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3 and 4 .Determinants and Matrices
medium
$\left| {\,\begin{array}{*{20}{c}}a&{a + b}&{a + 2b}\\{a + 2b}&a&{a + b}\\{a + b}&{a + 2b}&a\end{array}\,} \right|$ =
A
$9{a^2}(a + b)$
B
$9{b^2}(a + b)$
C
${a^2}(a + b)$
D
${b^2}(a + b)$
Solution
(b) संक्रिया ${C_1} \to {C_1} + {C_2} + {C_3}$.से $\left| {\,\begin{array}{*{20}{c}}{3a + 3b}&{a + b}&{a + 2b}\\{3a + 3b}&a&{a + b}\\{3a + 3b}&{a + 2b}&a\end{array}\,} \right|$
= $3\,(a + b)\,\left| {\,\begin{array}{*{20}{c}}1&{a + b}&{a + 2b}\\1&a&{a + b}\\1&{a + 2b}&a\end{array}\,} \right|$
अब संक्रिया ${R_3} \to {R_3} – {R_1}$, ${R_2} \to {R_2} – {R_1}$
= $3\,(a + b)\,\left| {\,\begin{array}{*{20}{c}}1&{a + b}&{a + 2b}\\0&{ – b}&{ – b}\\0&b&{ – 2b}\end{array}\,} \right|$
$ = 3(a + b)\,(2{b^2} + {b^2}) = 9{b^2}(a + b)$.
Standard 12
Mathematics