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3 and 4 .Determinants and Matrices
easy
$\left| {\,\begin{array}{*{20}{c}}{{1^2}}&{{2^2}}&{{3^2}}\\{{2^2}}&{{3^2}}&{{4^2}}\\{{3^2}}&{{4^2}}&{{5^2}}\end{array}\,} \right|$=
A
$8$
B
$-8$
C
$400$
D
$1$
Solution
(b) $\left| {\,\begin{array}{*{20}{c}}{{1^2}}&{{2^2}}&{{3^2}}\\{{2^2}}&{{3^2}}&{{4^2}}\\{{3^2}}&{{4^2}}&{{5^2}}\end{array}\,} \right|${Operate ${R_3}\, \to {R_3} – {R_2}$,${R_2} \to {R_2} – {R_1}$}
= $\left| {\,\begin{array}{*{20}{c}}1&4&9\\3&5&7\\5&7&9\end{array}\,} \right|$$ = 1(45 – 49) – 4\,(27 – 35) + 9\,(21 – 25)$
= $ – 4 + 32 – 36 = – 8$.
Standard 12
Mathematics
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