Question

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

Answer 1

Is this the system you are trying to solve?

$$x_1 + 4x_2 = 1$$
$$3x_1 + 12 x_2 = k$$

Answer 2

Yes

Answer 3

Is it any number except 3?

Answer 4

because if its 3 then there will be infinite solutions

Answer 5

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.

Answer 6

What do you mean by determinant of the coefficient matrix?

Answer 7

The coefficients infront of each x?

Answer 8

Yes. The determinant of a matrix is a complicated beast, but for a 2x2 matrix is as below:

$$
\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc
$$

Answer 9

doesnt the answer of each equation go into the matrix as well?

Answer 10

like for the equations shouldn't it be like

1 4 1
3 12 k

Answer 11

sorry I'm new to matrix

Answer 12

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:

$$\begin{bmatrix}1 & 4 \\ 3 & 12\end{bmatrix}$$