Answer to Question #243699 in Linear Algebra for dahara

Question

A system of equations is given below, 𝑡𝑥 + 2𝑦 + 3𝑧 = 𝑎 2𝑥 + 3𝑦 − 𝑡𝑧 = 𝑏 3𝑥 + 5𝑦 + (𝑡 + 1)𝑧 = 𝑐 Where 𝑡 is an integer and 𝑎, 𝑏, 𝑐 are real constants. The system does not have a unique solution, but it is consistent. Show that 𝑎 + 𝑏 = 𝑐

Answer 1

Solution:

Given,

$𝑡𝑥 + 2𝑦 + 3𝑧 = 𝑎 \\2𝑥 + 3𝑦 − 𝑡𝑧 = 𝑏 \\3𝑥 + 5𝑦 + (𝑡 + 1)𝑧 = 𝑐$

The system does not have a unique solution, but it is consistent.

It means the system has infinite number of solutions.

Augmented matrix is $\begin{bmatrix}t&2&3&|a\\ 2&3&-t&|b\\ 3&5&t+1&|c\end{bmatrix}$

We need to make all elements of row 3 zero as the system has infinite number of solutions.

Reduce matrix to row echelon form.

$=\begin{bmatrix}t&2&3&a\\ 0&\frac{5t-6}{t}&\frac{t^2+t-9}{t}&\frac{tc-3a}{t}\\ 0&0&\frac{-8t^2+7t+1}{5t-6}&\frac{5tb-3tc-a-6b+4c}{5t-6}\end{bmatrix}$

Now, we must have $\dfrac{-8t^2+7t+1}{5t-6}=0$ and $\dfrac{5tb-3tc-a-6b+4c}{5t-6}=0$ ...(i)

From $\dfrac{-8t^2+7t+1}{5t-6}=0$

$\Rightarrow -8t^2+7t+1=0 \\ \Rightarrow t=1, t=-\dfrac18$

Given that t is an integer, so t must be equal to 1, not $-\dfrac18$ .

Put t=1 in (i)

$\dfrac{5b-3c-a-6b+4c}{5-6}=0 \\\Rightarrow -b-a+c=0 \\\Rightarrow a+b=c$

Hence, proved.