નિશ્ચાયક left|begin{array}{ccc}2 & -3 & 5 6 & 0 & 4 1 & 5 &

Hindi

3 and 4 .Determinants and Matrices

medium

નિશ્ચાયક $\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$ ના ઘટકોના ઉપનિશ્ચાયક અને સહઅવયવ શોધો તથા ચકાસો કે $a_{11} A_{31}+a_{12} A_{32}+a_{13} A_{33}=0$

$0$

Solution

we have $\mathrm{M}_{11}=\left|\begin{array}{cc}0 & 4 \\ 5 & -7\end{array}\right|=0-20=-20 ; \mathrm{A}_{11}=(-1)^{1+1}(-20)=-20$

$\mathrm{M}_{12}=\left|\begin{array}{cc}6 & 4 \\ 1 & -7\end{array}\right|=-42-4=-46 ; \quad \mathrm{A}_{12}=(-1)^{1+2}(-46)=46$

$\mathrm{M}_{13}=\left|\begin{array}{ll}6 & 0 \\ 1 & 5\end{array}\right|=30-0=30 ; \quad \mathrm{A}_{13}=(-1)^{1+3}(30)=30$

$\mathrm{M}_{21}=\left|\begin{array}{cc}-3 & 5 \\ 5 & -7\end{array}\right|=21-25=-4 ; \quad \mathrm{A}_{21}=(-1)^{2+1}(-4)=4$
$\mathrm{M}_{22}=\left|\begin{array}{cc}2 & 5 \\ 1 & -7\end{array}\right|=-14-5=-19 ; \quad \mathrm{A}_{22}= (-1)^{2+2}(-19)=-19$

$\mathrm{M}_{23}=\left|\begin{array}{cc}2 & -3 \\ 1 & 5\end{array}\right|=10+3=13 ; \quad \quad \mathrm{A}_{23}=(-1)^{2 * 3}(13)=-13$

$\mathrm{M}_{31}=\left|\begin{array}{cc}-3 & 5 \\ 0 & 4\end{array}\right|=-12-0=-12 ; \quad \mathrm{A}_{31}=(-1)^{3+1}(-12)=-12$

$M_{32}=\left|\begin{array}{ll}2 & 5 \\ 6 & 4\end{array}\right|=8-30=-22 ; \quad \quad A_{32}=(-1)^{3+2}(-22)=22$

and $\mathrm{M}_{33}=\left|\begin{array}{cc}2 & -3 \\ 6 & 0\end{array}\right|=0+18=18 ; \quad \quad \mathrm{A}_{33}=(-1)^{3+3}(18)=18$

Now $\quad a_{11}=2, a_{12}=-3, a_{13}=5 ; \mathrm{A}_{31}=-12, \mathrm{A}_{32}=22, \mathrm{A}_{33}=18$

So $\quad a_{11} A_{31}+a_{12} A_{32}+a_{13} A_{33}$

$=2(-12)+(-3)(22)+5(18)=-24-66+90=0$

Standard 12

Mathematics

નિશ્ચાયક $\left|\begin{array}{rr}1 & -2 \\ 4 & 3\end{array}\right|$ ના બધા જ ઘટકના ઉપનિશ્ચાયક અને સહઅવયવ શોધો.

easy

જો $\left| {\begin{array}{*{20}{c}}
{{{\left( {a – x} \right)}^2}}&{{{\left( {a – y} \right)}^2}}&{{{\left( {a – z} \right)}^2}} \\
{{{\left( {b – x} \right)}^2}}&{{{\left( {b – y} \right)}^2}}&{{{\left( {b – z} \right)}^2}} \\
{{{\left( {c – x} \right)}^2}}&{{{\left( {c – y} \right)}^2}}&{{{\left( {c – z} \right)}^2}}
\end{array}} \right| = \frac{{ – 351}}{8}$ અને $x, y , z$ એ સમીકરણ $8t^3 – 62t^2 + 43t -7 = 0$ ના બીજ હોય અને ઉપરોક્ત નિશ્ચયકનું પાલન કરે છે અને $a, b, c$ એ ભિન્ન સંખ્યા હોય તો $|(a – b) (b – c) (c – a)|$ મેળવો.

normal

જો $A = {[{a_{ij}}]_{n \times n}}$ એ ચોરસ શ્રેણિક છે અને ${c_{ij}}$ એ ${a_{ij}}$ નો સહઅવય છે . જો $C = [{c_{ij}}]$ તો . . . .

normal

જો $\Delta=\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|$ અને $a_{i j}$ નો સહઅવયવ $\mathrm{A}_{i j}$ હોય, તો $\Delta$ નું મૂલ્ય ……… .

easy

જો નિશ્રાયક $\Delta = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}} \right|$, ${A_1},{B_1},{C_1}$ એ ${a_1},{b_1},{c_1}$ ના સહઅવયવ દર્શાવે છે તો આપેલ પૈકી . . . . સંબંધ અસત્ય છે.

easy