmatrix} \frac{3}{2} & \frac{\sqrt{11}}{2} \\ -\frac{\sqrt{11}}{2} & \frac{3}{2} \end{pmatrix} \]

2. For \(b = -\frac{\sqrt{11}}{2}\):
\[ A_2 = \begin{pmatrix} \frac{3}{2} & -\frac{\sqrt{11}}{2} \\ \frac{\sqrt{11}}{2} & \frac{3}{2} \end{pmatrix} \]

Thus, the total number of such matrices \(A\) is **2**.

### Final Answer

The number of \(2 \times 2\) matrices \(A\) with real entries that satisfy the given conditions is **2**.