How many 3 × 3 matrices M with entries from { 0 , 1 , 2 } are there, for which the sum of the diagonal entries of M T M is 5? [2017]

Question

How many $3 \times 3$ matrices $M$ with entries from ${0, 1, 2}$ are there, for which the sum of the diagonal entries of $M^{T} M$ is 5? [2017]

Smart Achievers · Accepted Answer

(2)

$Let M = [\begin{matrix} a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9} \end{matrix}],$ $where a_{i} \in {0, 1, 2}$

$Then M^{T} M = [\begin{matrix} a_{1} & a_{4} & a_{7} \\ a_{2} & a_{5} & a_{8} \\ a_{3} & a_{6} & a_{9} \end{matrix}] [\begin{matrix} a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9} \end{matrix}]$

$Sum of the diagonal entries in M^{T} M = 5$

$\Rightarrow (a_{1}^{2} + a_{4}^{2} + a_{7}^{2}) + (a_{2}^{2} + a_{5}^{2} + a_{8}^{2}) + (a_{3}^{2} + a_{6}^{2} + a_{9}^{2}) = 5,$

$which is possible when$

$Case I: Five$ $a_{i}' s$ $are$ 1 $and four$ $a_{i}' s$ $are zero.$

$Which can be done in$ ${}^{9}C_{4} ways$

$= \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$

$Case II: One$ $a_{i}$ $is$ $1, one$ $a_{i}$ $is 2 and the remaining seven$ $a_{i}' s$ $are zero$

$It can be done in$ ${}^{9}C_{1} \times {}^{8}C_{1} = 9 \times 8 = 72 ways$

$∴$ $Total number of ways = 126 + 72 = 198$

Solution

Q.
How many $3 \times 3$ matrices $M$ with entries from ${0, 1, 2}$ are there, for which the sum of the diagonal entries of $M^{T} M$ is 5 [2017]

1 126

2 198

3 162

4 135

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