નીચે આપેલા નિશ્ચાયકના ઘટકોના ઉપનિશ્ચાયક (Minors) અને સહઅવયવ (Cofactors) લખો: $\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$
(N/A) આપેલ નિશ્ચાયક $A = \left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$ છે.
ઉપનિશ્ચાયક $(M_{ij})$ અને સહઅવયવ $(A_{ij})$ ની વ્યાખ્યા મુજબ,આપણે દરેક ઘટક $a_{ij}$ માટે તેની ગણતરી કરીએ છીએ:
$M_{11} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1, A_{11} = (-1)^{1+1} M_{11} = 1$
$M_{12} = \left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right| = 0, A_{12} = (-1)^{1+2} M_{12} = 0$
$M_{13} = \left|\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right| = 0, A_{13} = (-1)^{1+3} M_{13} = 0$
$M_{21} = \left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right| = 0, A_{21} = (-1)^{2+1} M_{21} = 0$
$M_{22} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1, A_{22} = (-1)^{2+2} M_{22} = 1$
$M_{23} = \left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right| = 0, A_{23} = (-1)^{2+3} M_{23} = 0$
$M_{31} = \left|\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right| = 0, A_{31} = (-1)^{3+1} M_{31} = 0$
$M_{32} = \left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right| = 0, A_{32} = (-1)^{3+2} M_{32} = 0$
$M_{33} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1, A_{33} = (-1)^{3+3} M_{33} = 1$
ઉપનિશ્ચાયક $(M_{ij})$ અને સહઅવયવ $(A_{ij})$ ની વ્યાખ્યા મુજબ,આપણે દરેક ઘટક $a_{ij}$ માટે તેની ગણતરી કરીએ છીએ:
$M_{11} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1, A_{11} = (-1)^{1+1} M_{11} = 1$
$M_{12} = \left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right| = 0, A_{12} = (-1)^{1+2} M_{12} = 0$
$M_{13} = \left|\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right| = 0, A_{13} = (-1)^{1+3} M_{13} = 0$
$M_{21} = \left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right| = 0, A_{21} = (-1)^{2+1} M_{21} = 0$
$M_{22} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1, A_{22} = (-1)^{2+2} M_{22} = 1$
$M_{23} = \left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right| = 0, A_{23} = (-1)^{2+3} M_{23} = 0$
$M_{31} = \left|\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right| = 0, A_{31} = (-1)^{3+1} M_{31} = 0$
$M_{32} = \left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right| = 0, A_{32} = (-1)^{3+2} M_{32} = 0$
$M_{33} = \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| = 1, A_{33} = (-1)^{3+3} M_{33} = 1$
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Similar Questions
શ્રેણિક $A = \left[\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 4 & 2 \\ 3 & -4 & 6 \end{array}\right]$ ના નીચેના ઘટકોને તેમના સહ-અવયવો (co-factors) સાથે જોડો અને સાચો જવાબ પસંદ કરો.
ઘટક સહ-અવયવ $A$. $-1$ $(1)$ $-2$ $B$. $1$ $(2)$ $32$ $C$. $3$ $(3)$ $4$ $D$. $6$ $(4)$ $6$ $(5)$ $-6$
| ઘટક | સહ-અવયવ |
| $A$. $-1$ | $(1)$ $-2$ |
| $B$. $1$ | $(2)$ $32$ |
| $C$. $3$ | $(3)$ $4$ |
| $D$. $6$ | $(4)$ $6$ |
| $(5)$ $-6$ |
જો ${A_1}, {B_1}, {C_1}, \dots$ એ નિશ્ચાયક $\Delta = \begin{vmatrix} {a_1} & {b_1} & {c_1} \\ {a_2} & {b_2} & {c_2} \\ {a_3} & {b_3} & {c_3} \end{vmatrix}$ ના ઘટકો ${a_1}, {b_1}, {c_1}, \dots$ ના સહ-અવયવો (co-factors) હોય,તો $\begin{vmatrix} {B_2} & {C_2} \\ {B_3} & {C_3} \end{vmatrix} = $
Easy
View Solutionનીચે આપેલા નિશ્ચાયકના ઘટકોના ઉપનિશ્ચાયક (Minors) અને સહઅવયવ (Cofactors) લખો: $\left|\begin{array}{rr}2 & -4 \\ 0 & 3\end{array}\right|$
Easy
View Solutionજો $A = [a_{ij}]_{3 \times 3} = \begin{bmatrix} 3 & 2 & 4 \\ 1 & 4 & 1 \\ 2 & 6 & 3 \end{bmatrix}$ અને $A_{ij}$ એ $a_{ij}$ નો સહઅવયવ (cofactor) હોય,તો $a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}$ નું મૂલ્ય કેટલું થાય?
નિશ્ચાયક $\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$ માં ઘટક $6$ નો ઉપનિશ્ચાયક શોધો.
Easy
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