Question

Augmented matrix of equations $x - y + z = 9; 2x - 3y + 4z = 25; 2x + 6y + 4z = 10$ is reduced to Echelon form as $\begin{bmatrix} 1 & -1 & 1 & 9 \\ 0 & -2 & 3 & 16 \\ 0 & 0 & 7 & 28 \end{bmatrix}$. Then the solution set $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is

• A. \( \begin{bmatrix} 3
  -2
  4 \end{bmatrix} \)
• B. \( \begin{bmatrix} -4
  2
  3 \end{bmatrix} \)
• C. \( \begin{bmatrix} 3
  4
  -2 \end{bmatrix} \)
• D. \( \begin{bmatrix} -2 3 4 \end{bmatrix} \)

Hint:

When a system of linear equations is in echelon form, you can solve it using back-substitution, starting from the last equation.

Answer 1

The Correct Option is A

The given echelon form of the augmented matrix represents the following system of linear equations: \begin{align*} x - y + z &= 9 \quad &(1)
-2y + 3z &= 16 \quad &(2)
7z &= 28 \quad &(3) \end{align*} From equation (3), we can find the value of $z$: $$7z = 28 \implies z = \frac{28}{7} = 4$$ Substitute the value of $z$ into equation (2): $$-2y + 3(4) = 16$$ $$-2y + 12 = 16$$ $$-2y = 16 - 12$$ $$-2y = 4$$ $$y = \frac{4}{-2} = -2$$ Substitute the values of $y$ and $z$ into equation (1): $$x - (-2) + 4 = 9$$ $$x + 2 + 4 = 9$$ $$x + 6 = 9$$ $$x = 9 - 6 = 3$$ Thus, the solution set is $x = 3, y = -2, z = 4$, which can be written in matrix form as $X = \begin{bmatrix} 3
-2
4 \end{bmatrix}$.