The sum of three numbers is 6. If we multiply | vedclass

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(C) Let the first,second,and third numbers be $x, y,$ and $z$ respectively.According to the given conditions,we have the following system of linear eq

Vedclass · Accepted Answer

$x=1, y=2, z=3$

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(C) Let the first,second,and third numbers be $x, y,$ and $z$ respectively.According to the given conditions,we have the following system of linear equations:$x + y + z = 6$$y + 3z = 11$$x + z = 2y \implies x - 2y + z = 0$This system can be written as $AX = B$,where$A = \begin{bmatrix} 1 & 1 & 1 \ 0 & 1 & 3 \ 1 & -2 & 1 \end{bmatrix}, X = \begin{bmatrix} x \ y \ z \end{bmatrix}, B = \begin{bmatrix} 6 \ 11 \ 0 \end{bmatrix}$First,we calculate the determinant $|A|$:$|A| = 1(1 - (-6)) - 1(0 - 3) + 1(0 - 1) = 1(7) + 3 - 1 = 9 
eq 0$.Since $|A| 
eq 0$,the system has a unique solution $X = A^{-1}B$.Calculating the cofactor matrix $C$:$C_{11} = (1 - (-6)) = 7, C_{12} = -(0 - 3) = 3, C_{13} = (0 - 1) = -1$$C_{21} = -(1 - (-2)) = -3, C_{22} = (1 - 1) = 0, C_{23} = -(-2 - 1) = 3$$C_{31} = (3 - 1) = 2, C_{32} = -(3 - 0) = -3, C_{33} = (1 - 0) = 1$$adj(A) = C^T = \begin{bmatrix} 7 & -3 & 2 \ 3 & 0 & -3 \ -1 & 3 & 1 \end{bmatrix}$$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{9} \begin{bmatrix} 7 & -3 & 2 \ 3 & 0 & -3 \ -1 & 3 & 1 \end{bmatrix}$$X = A^{-1}B = \frac{1}{9} \begin{bmatrix} 7 & -3 & 2 \ 3 & 0 & -3 \ -1 & 3 & 1 \end{bmatrix} \begin{bmatrix} 6 \ 11 \ 0 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 42 - 33 + 0 \ 18 + 0 + 0 \ -6 + 33 + 0 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 9 \ 18 \ 27 \end{bmatrix} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}$Thus,$x = 1, y = 2, z = 3$.

The sum of three numbers is $6$. If we multiply the third number by $3$ and add the second number to it,we get $11$. By adding the first and third numbers,we get double the second number. Represent this algebraically and find the numbers using the matrix method.

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