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Find the value of parameter $a$, so that the system has no solution.

\begin{bmatrix} a & 3 & 2 & & 5 \\ 1 & 7 & 3 & = & 13 \\ 3 & 1 & a & & 3 \\ \end{bmatrix}

How to do that? I know I am supposed to get 0 0 0 = non-zero value, but how to actually get to the solution? The answer is 13/7.

Sriotchilism O'Zaic
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Juraj Mlich
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    Apply Gauss-Elimination and determine for which $a$ we get a row with $3$ zeros on the left. – Peter Oct 17 '17 at 18:31
  • Another possibility is to determine for which $a$ the determinant is $0$. But dom't forget to verify for which $a$ there is actually no solution. – Peter Oct 17 '17 at 18:34
  • Would you be so kind and actually show me the Gauss-Elimination in this example? – Juraj Mlich Oct 17 '17 at 18:39
  • First of all, swap the first and second column to have no $a's$ in the first column. – Peter Oct 17 '17 at 18:43
  • Can I just swap columns? – Juraj Mlich Oct 17 '17 at 18:55
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    Yes, you may swap columns. For there to be a solution, $[5,13,3]^T$ must lie in the column space of the coefficient matrix, for which the order of columns is irrelevant. – amd Oct 17 '17 at 19:21

1 Answers1

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from equation (2) we get $$x_1=13-7x_2-3x_3$$ plugging this in the first and third equation and we have $$x_2(3-7a)+x_3(2-3a)=5-13a$$ $$-20x_2+x_3(a-9)=-36$$ from our last equation $$x_2=\frac{9}{5}+\left(\frac{a-9}{20}\right)x_3$$ and now we can eliminate $$x_2$$ $$x_3((a+1)(7a-13)=8(1+a)$$ can you finish? if $$7a-13=0$$ so $$a=\frac{13}{7}$$ and we get no solutions.

Dr. Sonnhard Graubner
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