The number of 3 × 2 matrices A, which can be formed using the elements of the set {-2, -1, 0, 1, 2}

Q: Wait, why is the trace of $A^T A$ equal to the sum of squares of all elements of $A$?

That is a fantastic question! Let $A$ be a $3 \times 2$ matrix with elements $a_{ij}$. The matrix $A^T$ is $2 \times 3$. The product $A^T A$ is a $2 \times 2$ matrix. The diagonal elements of $A^T A$ are the dot product of the columns of $A$ with themselves. Specifically, the first diagonal element is $\sum_{i=1}^3 a_{i1}^2$ and the second is $\sum_{i=1}^3 a_{i2}^2$. Adding these together gives $\sum_{i=1}^3 a_{i1}^2 + \sum_{i=1}^3 a_{i2}^2$, which is exactly the sum of the squares of all six elements of $A$. It is a beautiful property of matrix multiplication!

Q: Why do we treat the sign variations separately from the arrangement of squares?

Think of it as a two-step process. First, we decide which squares go into which of the six positions in the matrix. This is a combinatorial problem of placing values like 4, 1, or 0. Once the squares are fixed, we then decide the actual value of the element. For a square of 4, the element could be 2 or -2. For a square of 1, it could be 1 or -1. By separating these, we ensure we do not double-count or miss any possibilities. It makes the counting systematic and error-free.

Q: Are there any other combinations of squares that sum to 5?

To be sure, let's check. We need six numbers from $\{0, 1, 4\}$ that sum to 5. If we use two 4s, the sum is at least 8, which is too much. If we use no 4s, we must use five 1s and one 0 to get 5. If we use one 4, we need the remaining five numbers to sum to 1, which means one 1 and four 0s. These are the only two partitions of 5 into six parts using the set $\{0, 1, 4\}$. You have covered all the ground!

Question

The number of $3 	imes 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a $3 	imes 2$ matrix $A$. The problem requires us to find the number of such matrices where the sum of the diagonal elements of $A^T A$ is exactly $5$.

The expression $A^T A$ results in a $2 	imes 2$ matrix. The diagonal elements of this product are the sums of the squares of the elements in each column of $A$.

Specifically, if $A = [a_{ij}]$, the trace of $A^T A$ is given by:

$$	ext{Tr}(A^T A) = \sum_{i=1}^{3} a_{i1}^2 + \sum_{i=1}^{3} a_{i2}^2 = \sum_{i=1}^{3} \sum_{j=1}^{2} a_{ij}^2$$

Thus, the condition simplifies to the requirement that the sum of the squares of all six elements of $A$ must equal $5$.

### The Combinatorial Hunt

The elements $a_{ij}$ are chosen from the set $\{-2, -1, 0, 1, 2\}$. Squaring these values, we obtain the set of possible squares: $\{0, 1, 4\}$.

We must find combinations of six squares from this set that sum to $5$. We identify two distinct cases:

**Case 1:** We use one $4$, one $1$, and four $0$s. The sum is $4 + 1 + 0 + 0 + 0 + 0 = 5$.

**Case 2:** We use five $1$s and one $0$. The sum is $1 + 1 + 1 + 1 + 1 + 0 = 5$.

No other combinations are possible, as two $4$s would exceed the sum of $5$, and any other combination of the available squares fails to reach the target sum.

### The Art of Arrangement

Now, we calculate the number of matrices for each case by considering both the placement of the values and their possible signs.

**For Case 1:**
We have $6$ positions to fill. We choose one position for the $4$ in $\binom{6}{1} = 6$ ways and one position for the $1$ in $\binom{5}{1} = 5$ ways. The remaining four positions are filled with $0$.

The value $4$ can be represented by $2$ or $-2$ ($2$ choices), and the value $1$ can be represented by $1$ or $-1$ ($2$ choices). The total number of matrices for Case 1 is:

$$6 	imes 5 	imes 2 	imes 2 = 120$$

**For Case 2:**
We have $6$ positions and must choose one position for the $0$ in $\binom{6}{1} = 6$ ways. The remaining five positions are filled with $1$s.

Each of the five $1$s can be either $1$ or $-1$, providing $2^5 = 32$ sign variations. The total number of matrices for Case 2 is:

$$6 	imes 32 = 192$$

### Final Calculation

To find the total number of matrices satisfying the condition, we sum the results from both cases:

$$	ext{Total} = 120 + 192 = 312$$

The total number of such matrices is **312**.

Animated Solution for Mathematics - Matrices and Determinants: The number of $3 \times 2$ matrices A, which can be formed using the elements of the set ${- 2, - 1, 0, 1, 2}$ such that the sum of all the diagonal elements of $A^{T} A$ is 5, is .........

Enter Numerical Value:

Visualized Solution

Visualizing Matrix $A$

Understanding $A^{T} A$

The Trace Property

Possible Element Values

Identifying Case 1

Case 1: Position Selection

Case 1: Sign Variations

Identifying Case 2

Case 2: Position Selection

Case 2: Sign Variations

Final Summation

The Sigma Insight: Types of Matrices

Analyzing the Setup

The Combinatorial Hunt

The Art of Arrangement

Final Calculation

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