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3 and 4 .Determinants and Matrices
normal
$\left| {\,\begin{array}{*{20}{c}}{41}&{42}&{43}\\{44}&{45}&{46}\\{47}&{48}&{49}\end{array}\,} \right| = $
A
$2$
B
$4$
C
$0$
D
$1$
Solution
(c)Given, $\Delta = \left| {\,\begin{array}{*{20}{c}}{41}&{42}&{43}\\{44}&{45}&{46}\\{47}&{48}&{49}\end{array}\,} \right|.$
Applying ${C_2} \to {C_2} – {C_1},\,{C_3} \to {C_3} – {C_2}$ we get $\Delta = \left| {\,\begin{array}{*{20}{c}}{41}&1&1\\{44}&1&1\\{47}&1&1\end{array}\,} \right|$. Since two columns (${C_2}$ and ${C_3}$) are identical, therefore $\Delta = 0$.
Standard 12
Mathematics
