Question #243699

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


1
Expert's answer
2021-09-30T00:46:46-0400

Solution:

Given,

𝑡𝑥+2𝑦+3𝑧=𝑎2𝑥+3𝑦𝑡𝑧=𝑏3𝑥+5𝑦+(𝑡+1)𝑧=𝑐𝑡𝑥 + 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 [t23a23tb35t+1c]\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.

=[t23a05t6tt2+t9ttc3at008t2+7t+15t65tb3tca6b+4c5t6]=\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 8t2+7t+15t6=0\dfrac{-8t^2+7t+1}{5t-6}=0 and 5tb3tca6b+4c5t6=0\dfrac{5tb-3tc-a-6b+4c}{5t-6}=0 ...(i)

From 8t2+7t+15t6=0\dfrac{-8t^2+7t+1}{5t-6}=0

8t2+7t+1=0t=1,t=18\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 18-\dfrac18 .

Put t=1 in (i)

5b3ca6b+4c56=0ba+c=0a+b=c\dfrac{5b-3c-a-6b+4c}{5-6}=0 \\\Rightarrow -b-a+c=0 \\\Rightarrow a+b=c

Hence, proved.


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