The system of equations x+y+z=5, x+2y+az=9, x | vedclass

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(B) The given system of equations is: $x+y+z=5$ $x+2y+az=9$ $x+2y+z=b$ We can write this in matrix form $AX=B$,where $A = \begin{bmatrix} 1 & 1 & 1 \

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$a=1, b 
eq 9$

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(B) The given system of equations is: $x+y+z=5$ $x+2y+az=9$ $x+2y+z=b$ We can write this in matrix form $AX=B$,where $A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & a \ 1 & 2 & 1 \end{bmatrix}$. For the system to be inconsistent,the determinant $|A|$ must be $0$ and the system must have no solution. $|A| = 1(2-2a) - 1(1-a) + 1(2-2) = 2-2a-1+a = 1-a$. Setting $|A|=0$,we get $1-a=0$,so $a=1$. Now,substitute $a=1$ into the equations: $x+y+z=5$ $x+2y+z=9$ $x+2y+z=b$ Comparing the second and third equations,we see that if $b 
eq 9$,the system represents two parallel planes,which means there is no solution (inconsistent). Therefore,the system is inconsistent when $a=1$ and $b 
eq 9$.

The system of equations $x+y+z=5, x+2y+az=9, x+2y+z=b$ is inconsistent if

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