Left| {,begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}{{b^2}}&{{c^2} + {a^2}}&amp

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3 and 4 .Determinants and Matrices

hard

$\left| {\,\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right| = $

$abc$

$4abc$

$4{a^2}{b^2}{c^2}$

${a^2}{b^2}{c^2}$

(IIT-1980)

Solution

(c) $\Delta = \left| {\,\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right|$

=$ – 2\,\left| {\,\begin{array}{*{20}{c}}0&{{c^2}}&{{b^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right|$,by ${R_1}→{R_1}-(R_2 + R_3)$

= $ – 2\,\left| {\,\begin{array}{*{20}{c}}0&{{c^2}}&{{b^2}}\\{{b^2}}&{{a^2}}&0\\{{c^2}}&0&{{a^2}}\end{array}\,} \right|$, by $\begin{array}{l}{R_2} \to {R_2} – {R_1}\\{R_3} \to {R_3} – {R_1}\end{array}$

= $ – 2\{ – {c^2}({b^2}{a^2}) + {b^2}( – {c^2}{a^2})\} = 4{a^2}{b^2}{c^2}$.

Trick: Put $a=1,b=2,c=3$ so that the option give different values.

Standard 12

Mathematics

નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

hard

જો $x$ એ ધન પૂર્ણાંક હોય તો $\Delta = \left| {\,\begin{array}{*{20}{c}}{x!}&{(x + 1)!}&{(x + 2)!}\\{(x + 1)!}&{(x + 2)!}&{(x + 3)!}\\{(x + 2)!}&{(x + 3)!}&{(x + 4)!}\end{array}\,} \right|$= . . .

hard

જો સમીકરણ સંહતિ $ax + y + z = 0$, $x + by + z = 0$ અને $x + y + cz = 0 $ ને શૂન્યતર ઉકેલ હોય તો $\frac{1}{{1 – a}} + \frac{1}{{1 – b}} + \frac{1}{{1 – c}} = . .. . $ (કે જ્યાં $a,b,c \ne 1$ )

medium

નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a+b)(b-c)(c-a)$

medium

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\{b + c}&{c + a}&{a + b}\\{b + c – a}&{c + a – b}&{a + b – c}\end{array}\,} \right|$ = . .

easy

$\left| {\,\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right| = $

Solution

Similar Questions

નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

જો $x$ એ ધન પૂર્ણાંક હોય તો $\Delta = \left| {\,\begin{array}{*{20}{c}}{x!}&{(x + 1)!}&{(x + 2)!}\\{(x + 1)!}&{(x + 2)!}&{(x + 3)!}\\{(x + 2)!}&{(x + 3)!}&{(x + 4)!}\end{array}\,} \right|$= . . .

જો સમીકરણ સંહતિ $ax + y + z = 0$, $x + by + z = 0$ અને $x + y + cz = 0 $ ને શૂન્યતર ઉકેલ હોય તો $\frac{1}{{1 – a}} + \frac{1}{{1 – b}} + \frac{1}{{1 – c}} = . .. . $ (કે જ્યાં $a,b,c \ne 1$ )

નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a+b)(b-c)(c-a)$

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\{b + c}&{c + a}&{a + b}\\{b + c – a}&{c + a – b}&{a + b – c}\end{array}\,} \right|$ = . .

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