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The question:

I know that the answers is b) ab≠1, but I have no clue how to get to that answer... Can someone help me?

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    Do you know what a determinant is? – 2017-02-15
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    Compute the determinant $D$ of the system. You get $D=\left|\matrix{1&b\cr 2a&2}\right|$ and you want the condition $D\neq0$ to hold. – 2017-02-15

4 Answers 4

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The coefficient matrix has to have non-zero determinant in order to have a unique solution. This yields $$ 2-2ab\neq 0 $$ which is equivalent to $$ ab\neq 1 $$

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We can proceed by finding the rank of this system as:

$$\begin{array}{ccc} 1 & b & | -1\\ 2a & 2 & | 5\end{array}$$ $$\Rightarrow \begin{array}{ccc} 1 & b & | -1\\ 0 & 2-2ab & | 5 + 2a\end{array}$$ performing the operation $R_2 \to R_2-2aR_1$.

Now, if this system is to have an unique solution, we must have that Rank$_A =$ Rank$_{A:b} =$ order of matrix $A = 2$. Thus, $2-2ab \neq 0 \Rightarrow ab\neq 1$.

Also note, that this is the same condition of $|A| \neq 0$. Hope it helps.

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Thank to all for the responses. Also these guys helped me so answer is that this can easily be done with determinants. If a square matrix's determinant does not equal zero, then that square matrix will have an inverse hence having a unique solution. Since this is a 2x2 matrix, just compute the determinant with the condition that it cannot equal zero: (1)(2)-(2ab) =/= 0 2 =/= 2ab 1=/= ab

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For two simultaneous equations, a1x+b1y=c1 a2x+b2y=c2 The condition for then having one solution is a1/a2 not equal to b1/b2

So 1/2a not equal to b/2 Which gives you , ab not equal to 1