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Matrices and Determinants question

2022 · 27 Jul · Shift 1 · Q44

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Matrices and Determinants
2022 · 27 Jul · Shift 1 · Q44

Matrices and Determinants question

2022 · 27 Jul · Shift 1 · Q44

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1

Let

S

be the set containing all

3 \times 3

matrices with entries from

\{-1,0,1\}

. The total number of matrices

A \in S

such that the sum of all the diagonal elements of

A^{\mathrm{T}} A

is 6 is

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 5376

Interpret the condition

For any matrix $A=(a_{ij})$ , the matrix $A^{\mathrm T}A$ has diagonal entries

(A^{\mathrm T}A)_{jj} = \sum_{i=1}^3 a_{ij}^2.

So the sum of all diagonal elements of $A^{\mathrm T}A$ is

\operatorname{tr}(A^{\mathrm T}A)=\sum_{i=1}^3\sum_{j=1}^3 a_{ij}^2.

This is just the sum of squares of all entries of $A$ .

Since each entry of $A$ belongs to $\{-1,0,1\}$ , we have

a_{ij}^2 = \begin{cases} 1,& a_{ij}=\pm 1,\\ 0,& a_{ij}=0. \end{cases}

Hence,

\operatorname{tr}(A^{\mathrm T}A)=6

means exactly that among the 9 entries of $A$ , exactly 6 are nonzero (i.e. equal to $\pm1$ ), and the remaining 3 are zero.

Count ways to choose positions of the 6 nonzero entries

We must choose 6 positions out of 9 for the entries that are $\pm1$ :

\binom{9}{6} = 84.

Assign signs to the nonzero entries

Each of the 6 chosen nonzero entries can independently be either $1$ or $-1$ . So the number of sign assignments is

2^6 = 64.

Total number of matrices

Therefore the required number of matrices is

\binom{9}{6}\cdot 2^6 = 84\cdot 64.

Now,

84\cdot 64 = 5376.

Comparison with stored answer

Our derived answer is

5376.

This matches the stored correct answer.

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