Question

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

Answer 1

Answer on Question #46774 – Math - Linear Algebra

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

Solution

$$\left\{ \begin{array}{c} a + 3b + 2c = 3 \\ 2a + (-1)b + (-3)c = -8 \\ 5a + 2b + c = 9 \end{array} \right.$$

First let

$$A = \begin{pmatrix} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{pmatrix}.$$

This is the matrix formed by the coefficients of the given system of equations. The matrix method gives

$$\begin{pmatrix} a \\ b \\ c \end{pmatrix} = A^{-1} \begin{pmatrix} 3 \\ -8 \\ 9 \end{pmatrix}.$$

Take note that the right hand values of the system are 3, -8, and 9 and they are highlighted here:

$$\left\{ \begin{array}{c} a + 3b + 2c = 3 \\ 2a + (-1)b + (-3)c = -8 \\ 5a + 2b + c = 9 \end{array} \right.$$

These values are important as they will be used to replace the columns of the matrix $A$.

Now let's calculate the determinant of the matrix $A$

$$\det A = \left| \begin{array}{ccc} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array} \right| = 1 \cdot 1 \cdot (-1) + 3 \cdot 5 \cdot (-3) + 2 \cdot 2 \cdot 2 - 2 \cdot 5 \cdot (-1) - 2 \cdot 1 \cdot 3 - 1 \cdot 2 \cdot (-3) = -28.$$

Now replace the third column of $A$ (that corresponds to the variable 'c') with the values that form the right hand side of the system of equations. We will denote this new matrix $A_c$ (since we're replacing the 'c' column so to speak).

Now compute the determinant of $A_c$

$$\det A_c = \left| \begin{array}{ccc} 1 & 3 & 3 \\ 2 & -1 & -8 \\ 5 & 2 & 9 \end{array} \right| = 1 \cdot 9 \cdot (-1) + 3 \cdot 5 \cdot (-8) + 2 \cdot 2 \cdot 3 - 3 \cdot 5 \cdot (-1) - 2 \cdot 9 \cdot 3 - 1 \cdot 2 \cdot (-8) = -140.$$

To find the solution for $c$, divide the determinant of $A_c$ by the determinant of $A$ to get:

$$c = \frac{\det A_c}{\det A} = \frac{-140}{-28} = 5.$$

Answer: 5.

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