સમીકરણોની સિસ્ટમ ઉકેલવા માટે ગુણાકાર left[beg | vedclass

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(A) ધારો કે $A = \left[\begin{array}{ccc}1 & -1 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4\end{array}ight]$ અને $B = \left[\begin{array}{ccc}-2 & 0 & 1 \ 9 &

Vedclass · Accepted Answer

$x=0, y=5, z=3$

Vedclass · Answer

(A) ધારો કે $A = \left[\begin{array}{ccc}1 & -1 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4\end{array}ight]$ અને $B = \left[\begin{array}{ccc}-2 & 0 & 1 \ 9 & 2 & -3 \ 6 & 1 & -2\end{array}ight]$.પ્રથમ,આપણે ગુણાકાર $AB$ શોધીએ:$AB = \left[\begin{array}{ccc}1(-2)+(-1)(9)+2(6) & 1(0)+(-1)(2)+2(1) & 1(1)+(-1)(-3)+2(-2) \ 0(-2)+2(9)+(-3)(6) & 0(0)+2(2)+(-3)(1) & 0(1)+2(-3)+(-3)(-2) \ 3(-2)+(-2)(9)+4(6) & 3(0)+(-2)(2)+4(1) & 3(1)+(-2)(-3)+4(-2)\end{array}ight]$$= \left[\begin{array}{ccc}-2-9+12 & 0-2+2 & 1+3-4 \ 0+18-18 & 0+4-3 & 0-6+6 \ -6-18+24 & 0-4+4 & 3+6-8\end{array}ight] = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight] = I$.કારણ કે $AB = I$,તેથી $A^{-1} = B = \left[\begin{array}{ccc}-2 & 0 & 1 \ 9 & 2 & -3 \ 6 & 1 & -2\end{array}ight]$.સમીકરણોની સિસ્ટમને $AX = C$ તરીકે લખી શકાય,જ્યાં $X = \left[\begin{array}{c}x \ y \ z\end{array}ight]$ અને $C = \left[\begin{array}{c}1 \ 1 \ 2\end{array}ight]$.તેથી $X = A^{-1}C = \left[\begin{array}{ccc}-2 & 0 & 1 \ 9 & 2 & -3 \ 6 & 1 & -2\end{array}ight] \left[\begin{array}{c}1 \ 1 \ 2\end{array}ight]$.$X = \left[\begin{array}{c}-2(1)+0(1)+1(2) \ 9(1)+2(1)+(-3)(2) \ 6(1)+1(1)+(-2)(2)\end{array}ight] = \left[\begin{array}{c}-2+0+2 \ 9+2-6 \ 6+1-4\end{array}ight] = \left[\begin{array}{c}0 \ 5 \ 3\end{array}ight]$.આમ,$x=0, y=5, z=3$.

Similar Questions

સમીકરણો $x-y+2z=4$,$3x+y+4z=6$ અને $x+y+z=1$ ના

જો $x=\alpha, y=\beta, z=\gamma$ એ સમીકરણ સંહતિ:
$\begin{aligned} 2x-y+8z &= 13 \\ 3x+4y+5z &= 18 \\ 5x-2y+7z &= 20 \end{aligned}$
નો ઉકેલ હોય,તો $\alpha\beta+\beta\gamma+\gamma\alpha=$ શોધો.

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