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Matrices and Determinants question

2023 · 13 Apr · Shift 1 · Q37

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2023 · 13 Apr · Shift 1 · Q37

Matrices and Determinants question

2023 · 13 Apr · Shift 1 · Q37

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1

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 :

A
$10^{9}$
B
$9^{10}$
C
$10^{6}$
D
$6^{10}$

View written solutionFree

Correct answer: C

A symmetric matrix $A$ of order $3$ has the form

A=\begin{pmatrix} a & b & c\\ b & d & e\\ c & e & f \end{pmatrix}

Because the matrix is symmetric, we must have $a_{ij}=a_{ji}.$

Count the independent entries.

For a $3\times 3$ symmetric matrix, the entries are determined by:

$3$ diagonal entries: $a,d,f$
$3$ entries above the diagonal: $b,c,e$

The entries below the diagonal are then fixed by symmetry.

So total number of independent choices is $3+3=6.$

Each independent entry can be chosen from the set $\{0,1,2,3,4,5,6,7,8,9\},$ which has $10$ elements.

Hence total number of symmetric matrices is $10^6.$

Compare with options:

A: $10^9$
B: $9^{10}$
C: $10^6$
D: $6^{10}$

Therefore, the correct option is $\boxed{\text{C }(10^6)}.$

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