How to solve the system of equations $a+b+c=3,2a+2b-c=1,5a-4b-3c=-7$?

Question

So I was bored, and decided to start solving systems of equations for fun. After a while, I came up with this.$$a+b+c=3\\2a+2b-c=1\\5a-4b-3c=-7$$which I thought that I might be able to solve. Here is my attempt at doing so:

We can subtract equation $1$ from equation $2$ to get$$a+b-2c=-2\\\implies a+b=2c-2\\\implies 2c-2+c=3\\\implies3c=5\implies c=\dfrac53$$Now we have$$5a-4b-5=-7\implies5a-4b=-2\\\implies5a-4b+3=\dfrac{10}3+2a+2b\\\implies3a-6b=\dfrac13\implies a=\dfrac19+2b\\\implies\dfrac19+3b+\dfrac53=3\implies\dfrac{16}9+3b=3\\\implies3b=\dfrac{11}9\implies b=\dfrac{11}{27}\\\implies a=\dfrac{25}{27}$$Now, I plugged each value in, and it turned out that sadly, I am incorrect. So is there another way that I could more easily solve this? Yes! We can put this into a $3\times3$ matrix, although the strategy I don't exactly remember. Here is my attempt to remember the strategy of solving systems of equations using a $3\times3$ matrix:

So we have the three by three matrix$$A=\begin{bmatrix}1&1&1\\2&2&-1\\5&-4&-3\end{bmatrix}$$and we know that this system of equations has a solution if our matrix $A$ has an inverse. ($\det(A)\ne0$) So, we have$$\det(A)=1(-6-4)-1(-6+5)+1(-8-10)=-10+1-18=-27$$although I am unsure what to do from here.

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My question

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What could I do to solve this system of equations? I feel like I know a viable strategy, but I'm having trouble using it.

Answer 1

The approach with determinants is called Cramer's rule. The solution is the triplet $$(a,b,c)=\left(\frac{{\rm det}({\rm A_a})}{{\rm det}({\rm A})},\frac{{\rm det}({\rm A_b})}{{\rm det}({\rm A})},\frac{{\rm det}({\rm A_c})}{{\rm det}({\rm A})}\right),$$ where for the present system \begin{align*} {\rm det}({\rm A_a})= \left|\begin{matrix} \color{red}{3} & 1 & 1\\ \color{red}{1} & 2 & -1\\ \color{red}{-7} & -4 & -3 \end{matrix}\right|,\;\; {\rm det}({\rm A_b})= \left|\begin{matrix} 1 & \color{red}{3} & 1\\ 2 & \color{red}{1} & -1\\ 5 & \color{red}{-7} & -3 \end{matrix}\right|\quad \text{and} \quad {\rm det}({\rm A_c})= \left|\begin{matrix} 1 & 1 & \color{red}{3}\\ 2 & 2 & \color{red}{1}\\ 5 & -4 & \color{red}{-7} \end{matrix}\right|. \end{align*} The red column is the right side of the system.

Answer 2

Elementary methods often involve eliminating one free variable from consideration, by adding/substracting equations appropriately. In this case, let us try to form a pair of equations in terms of $a$ and $b$.

Summing $(1)$ and $(2)$, we obtain $3a+3b=4 \implies a+b = \dfrac{4}{3}$. Multiplying equation $(1)$ by $3$ throughout and then adding it to equation $(3),$ we obtain $8a-b=2$. But we can easily solve these two equations to get $a=\dfrac{10}{27}$. Subsequently, we can use backwards substitution to find the values of $b$ and $c$.

Of course, there are many viable ways to carry out this idea, which is ultimately just Gaussian elimination in disguise.