Question

Consider the following set of equations x + 2y = 5, 4x + 8y = 12, 3x +6y + 3z = 15. This set

• A. Has unique solution
• B. Has no solution
• C. Has infinite number of solutions
• D. Has 3 solutions

Answer 1

The correct option is B Has no solution

Given AX = B

$\Rightarrow \left[\begin{array}{ccc}1& 2& 0\\ 4& 8& 0\\ 3& 6& 3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}5\\ 12\\ 15\end{array}\right]$

Augmented matrix

[A:B] = $\left[\begin{array}{ccccc}1& 2& 0& :& 5\\ 4& 8& 0& :& 12\\ 3& 6& 3& :& 15\end{array}\right]$

$R_2\to R_2-4R_1;R_3\to R_3-3R_1$

[A:B] $\sim$ $\left[\begin{array}{ccccc}1& 2& 0& :& 5\\ 0& 0& 0& :& -8\\ 0& 0& 3& :& 0\end{array}\right]$

$R_2\leftrightarrow R_3$

[A:B] $\sim$ $\left[\begin{array}{ccccc}1& 2& 0& :& 5\\ 0& 0& 3& :& 0\\ 0& 0& 0& :& -8\end{array}\right]$

$\Rightarrow \rho \left(A\right)=2,\rho (A:B)=3$

So $\rho \left(A\right)$ $\ne$ $\rho (A:B)$

$\therefore$ No solution exist.