If the system of linear equations $x+y+2z=6$, $2x+3y+az=a+1$, $-x-3y+bz=2b$, where $a,b\in\mathbf R$, has infinitely many solutions, then $7a+3b$ is equal to :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Eliminating: $R_2-2R_1$ gives $y+(a-4)z=a-11$; $R_3+R_1$ gives $-2y+(b+2)z=2b+6$. Adding twice the first to the second: $(b+2a-6)z=2a+2b-16$. For infinitely many solutions, $2a+b=6$ and $a+b=8$, so $a=-2,\ b=10$ and $7a+3b=16$. Answer (1).

Q. If the system of linear equations $x+y+2z=6$, $2x+3y+az=a+1$, $-x-3y+bz=2b$, where $a,b\in\mathbf R$, has infinitely many solutions, then $7a+3b$ is equal to :

Explanation

Reviewed by Dr. Abubacker

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