ધારો કે A = left[ {begin{array}{*{20}{c}}1&0& | vedclass

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(D) આપેલ છે કે $A = \left[ {\begin{array}{*{20}{c}}1&0&0\2&1&0\3&2&1\end{array}} ight]$,$A{u_1} = \left[ {\begin{array}{*{20}{c}}1\0\0\end{array

Vedclass · Accepted Answer

$\left[ {\begin{array}{*{20}{c}}1\{ - 1}\{ - 1}\end{array}} ight]$

Vedclass · Answer

(D) આપેલ છે કે $A = \left[ {\begin{array}{*{20}{c}}1&0&0\2&1&0\3&2&1\end{array}} ight]$,$A{u_1} = \left[ {\begin{array}{*{20}{c}}1\0\0\end{array}} ight]$,અને $A{u_2} = \left[ {\begin{array}{*{20}{c}}0\1\0\end{array}} ight]$.બંને સમીકરણોનો સરવાળો કરતા:$A{u_1} + A{u_2} = \left[ {\begin{array}{*{20}{c}}1\0\0\end{array}} ight] + \left[ {\begin{array}{*{20}{c}}0\1\0\end{array}} ight]$$A(u_1 + u_2) = \left[ {\begin{array}{*{20}{c}}1\1\0\end{array}} ight]$તેથી,$u_1 + u_2 = A^{-1} \left[ {\begin{array}{*{20}{c}}1\1\0\end{array}} ight]$.પ્રથમ,$|A| = 1(1-0) - 0 + 0 = 1$ શોધો.ત્યારબાદ,$adj(A)$ શોધો. સહ-અવયવો નીચે મુજબ છે:$C_{11} = 1, C_{12} = -2, C_{13} = 1$$C_{21} = 0, C_{22} = 1, C_{23} = -2$$C_{31} = 0, C_{32} = 0, C_{33} = 1$તેથી,$adj(A) = \left[ {\begin{array}{*{20}{c}}1&0&0\-2&1&0\1&-2&1\end{array}} ight]$.કારણ કે $|A| = 1$,તેથી $A^{-1} = adj(A)$.$u_1 + u_2 = \left[ {\begin{array}{*{20}{c}}1&0&0\-2&1&0\1&-2&1\end{array}} ight] \left[ {\begin{array}{*{20}{c}}1\1\0\end{array}} ight] = \left[ {\begin{array}{*{20}{c}}1(1)+0(1)+0(0)\-2(1)+1(1)+0(0)\1(1)-2(1)+1(0)\end{array}} ight] = \left[ {\begin{array}{*{20}{c}}1\-1\-1\end{array}} ight]$.

Similar Questions

ધારો કે $A$ એ $2 \times 2$ શ્રેણિક છે.
$\text{વિધાન}-1: adj(adj A) = A$
$\text{વિધાન}-2: |adj A| = |A|$

જો $A = \begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix}$ હોય,તો $2A - 3A^{-1} = $

શ્રેણિક $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ નો વ્યસ્ત શ્રેણિક શોધો.

જો $A$ એક એવો ચોરસ શ્રેણિક હોય કે જેથી $A(\operatorname{adj} A) = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}$ થાય,તો $\operatorname{det}(\operatorname{adj} A)$ ની કિંમત શોધો.

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