If AX=D represents the system of linear equat | vedclass

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(C) The given system of equations is:$3x - 4y + 7z = -6$$5x + 2y - 4z = -9$$8x - 6y - z = -5$This can be written as $AX = D$,where $A = \begin{bmatrix

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$\operatorname{Rank}(A)=\operatorname{Rank}([A|D])=3$

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(C) The given system of equations is:$3x - 4y + 7z = -6$$5x + 2y - 4z = -9$$8x - 6y - z = -5$This can be written as $AX = D$,where $A = \begin{bmatrix} 3 & -4 & 7 \ 5 & 2 & -4 \ 8 & -6 & -1 \end{bmatrix}$ and $D = \begin{bmatrix} -6 \ -9 \ -5 \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = 3(-2 - 24) + 4(-5 + 32) + 7(-30 - 16)$$|A| = 3(-26) + 4(27) + 7(-46)$$|A| = -78 + 108 - 322 = -292 
eq 0$.Since $|A| 
eq 0$,the rank of matrix $A$ is $3$.For the augmented matrix $[A|D] = \begin{bmatrix} 3 & -4 & 7 & | & -6 \ 5 & 2 & -4 & | & -9 \ 8 & -6 & -1 & | & -5 \end{bmatrix}$,the rank is also $3$ because the determinant of the $3 	imes 3$ submatrix $A$ is non-zero.Thus,$\operatorname{Rank}(A) = \operatorname{Rank}([A|D]) = 3$.

If $AX=D$ represents the system of linear equations $3x-4y+7z+6=0$,$5x+2y-4z+9=0$ and $8x-6y-z+5=0$,then

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