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A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The…
A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the determinant chosen is non-zero is .........
$\frac{4}{8}$ $\frac{3}{8}$ $\frac{2}{8}$ $\frac{5}{8}$ Solution
Total possible $2 \times 2$ matrices with only elements 0 and 1 are
$
\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 0 \\
1 & 1
\end{array}\right]
$
$
\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 1
\end{array}\right]
$
$
\left[\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right]
$
$
\left[\begin{array}{ll}
0 & 1 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right]
$
$\therefore$ Total possible number of matrices $=16$
Number of matrices whose determinant non zero $=6$
$\therefore$ Required probability $=\frac{6}{16}=\frac{3}{8}$
Hence, option (2) is correct
Asked in: AP EAMCET 2020 (22 Sep Shift 2)
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