Q.4 (a) Test the consistency and hence solve the following system of linear equations, \(-2y+3t=2,\ 2x+y+z+t=-4,\ 4x-3y+z+7t=8\). [As printed; the first equation is very likely \(x-2y+3t=2\), with the \(x\) dropped in print.]
Best answer
The first equation is printed as \(-2y+3t=2\), with no x-term visible. The standard textbook problem is \(x-2y+3t=2\), so that is solved first; the as-printed version is solved after it.
(A) With \(x-2y+3t=2\):
\([A|B]=\left[\begin{array}{cccc|c}1&-2&0&3&2\\2&1&1&1&-4\\4&-3&1&7&8\end{array}\right]\)
\(R_2\to R_2-2R_1,\ R_3\to R_3-4R_1\): \(\left[\begin{array}{cccc|c}1&-2&0&3&2\\0&5&1&-5&-8\\0&5&1&-5&0\end{array}\right]\)
\(R_3\to R_3-R_2\): \(\left[\begin{array}{cccc|c}1&-2&0&3&2\\0&5&1&-5&-8\\0&0&0&0&8\end{array}\right]\)
\(\rho(A)=2\ne\rho([A|B])=3\). The system is inconsistent and has no solution.
(B) As printed (\(-2y+3t=2\)): \([A|B]=\left[\begin{array}{cccc|c}2&1&1&1&-4\\4&-3&1&7&8\\0&-2&0&3&2\end{array}\right]\). \(R_2\to R_2-2R_1\): \((0,-5,-1,5\mid16)\); with \((0,-2,0,3\mid2)\) the rank of A is 3 and the rank of [A|B] is 3 < 4 unknowns, so the system is consistent with infinitely many solutions (one free variable). Let \(t=k\): \(y=\frac{3k-2}2\), \(x=4\), \(z=-11-\frac{5k}2\). (Check: \(16-3\cdot\frac{3k-2}2-11-\frac{5k}2+7k=8\) ✓.)
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