The number of symmetric matrices of order 3, with all the entries from the set {0, 1, 2, 3, 4, 5, 6,

Q: Wait, why do we count the diagonal elements? Don't they have to be zero for a symmetric matrix?

That is a common confusion! You are likely thinking of a *skew-symmetric* matrix, where $A = -A^T$, which forces the diagonal elements to be zero. For a standard *symmetric* matrix, the condition is simply $A = A^T$, which imposes no restrictions on the diagonal elements. They are perfectly free to be any value from your set.

Q: Why is the answer $10^6$ and not $10^9$?

Great question! If the matrix were not symmetric, every one of the 9 positions in a $3 \times 3$ matrix would be independent, leading to $10^9$ possibilities. However, symmetry acts as a constraint. Because $a_{ij}$ must equal $a_{ji}$, choosing the upper triangle automatically fixes the lower triangle. We lose the freedom to choose those lower elements, reducing our total independent choices to 6.

Q: Does the set of digits $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ change the calculation?

Absolutely. The set defines the number of choices ($k$) for each independent position. Since there are 10 digits in that set, $k=10$. If the set were different—say, only $\{1, 2, 3\}$—then $k$ would be 3, and the answer would be $3^6$.

Question

The number of symmetric matrices of order 3, with all the entries from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ is

Options:

(A) $6^{10}$

(B) $10^6$

(C) $9^{10}$

(D) $10^9$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are an architect designing a structure, but instead of steel and glass, you are building with numbers. You have a $3 	imes 3$ grid, and your task is to fill it with digits from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

The structure must be symmetric. In the world of linear algebra, a symmetric matrix is one that remains unchanged when you flip it over its main diagonal. Mathematically, this is the beautiful condition:

$$A = A^T \quad 	ext{or} \quad a_{ij} = a_{ji}$$

### Visualizing the Mirror

Let us look at our $3 	imes 3$ grid. We have nine slots in total. If we were filling this grid without any rules, we would have $10^9$ possible matrices.

Symmetry acts like a mirror placed along the principal diagonal—the line running from the top-left to the bottom-right. The diagonal elements, $a_{11}, a_{22},$ and $a_{33}$, sit right on this mirror line. They do not have a reflection; they are independent. We can choose any of the 10 digits for each of these three spots.

### The Dance of the Off-Diagonals

Now, consider the elements above the diagonal: $a_{12}, a_{13},$ and $a_{23}$. These are our "free" variables. We can pick any digit for these three positions.

Because the matrix is symmetric, the elements below the diagonal—$a_{21}, a_{31},$ and $a_{32}$—are not free at all. They are slaves to the choices we made above.

If we choose $a_{12} = 5$, then $a_{21}$ *must* be 5. If we choose $a_{13} = 2$, then $a_{31}$ *must* be 2. The mirror effect forces these values upon us; we have no choice in the matter.

### Counting the Freedom

We have 3 independent choices on the diagonal and 3 independent choices in the upper triangle. That gives us a total of $3 + 3 = 6$ independent entries.

For each of these 6 positions, we have 10 possible digits to choose from. By the fundamental principle of counting, the total number of such matrices is:

$$10 	imes 10 	imes 10 	imes 10 	imes 10 	imes 10 = 10^6$$

### The Generalization

This logic is a universal truth. For any $n 	imes n$ symmetric matrix, the number of independent entries is given by the sum of the diagonal ($n$) and the upper triangle:

$$\frac{n^2 - n}{2}$$

Adding these together, we get:

$$\frac{n^2 - n}{2} + n = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}$$

For our case, $n=3$, so we have $\frac{3(4)}{2} = 6$ independent spots. If you have $k$ choices for each entry, the total number of symmetric matrices is:

$$k^{\frac{n(n+1)}{2}}$$

It is a simple, elegant result. We started with a complex-looking grid and realized that symmetry reduces the problem to its core essence. Keep this perspective in your toolkit—whenever you see symmetry in a problem, look for the "mirror" and identify what is truly independent. You have mastered the logic; the final answer is **$1,000,000$**.

Animated Solution for Mathematics - Matrices and Determinants: The number of symmetric matrices of order 3, with all the entries from the set ${0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$ is

Select Answer:

Visualized Solution

The Problem Setup

What is a Symmetric Matrix?

The Principal Diagonal

Upper Triangular Elements

The Mirror Effect

Filling the Lower Triangle

Counting Independent Entries

Choices per Entry

Total Number of Matrices

General Formula

Final Conclusion

The Sigma Insight: Types of Matrices

Analyzing the Setup

Visualizing the Mirror

The Dance of the Off-Diagonals

Counting the Freedom

The Generalization

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of symmetric matrices of order 3, with all the entries from the set {0,1,2,3,4,5,6,7,8,9} is