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Probability Part 2

Section: Mathematics

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Question : 1 of 100

Marks: +1, -0

Let

A = [a_{i j}]

be a square matrix of order 2 with entries either 0 or 1 . Let

E

be the event that

A

is an invertible matrix. Then the probability

P(E)

[24 Jan 2025 Shift 2]

$\; \frac{3}{16}$
$\; \frac{3}{8}$
$\; \frac{1}{8}$
$\; \frac{5}{8}$

Solution:

A = [a_{i j}] 2 \times 2

and entries are 0 or 1.

\; \therefore \begin{vmatrix} a & b \\ c & d \end{vmatrix} = 0

\Rightarrow a d - b c = 0

Case I:

a d = b c = 1

\therefore \;\; a = b = c = d = 1

Case II:

a d = b c = 0

a = 0, d = 0 \;\;\;\;\; b = 0, c = 0

a = 0, d = 1 \;\;\;\; b = 0, c = 1

a = 1, d = 0 \;\;\;\;\; b = 1, c = 0

∴ Total 10 cases when matrix is non invertible

Total possible matrix

= 2^{4} = 16

Required probability of invertible

= \frac{16-10}{16} = \frac{6}{16} = \frac{3}{8}

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