નિશ્રાયક Delta = left| {,begin{array}{*{20}{c}}{{a₁}}&{{b₁}}&{{c₁}}{{a

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3 and 4 .Determinants and Matrices

easy

નિશ્રાયક $\Delta = \left| {\,\begin{array}{{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ માં જો ${A_1},{B_1},{C_1}$....એ અનુક્રમે ${a_1},{b_1},{c_1}$,......ના સહઅવયવ દર્શાવે છે તો $\left| {\begin{array}{{20}{c}}{{B_2}}&{{C_2}}\\{{B_3}}&{{C_3}}\end{array}} \right| = . . . .$

${a_1}\Delta $

${a_1}{a_3}\Delta $

$({a_1} + {b_1})\Delta $

એકપણ નહી.

Solution

(a) ${B_2} = \left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{c_1}}\\{{a_3}}&{{c_3}}\end{array}} \right| = {a_1}{c_3} – {c_1}{a_3}$

${C_2} = – \left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}\\{{a_3}}&{{b_3}}\end{array}\,} \right| = – ({a_1}{b_3} – {a_3}{b_1})$

${B_3} = – \left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{c_1}}\\{{a_2}}&{{c_2}}\end{array}\,} \right| = – ({a_1}{c_2} – {a_2}{c_1})$

${C_3} = \left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}\\{{a_2}}&{{b_2}}\end{array}\,} \right| = {a_1}{b_2} – {a_2}{b_1}$

$\left| {\,\begin{array}{*{20}{c}}{{B_2}}&{{C_2}}\\{{B_3}}&{{C_3}}\end{array}\,} \right| = \left| {\begin{array}{*{20}{c}}{{a_1}{c_3} – {a_3}{c_1}}&{ – ({a_1}{b_3} – {a_3}{b_1})}\\{ – ({a_1}{c_2} – {a_2}{c_1})}&{{a_1}{b_2} – {a_2}{b_1}}\end{array}\,} \right|$

$ = \left| {\,\begin{array}{*{20}{c}}{{a_1}{c_3}}&{ – {a_1}{b_3}}\\{ – {a_1}{c_2}}&{{a_1}{b_2}}\end{array}\,} \right| + \left| {\,\begin{array}{*{20}{c}}{{a_1}{c_3}}&{{a_3}{b_1}}\\{ – {a_1}{c_2}}&{ – {a_2}{b_1}}\end{array}\,} \right|$

$ + \left| {\,\begin{array}{*{20}{c}}{ – {a_3}{c_1}}&{ – {a_1}{b_3}}\\{\,\,\,{a_2}{c_1}}&{{a_1}{b_2}}\end{array}\,} \right| + \left| {\,\begin{array}{*{20}{c}}{ – {a_3}{c_1}}&{{a_3}{b_1}}\\{{a_2}{c_1}}&{ – {a_2}{b_1}}\end{array}\,} \right|$

$ = a_1^2({b_2}{c_3} – {b_3}{c_2}) + {a_1}{b_1}( – {c_3}{a_2} + {a_3}{c_2})$

$ + {a_1}{c_1}( – {a_3}{b_2} + {a_2}{b_3}) + {c_1}{b_1}({a_3}{a_2} – {a_2}{a_3})$

$ = {a_1}\Delta $.

Standard 12

Mathematics

જો $P = \left[ {\begin{array}{*{20}{c}}1&\alpha &3\\1&3&3\\2&4&4\end{array}} \right]$ એ $3×3 $ શ્રેણિક $A$ નો સહઅવયવજ હોય અને $ |A|=4$ તો $\alpha $ મેળવો.

medium

(JEE MAIN-2013)

$\left| {\,\begin{array}{{20}{c}}{{{\log }_3}512}&{{{\log }_4}3}\\{{{\log }_3}8}&{{{\log }_4}9}\end{array}\,} \right| \times \left| {\,\begin{array}{{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\\{{{\log }_3}4}&{{{\log }_3}4}\end{array}\,} \right|=$

medium

જો ${\Delta _1} = \left| {\begin{array}{{20}{c}}
{{b^5}{c^6}\left( {{c^3} – {b^3}} \right)}&{{a^4}{c^6}\left( {{a^3} – {c^3}} \right)}&{{a^4}{b^5}\left( {{b^3} – {a^3}} \right)} \\
{{b^2}{c^3}\left( {{b^6} – {c^6}} \right)}&{a{c^3}\left( {{c^6} – {a^6}} \right)}&{a{b^2}\left( {{a^6} – {b^6}} \right)} \\
{{b^2}{c^3}\left( {{c^3} – {b^3}} \right)}&{a{c^3}\left( {{a^3} – {c^3}} \right)}&{a{b^2}\left( {{b^3} – {a^3}} \right)}
\end{array}} \right|$ અને ${\Delta _2} = \left| {\begin{array}{{20}{c}}
a&{{b^2}}&{{c^3}} \\
{{a^4}}&{{b^5}}&{{c^6}} \\
{{a^7}}&{{b^8}}&{{c^9}}
\end{array}} \right|$ તો ${\Delta _1}{\Delta _2}$ મેળવો.

normal

બીજી હારના ઘટકોના સહઅવયવના ઉપયોગથી $\Delta=\left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|$ નું મૂલ્ય શોધો.

easy

નીચે આપેલા નિશ્ચાયકના પ્રત્યેક ઘટકના ઉપનિશ્ચાયક અને સહઅવયવ લખો : $\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$

easy

Solution

Similar Questions

જો $P = \left[ {\begin{array}{*{20}{c}}1&\alpha &3\\1&3&3\\2&4&4\end{array}} \right]$ એ $3×3 $ શ્રેણિક $A$ નો સહઅવયવજ હોય અને $ |A|=4$ તો $\alpha $ મેળવો.

$\left| {\,\begin{array}{*{20}{c}}{{{\log }_3}512}&{{{\log }_4}3}\\{{{\log }_3}8}&{{{\log }_4}9}\end{array}\,} \right| \times \left| {\,\begin{array}{*{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\\{{{\log }_3}4}&{{{\log }_3}4}\end{array}\,} \right|=$

બીજી હારના ઘટકોના સહઅવયવના ઉપયોગથી $\Delta=\left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|$ નું મૂલ્ય શોધો.

નીચે આપેલા નિશ્ચાયકના પ્રત્યેક ઘટકના ઉપનિશ્ચાયક અને સહઅવયવ લખો : $\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$

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$\left| {\,\begin{array}{{20}{c}}{{{\log }_3}512}&{{{\log }_4}3}\\{{{\log }_3}8}&{{{\log }_4}9}\end{array}\,} \right| \times \left| {\,\begin{array}{{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\\{{{\log }_3}4}&{{{\log }_3}4}\end{array}\,} \right|=$