Let A be a `3 xx 3` matrix having entries from the set {-1,0,1}. The number of all such matrices A having sum of all the entries equal to 5, is _____

Answer

To solve the problem of finding the number of 3x3 matrices with entries from the set {-1, 0, 1} such that the sum of all entries equals 5, we can follow these steps:

### Step 1: Understand the Matrix Structure
A 3x3 matrix has 9 entries. The possible entries are -1, 0, and 1. We need to find combinations of these entries such that their total sum equals 5.

### Step 2: Set Up the Equation
Let:
- \( x_1 \) = number of 1's in the matrix
- \( x_0 \) = number of 0's in the matrix
- \( x_{-1} \) = number of -1's in the matrix

From the matrix, we know:
\[ x_1 + x_0 + x_{-1} = 9 \] (total number of entries)
And the sum of the entries must equal 5:
\[ 1 \cdot x_1 + 0 \cdot x_0 - 1 \cdot x_{-1} = 5 \]
This simplifies to:
\[ x_1 - x_{-1} = 5 \]

### Step 3: Solve the System of Equations
We now have a system of equations:
1. \( x_1 + x_0 + x_{-1} = 9 \)
2. \( x_1 - x_{-1} = 5 \)

From the second equation, we can express \( x_{-1} \) in terms of \( x_1 \):
\[ x_{-1} = x_1 - 5 \]

Substituting this into the first equation:
\[ x_1 + x_0 + (x_1 - 5) = 9 \]
This simplifies to:
\[ 2x_1 + x_0 - 5 = 9 \]
\[ 2x_1 + x_0 = 14 \]
Now we can express \( x_0 \) in terms of \( x_1 \):
\[ x_0 = 14 - 2x_1 \]

### Step 4: Determine Valid Values for \( x_1 \)
Since \( x_0 \) must be non-negative:
\[ 14 - 2x_1 \geq 0 \]
\[ 2x_1 \leq 14 \]
\[ x_1 \leq 7 \]

Also, from \( x_{-1} = x_1 - 5 \), we need \( x_{-1} \) to be non-negative:
\[ x_1 - 5 \geq 0 \]
\[ x_1 \geq 5 \]

Thus, \( x_1 \) can take values in the range:
\[ 5 \leq x_1 \leq 7 \]

### Step 5: Calculate Possible Combinations
Now we calculate the combinations for \( x_1 = 5, 6, 7 \):

1. **For \( x_1 = 5 \)**:
- \( x_{-1} = 0 \)
- \( x_0 = 4 \)
- The number of arrangements is given by:
\[ \frac{9!}{5! \cdot 4!} = \frac{362880}{120 \cdot 24} = 126 \]

2. **For \( x_1 = 6 \)**:
- \( x_{-1} = 1 \)
- \( x_0 = 2 \)
- The number of arrangements is:
\[ \frac{9!}{6! \cdot 2! \cdot 1!} = \frac{362880}{720 \cdot 2 \cdot 1} = 252 \]

3. **For \( x_1 = 7 \)**:
- \( x_{-1} = 2 \)
- \( x_0 = 0 \)
- The number of arrangements is:
\[ \frac{9!}{7! \cdot 0! \cdot 2!} = \frac{362880}{5040 \cdot 1 \cdot 2} = 36 \]

### Step 6: Total the Combinations
Now, we sum all the combinations:
\[ 126 + 252 + 36 = 414 \]

### Final Answer
Thus, the total number of 3x3 matrices with entries from the set {-1, 0, 1} such that the sum of all entries equals 5 is **414**.