Question

Solve the set of linear equations by the matrix method : a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9. Sove for c

Answer 1

Answer on Question #40855 – Math - Linear Algebra

Question:

Solve the set of linear equations by the matrix method: $a + 3b + 2c = 3$, $2a - b - 3c = -8$, $5a + 2b + c = 9$. Sove for $c$.

Solution:

The matrix of this system is $A = \begin{pmatrix} 1 & 3 & 2 \\ 2 & -1 & -1 \\ 5 & 2 & 1 \end{pmatrix}$, the column of free terms is $b = \begin{pmatrix} 3 \\ -8 \\ 9 \end{pmatrix}$.

In matrix form this equation has presentation $Ax = b$, where $x = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$.

Let premultiply both sides of the equation by $A^{-1}$, the inverse of $A$.

Since $A^{-1}Ax = Ix = x$, we know the following.

$$x = A^{-1}b$$

So let's find the inverse of $A$. Compute the algebraic addition to the elements of the matrix $A$

$$\mathbf{M}_{1,1} = (-1)^{1+1} \left| \begin{array}{cc} -1 & -3 \\ 2 & 1 \end{array} \right| = 5$$ $$\mathbf{M}_{1,2} = (-1)^{1+2} \left| \begin{array}{cc} 2 & -3 \\ 5 & 1 \end{array} \right| = -17$$ $$\mathbf{M}_{1,3} = (-1)^{1+3} \left| \begin{array}{cc} 2 & -1 \\ 5 & 2 \end{array} \right| = 9$$ $$\mathbf{M}_{2,1} = (-1)^{2+1} \left| \begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array} \right| = 1$$ $$\mathbf{M}_{2,2} = (-1)^{2+2} \left| \begin{array}{cc} 1 & 2 \\ 5 & 1 \end{array} \right| = -9$$ $$\mathbf{M}_{2,3} = (-1)^{2+3} \left| \begin{array}{cc} 1 & 3 \\ 5 & 2 \end{array} \right| = 13$$ $$\mathbf{M}_{3,1} = (-1)^{3+1} \left| \begin{array}{cc} 3 & 2 \\ -1 & -3 \end{array} \right| = -7$$ $$\mathbf{M}_{3,2} = (-1)^{3+2} \left| \begin{array}{cc} 1 & 2 \\ 2 & -3 \end{array} \right| = 7$$ \begin{array}{l} \mathbf{M}_{3,3} = (-1)^{3+3} \left| \begin{array}{cc} 1 & 3 \\ 2 & -1 \end{array} \right| = -7 \\ C^* = \left( \begin{array}{ccc} 5 & -17 & 9 \\ 1 & -9 & 13 \\ -7 & 7 & -7 \end{array} \right) \\ C^*^T = \left( \begin{array}{ccc} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{array} \right) \\ \end{array}

So, an inverse of A is

A^{-1} = \frac{C^*^T}{\det A} = \left( \begin{array}{ccc} -\frac{5}{28} & -\frac{1}{28} & \frac{1}{4} \\ \frac{17}{28} & \frac{9}{28} & -\frac{1}{4} \\ -\frac{9}{28} & -\frac{13}{28} & \frac{1}{4} \end{array} \right)

Thus our solution is

$$x = A^{-1}b = \left( \begin{array}{ccc} -\frac{5}{28} & -\frac{1}{28} & \frac{1}{4} \\ \frac{17}{28} & \frac{9}{28} & -\frac{1}{4} \\ -\frac{9}{28} & -\frac{13}{28} & \frac{1}{4} \end{array} \right) * \left( \begin{array}{c} 3 \\ -8 \\ 9 \end{array} \right) = \left( \begin{array}{c} 2 \\ -3 \\ 5 \end{array} \right)$$

Answer: $a = 2, b = -3, c = 5$.