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The probability that a randomly chosen $2 \times 2$ matrix with all the entries from the set of first 10 primes, is singular, is equal to :
Option: 1
$\frac{133}{10^{4}}$

Option: 2
$\frac{18}{10^{3}}$

Option: 3
$\frac{19}{10^{3}}$

Option: 4
$\frac{271}{10^{4}}$

Answers (1)

best_answer

Let the matrix $\mathrm{A}$ be $\mathrm{\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]}$

If $\mathrm{A}$ is singular then $\mathrm{|A|=0}$

$\mathrm{\Rightarrow a d-b c=0} \\$

$\mathrm{\Rightarrow a d=b c}$

Case I : $\mathrm{a= d}$

$\mathrm{a}$ can take $\mathrm{10}$ values.

$\mathrm{d}$ can take $\mathrm{1}$ value which equals $\mathrm{a}$

Now for $\mathrm{\text { ad }=b c}$ to hold

$\mathrm{b c=a^{2} \Rightarrow b=a, c=a( as\: a\: is\: prime)}$

$\therefore$ only one option for $\mathrm{b\: and\: c}$

Total such matrices $\mathrm{=10\times1\times1\times1=10}$

Case II : $\mathrm{a\neq d}$

In this case $\mathrm{a}$ can take $\mathrm{10}$ values and $\mathrm{d}$ can take $\mathrm{9}$ values.

Now for $\mathrm{a d=b c, \quad(b=a \text { and } c=d)}$

$\mathrm{ \text { or } \quad(b=d \text { and } c=a) }\\$

$\mathrm{\therefore \quad 10 \times 9 \times 2=180} \\$

$\mathrm{\text {favourable }=10+180=190}$

$\mathrm{Total\: matrices =10^{4}}$

$\mathrm{Probability =\frac{190}{10^{4}}=\frac{19}{10^{3}}}$

Hence answer is option 3.

Posted by

SANGALDEEP SINGH

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