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(A) To find the number of solutions of the system of linear homogeneous equations:$x-y+z=0$ $(i)$$x+2y-z=0$ $(ii)$$2x+y+3z=0$ $(iii)$We can write the

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(A) To find the number of solutions of the system of linear homogeneous equations:$x-y+z=0$ $(i)$$x+2y-z=0$ $(ii)$$2x+y+3z=0$ $(iii)$We can write the system in matrix form $AX=0$,where $A = \begin{bmatrix} 1 & -1 & 1 \ 1 & 2 & -1 \ 2 & 1 & 3 \end{bmatrix}$ and $X = \begin{bmatrix} x \ y \ z \end{bmatrix}$.First,we calculate the determinant of matrix $A$:$|A| = 1(2 	imes 3 - (-1) 	imes 1) - (-1)(1 	imes 3 - (-1) 	imes 2) + 1(1 	imes 1 - 2 	imes 2)$$|A| = 1(6 + 1) + 1(3 + 2) + 1(1 - 4)$$|A| = 7 + 5 - 3 = 9$.Since $|A| 
eq 0$,the matrix $A$ is non-singular and invertible.For a homogeneous system $AX=0$,if $|A| 
eq 0$,the only solution is the trivial solution $X=0$ (i.e.,$x=0, y=0, z=0$).Therefore,there is exactly $1$ solution.

The number of solutions of the following system of linear homogeneous equations $x-y+z=0$,$x+2y-z=0$,and $2x+y+3z=0$ is

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