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

Solve the set of linear equations by the matrix method :

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

We need find vector $\left( \begin{array}{l}a\\ b\\ c \end{array} \right)$ such as

$$\mathrm {A} * \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = B,$$

where $A = \left( \begin{array}{rrr}1 & 3 & 2\\ 2 & -1 & -3\\ 5 & 2 & 1 \end{array} \right), B = \left( \begin{array}{c}3\\ -8\\ 9 \end{array} \right).$

Multiply (1) by $A^{-1}$ :

$$A ^ {- 1} A * \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = A ^ {- 1} B$$ $$E * \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = A ^ {- 1} B$$ $$\left( \begin{array}{c} a \\ b \\ c \end{array} \right) = A ^ {- 1} B$$

1) First find the **det $A$**

$$\det A = 1 * (- 1 + 6) - 2 * (3 - 4) + 5 * (- 9 + 2) = 5 + 2 - 35 = - 28$$

2) Calculating Matrix of Minors.

It is easy for example if we need find cell $M_{11}$ , I ignore the values in the current row and columns, and calculate the determinant using the remaining values

$$M _ {1 1} = \left( \begin{array}{c c} * & - 1 \\ & 2 \end{array} \right) = \left| \begin{array}{c c} - 1 & - 3 \\ 2 & 1 \end{array} \right| = - 1 * 1 - (- 3) * 2 = 5$$ $$\begin{array}{l} M = \left( \begin{array}{cccc|cc} -1 & -3 & |2 & -3 & |2 & -1 \\ 2 & 1 & |5 & 1 & |5 & 2 \\ |3 & 2 & |1 & 2 & |1 & 3 \\ |2 & 1 & |5 & 1 & |5 & 2 \\ |3 & 2 & |1 & 2 & |1 & 3 \\ -1 & -3 & |2 & -3 & |2 & -1 \end{array} \right) \\ = \left( \begin{array}{cccc} -1 * 1 - (-3) * 2 & 2 * 1 - (-3) * 5 & 2 * 2 - (-1) * 5 \\ 3 * 1 - 2 * 2 & 1 * 1 - 2 * 5 & 1 * 2 - 5 * 3 \\ 3 * (-3) - 2 * (-1) & -3 * 1 - 2 * 2 & 1 * (-1) - 2 * 3 \end{array} \right) \\ = \left( \begin{array}{ccc} 5 & 17 & 9 \\ -1 & -9 & -13 \\ -7 & -7 & -7 \end{array} \right) \end{array}$$

3) Matrix of Cofactors

Then find Matrix of Cofactors Just apply a "checkerboard" of minuses to the "Matrix of Minors". In other words, you need to change the sign of alternate cells, like this:

$$\left( \begin{array}{ccc} 5 & 17 & 9 \\ -1 & -9 & -13 \\ -7 & -7 & -7 \end{array} \right) \sim \left( \begin{array}{ccc} + & - & + \\ - & + & - \\ + & - & + \end{array} \right) \sim \left( \begin{array}{ccc} 5 & -17 & 9 \\ 1 & -9 & 13 \\ -7 & 7 & -7 \end{array} \right)$$ $$Matrix of Cofactors = \left( \begin{array}{ccc} 5 & -17 & 9 \\ 1 & -9 & 13 \\ -7 & 7 & -7 \end{array} \right)$$

4) Adjoint Matrix

Now "Transpose" all elements of the previous matrix. In other words swap their positions over the diagonal (the diagonal stays the same)

$$Adjoint Matrix = \left( \begin{array}{ccc} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{array} \right)$$

5) Inverse matrix

And now multiply the Adjoint Matrix by 1/Determinant, so the inverse matrix will be written as

$$A^{-1} = \det A * Adjoint Matrix = -\frac{1}{28} \left( \begin{array}{ccc} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{array} \right)$$

Solution can be find as:

$$\left( \begin{array}{c} a \\ b \\ c \end{array} \right) = A^{-1} * B = -\frac{1}{28} \left( \begin{array}{ccc} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{array} \right) * \left( \begin{array}{c} 3 \\ -8 \\ 9 \end{array} \right) =$$ $$= - \frac {1}{28} \left( \begin{array}{c} 5 * 3 + 1 * (-8) + (-7) * 9 \\ (-17) * 3 + (-9) * (-8) + 7 * 9 \\ 9 * 3 + 13 * (-8) + (-7) * 9 \end{array} \right) = - \frac {1}{28} \left( \begin{array}{c} -56 \\ 84 \\ -140 \end{array} \right) = \left( \begin{array}{c} - \frac {1}{28} * (-56) \\ - \frac {1}{28} * 84 \\ - \frac {1}{28} * (-140) \end{array} \right) = \left( \begin{array}{c} 2 \\ -3 \\ 5 \end{array} \right)$$

Answer: $\left( \begin{array}{l} 2 \\ -3 \\ 5 \end{array} \right)$