anonymous
  • anonymous
Which values of constant k does the system of linear equations give exactly one solution (x1)+(4x)(2)=1 and (3)(x1) + (12)(x2) = k
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Is this the system you are trying to solve? $$x_1 + 4x_2 = 1$$ $$3x_1 + 12 x_2 = k$$
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Is it any number except 3?

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anonymous
  • anonymous
because if its 3 then there will be infinite solutions
anonymous
  • anonymous
If so, the answer is that there is no such value. There is one value for which there is an infinite number of solutions, and an infinite number of values for which there are no solutions. To prove it, consider that two equations have a unique solution if and only if the determinant of the coefficient matrix is zero. But the determinant of the coefficient matrix in this case is \(12\times1 - 3\times 4 = 0\). Note that \(k\) plays no part in it, so no amount of tweaking \(k\) can change the determinant.
anonymous
  • anonymous
What do you mean by determinant of the coefficient matrix?
anonymous
  • anonymous
The coefficients infront of each x?
anonymous
  • anonymous
Yes. The determinant of a matrix is a complicated beast, but for a 2x2 matrix is as below: $$ \det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc $$
anonymous
  • anonymous
doesnt the answer of each equation go into the matrix as well?
anonymous
  • anonymous
like for the equations shouldn't it be like 1 4 1 3 12 k
anonymous
  • anonymous
sorry I'm new to matrix
anonymous
  • anonymous
Yes, but the coefficient matrix is the one that is made up by just the coefficients of the variables. So, in this case, it's: $$\begin{bmatrix}1 & 4 \\ 3 & 12\end{bmatrix}$$

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