If the system of equations \(x-2y+3z=9\), \(2x+y+z=b\), \(x-7y+az=24\) has infinitely many solutions then \(a-b\) is equal to
5
For infinitely many solutions, the third equation's coefficients must be a linear combination of the first two: Row3 = m·Row1 + n·Row2, and the same combination must hold for the constants.
Matching coefficients: for x: \(m+2n=1\); for y: \(-2m+n=-7\); for z: \(3m+n=a\).
Solving the first two: from \(m=1-2n\), substitute into \(-2m+n=-7\): \(-2(1-2n)+n=-7\Rightarrow -2+4n+n=-7\Rightarrow5n=-5\Rightarrow n=-1\), so \(m=1-2(-1)=3\).
Then \(a=3m+n=3(3)+(-1)=9-1=8\).
For the constant term: \(24=9m+bn=9(3)+b(-1)=27-b\Rightarrow b=27-24=3\).
Verification: with \(a=8\), the coefficient determinant \(\begin{vmatrix}1&-2&3\\2&1&1\\1&-7&8\end{vmatrix}=1(8+7)+2(16-1)+3(-14-1)=15+30-45=0\), confirming rank deficiency (consistent with infinite solutions).
Therefore \(a-b=8-3=5\).
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