Let A be a 3 x 3 matrix having entries from the set {-1, 0, 1}. The number of all such matrices A ha

Q: Wait, why can't we just use stars and bars to solve this?

Stars and bars is a technique for distributing identical items into distinct bins where order doesn't matter. Here, the matrix slots are distinct, and the arrangement of the numbers matters—a matrix with a $1$ in the top-left is different from one with a $1$ in the bottom-right. This is a permutation problem with identical items, which is why we use the multinomial coefficient $\frac{9!}{x!y!z!}$.

Q: Why are there only three cases for $z$?

Great question! We derived the constraint $2z+y=4$. Since $y$ represents a count of zeros, it must be greater than or equal to $0$. This implies $2z \le 4$, or $z \le 2$. Since $z$ must also be a non-negative integer, the only possible values for $z$ are $0, 1,$ and $2$.

Q: What if the sum of the entries was negative, like $-5$?

The beauty of this problem lies in its symmetry! If the sum were $-5$, you would simply swap the roles of $1$ and $-1$. The number of $1$s ($x$) would become the number of $-1$s ($z$), and vice versa. The total number of matrices would remain exactly the same.

Question

Let $A$ be a $3 	imes 3$ 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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a $3 	imes 3$ grid, a blank canvas of nine empty cells. You are tasked with filling these cells using only the values $\{-1, 0, 1\}$, with the strict condition that the sum of all nine entries must equal exactly $5$.

To solve this, we translate the physical grid into the language of algebra. Let $x$ be the number of times $1$ appears, $y$ be the number of times $0$ appears, and $z$ be the number of times $-1$ appears.

Since there are $9$ total slots, we have our first constraint:
$$x + y + z = 9$$

The sum of the entries gives us our second constraint:
$$1(x) + 0(y) + (-1)(z) = 5$$
This simplifies elegantly to $x - z = 5$.

### The Algebraic Reduction

We now have a system of two equations with three variables. We want to reduce this to a single, manageable constraint.

From $x - z = 5$, we know that $x = z + 5$. Substituting this into our first equation, we get:
$$(z + 5) + y + z = 9$$

Simplifying this, we arrive at the heartbeat of our problem:
$$2z + y = 4$$

Because $y$ and $z$ represent counts, they must be non-negative integers. This equation dictates that $2z \le 4$, which means $z$ can only be $0, 1,$ or $2$. We have reduced the problem into three distinct, solvable cases.

### The Case Analysis

**Case 1:** If $z = 0$, then $y = 4 - 2(0) = 4$. Consequently, $x = 0 + 5 = 5$.

We are arranging $5$ ones, $4$ zeros, and $0$ minus ones. The number of ways to arrange these is given by the multinomial coefficient:
$$\frac{9!}{5!4!0!} = 126$$

**Case 2:** If $z = 1$, then $y = 4 - 2(1) = 2$. Consequently, $x = 1 + 5 = 6$.

We are arranging $6$ ones, $2$ zeros, and $1$ minus one. The number of ways is:
$$\frac{9!}{6!2!1!} = 252$$

**Case 3:** If $z = 2$, then $y = 4 - 2(2) = 0$. Consequently, $x = 2 + 5 = 7$.

We are arranging $7$ ones, $0$ zeros, and $2$ minus ones. The number of ways is:
$$\frac{9!}{7!0!2!} = 36$$

### The Grand Total

We have methodically dismantled the problem. We found $126$ matrices for the first case, $252$ for the second, and $36$ for the third.

Since these cases are mutually exclusive, we simply sum them up:
$$126 + 252 + 36 = 414$$

There are exactly **414** such matrices. This problem teaches us that even when faced with a complex grid, the right choice of variables and a systematic reduction of constraints can turn a daunting task into a series of simple, satisfying steps.

Animated Solution for Mathematics - Permutations and Combinations: Let $A$ be a $3 \times 3$ 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

Enter Numerical Value:

Visualized Solution

Matrix Structure and Constraints

Defining Variables

Formulating Equations

Reducing to a Single Constraint

Finding Possible Values for $z$

Case 1: $z = 0$

Case 2: $z = 1$

Case 3: $z = 2$

Total Number of Matrices

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Algebraic Reduction

The Case Analysis

The Grand Total

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