Question

What is the solution to the system of linear equations represented by the matrix below?

Answer 1

$\begin{bmatrix}2&4\\1&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}6\\3\end{bmatrix}$

Notice that the second row is half the first, so the rows are dependent. This means there are infinitely many solutions to the system.

Answer 2

Answer:

D. The system of equations has infinite solutions.

Step-by-step explanation:

Given system of equations,

$\begin{bmatrix}2 & 4 \\ 1 & 2\end{bmatrix}.\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}6\\ 3\end{bmatrix}$

$\implies \begin{bmatrix}2x+4y\\ x+2y\end{bmatrix}=\begin{bmatrix}6\\ 3\end{bmatrix}$

By comparing both sides,

We get,

2x + 4y = 6,

x + 2y = 3,

We know that a system $a_1x+b_1y=c_1$, $a_2x+b_2y=c_2$

has a unique solution if,

$\frac{a_1}{a_2}\neq \frac{b_1}{b_2}\neq \frac{c_1}{c_2}$

No solution, if,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$

Infinitely many solution if,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Here,

$\frac{2}{1}=\frac{4}{2}=\frac{6}{3}=2$

Hence, the system of equation has infinite solutions.

Option D is correct.