The system of linear equations x+y+z=6, x+2y+ | vedclass

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(B) Given equations are:$x+y+z=6 \quad (1)$$x+2y+3z=10 \quad (2)$$x+2y+az=b \quad (3)$Representing the system in matrix form $AX=B$,the augmented matr

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$a=3, b 
eq 10$

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(B) Given equations are:$x+y+z=6 \quad (1)$$x+2y+3z=10 \quad (2)$$x+2y+az=b \quad (3)$Representing the system in matrix form $AX=B$,the augmented matrix $[A|B]$ is:$\begin{bmatrix} 1 & 1 & 1 & | & 6 \ 1 & 2 & 3 & | & 10 \ 1 & 2 & a & | & b \end{bmatrix}$Applying row operations:$R_2 ightarrow R_2 - R_1$ gives $\begin{bmatrix} 1 & 1 & 1 & | & 6 \ 0 & 1 & 2 & | & 4 \ 1 & 2 & a & | & b \end{bmatrix}$$R_3 ightarrow R_3 - R_1$ gives $\begin{bmatrix} 1 & 1 & 1 & | & 6 \ 0 & 1 & 2 & | & 4 \ 0 & 1 & a-1 & | & b-6 \end{bmatrix}$$R_3 ightarrow R_3 - R_2$ gives $\begin{bmatrix} 1 & 1 & 1 & | & 6 \ 0 & 1 & 2 & | & 4 \ 0 & 0 & a-3 & | & b-10 \end{bmatrix}$For the system to have no solutions,the rank of the coefficient matrix must be less than the rank of the augmented matrix. This occurs when the last row represents an impossible equation,i.e.,$0x + 0y + 0z = k$ where $k 
eq 0$.Therefore,$a-3 = 0 \Rightarrow a=3$ and $b-10 
eq 0 \Rightarrow b 
eq 10$.

The system of linear equations $x+y+z=6, x+2y+3z=10$ and $x+2y+az=b$ has no solutions when

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