The equations x+y+z=3, x+2y+2z=6 and x+ay+3z= | vedclass

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(C) The given system of linear equations is:$x+y+z=3$$x+2y+2z=6$$x+ay+3z=b$Representing the system in matrix form $AX=B$, where $A = \begin{bmatrix} 1

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Unique solution when $a 
eq 3$, $b$ is any value

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(C) The given system of linear equations is:$x+y+z=3$$x+2y+2z=6$$x+ay+3z=b$Representing the system in matrix form $AX=B$, where $A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 2 \ 1 & a & 3 \end{bmatrix}$.The determinant of the coefficient matrix $A$ is given by:$\Delta = |A| = 1(6-2a) - 1(3-2) + 1(a-2)$$\Delta = 6 - 2a - 1 + a - 2 = 3 - a$.For the system to have a unique solution, the determinant $\Delta$ must be non-zero, i.e.,$\Delta 
eq 0$.$3 - a 
eq 0 \Rightarrow a 
eq 3$.If $a 
eq 3$, the system has a unique solution regardless of the value of $b$.

The equations $x+y+z=3$, $x+2y+2z=6$ and $x+ay+3z=b$ have

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