Question

How do you solve 3a + 6b + c + 4d = 2, a – 2b + 3c – d = - 3, 2a - 4c + d = - 15 and a + 8b + 2c + d = 6 using matrices?

Answer 1

Hint: The given system of linear equations has 4 equations and 4 unknowns. We can write all the coefficients of variables in a matrix, it will be $4\times 4$ matrix lets call it matrix A . we can write all the variables in another matrix that will be $4\times 1$ matrix let’s call it matrix B. we can see that product of A and B will us a matrix that has all the constant of the equation. If we multiply the inverse of A both sides, we will get the value of all the variables.

Complete step by step answer:
The given equation is 3a + 6b + c + 4d = 2, a – 2b + 3c – d = - 3, 2a - 4c + d = - 15 and a + 8b + 2c + d = 6
Let’s form matrix A of all the coefficients of the variables.
So $A=\left[ \begin{matrix}
   3 & 6 & 1 & 4 \\
   1 & -2 & 3 & -1 \\
   2 & 0 & -4 & 1 \\
   1 & 8 & 2 & 1 \\
\end{matrix} \right]$
Let’s form matrix B of all variables
So $B=\left[ \begin{matrix}
   a \\
   b \\
   c \\
   d \\
\end{matrix} \right]$
We can write $\left[ \begin{matrix}
   3 & 6 & 1 & 4 \\
   1 & -2 & 3 & -1 \\
   2 & 0 & -4 & 1 \\
   1 & 8 & 2 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   a \\
   b \\
   c \\
   d \\
\end{matrix} \right]=\left[ \begin{matrix}
   2 \\
   -3 \\
   -15 \\
   6 \\
\end{matrix} \right]$ or $AB=\left[ \begin{matrix}
   2 \\
   -3 \\
   -15 \\
   6 \\
\end{matrix} \right]$
So B is equal to ${{A}^{-1}}\left[ \begin{matrix}
   2 \\
   -3 \\
   -15 \\
   6 \\
\end{matrix} \right]$. If A is equal to $\left[ \begin{matrix}
   3 & 6 & 1 & 4 \\
   1 & -2 & 3 & -1 \\
   2 & 0 & -4 & 1 \\
   1 & 8 & 2 & 1 \\
\end{matrix} \right]$ then ${{A}^{-1}}$ is equal to $\left[ \begin{matrix}
   -\dfrac{1}{73} & \dfrac{25}{73} & \dfrac{22}{73} & \dfrac{7}{73} \\
   -\dfrac{17}{292} & -\dfrac{13}{292} & \dfrac{9}{292} & \dfrac{23}{146} \\
   \dfrac{11}{146} & \dfrac{17}{146} & -\dfrac{23}{146} & -\dfrac{2}{73} \\
   \dfrac{24}{73} & -\dfrac{16}{73} & -\dfrac{17}{73} & -\dfrac{22}{73} \\
\end{matrix} \right]$
The value of ${{A}^{-1}}\left[ \begin{matrix}
   2 \\
   -3 \\
   -15 \\
   6 \\
\end{matrix} \right]$ is equal to $\left[ \begin{matrix}
   -5 \\
   \dfrac{1}{2} \\
   2 \\
   3 \\
\end{matrix} \right]$
So a = -5 , b= $\dfrac{1}{2}$ , c = 2 and d = 3

Note:
In the above solution if the det value of matrix A will equal to 0, then we can find the inverse of matrix A. In that case the system of equations will be inconsistent that means it can have infinite solution or no solution. The system is consistent if it has only a solution.