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# algebra??

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For a certain value of $$k$$, the system

\begin{align*} x + y + 3z &= 10, \\ -4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*}

has no solutions. What is this value of $$k$$?

Jul 8, 2024

#2
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To determine the value of k for which the system has no solution, we can first rewrite the system in matrix form:

| 1 1 3 | | x | | 10 |
| -4 2 5 | * | y | = | 7 |
| k 0 1 | | z | | 3 |

To check if the system has no solution, we need to calculate the determinant of the coefficient matrix and the determinant of the augmented matrix. If both determinants are nonzero and not equal to each other, then the system has no solution.

The determinant of the coefficient matrix is:
det = 1(2*1 - 0*5) - 1(-4*1 - 3*5) + 3(-4*0 - 2*5) = 2 + 17 - 30 = -11

The determinant of the augmented matrix is:
det_aug = 1(2*3 - 0*7) - 1(-4*3 - 5*7) + 3(-4*0 - 2*7) = 6 + 29 - 0 = 35

Since the determinants are not equal (det ≠ det_aug) and det is not equal to zero, there is no solution for the system.

Therefore, for the system to have no solution, the value of k must be such that the determinant of the coefficient matrix is equal to zero:

-11 + k(0+4) = 0

-11 + 4k = 0

4k = 11

k = 11/4

Therefore, the value of k for which the system has no solution is k = 11/4.

Jul 8, 2024