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Example 1.

For a matrix

    \[\mathbf{A} = \left[\begin{array}{cc} 2 & 3 \\ 1 & 2 \\ \end{array} \right]\]

find \mathbf{A^{-1}}.

Solution:

    \[\mathbf{[A|I]} =\left[\begin{array}{cc|cc} 2 & 3 & 1 & 0 \\ 1 & 2 & 0 & 1 \\ \end{array} \right] \sim^{1} \left[\begin{array}{cc|cc} 1 & 2 & 0 & 1 \\ 2 & 3 & 1 & 0 \\ \end{array} \right] \sim^{2} \left[\begin{array}{cc|cc} 1 & 2 & 0 & 1 \\ 0 & -1 & 1 & -2 \\ \end{array} \right] \sim^{3}\]

    \[\sim^{3} \left[\begin{array}{cc|cc} 1 & 0 & 2 & -3 \\ 0 & -1 & 1 & -2 \\ \end{array} \right] \sim^{4} \left[\begin{array}{cc|cc} 1 & 0 & 2 & -3 \\ 0 & 1 & -1 & 2 \\ \end{array} \right] = \mathbf{[I|A^{-1}]}.\]

In the order, we apply the following transformations:

(1.) interchange the first and the second row,

(2.) the first row multiplied by -2 added to second row.

(3.) the second row multiplied by 2 added to first row,

(4.) the first row multiplied by -1.

An inverse matrix of the given matrix \mathbf{A} is

    \[\mathbf{A^{-1}}= \left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \\ \end{array} \right].\]

Example 2.

Using the elementary transformations of matrices, find the inverse matrix of matrix

    \[\mathbf{A} = \left[\begin{array}{ccc} 1 & -1 & 2 \\ 1 & 2  & -1\\ -4 & 4  & 1\\ \end{array} \right].\]

Solution:

With R_i we will denote a row in which we perform elementary transformations.

    \[\mathbf{[A|I]} =\left[\begin{array}{ccc|ccc} 1 & -1 & 2 & 1 & 0 & 0 \\ 1 & 2 & -1 & 0 & 1 & 0 \\ -4 & 4 & 1 & 0 & 0 & 1 \end{array} \right] & \! \begin{aligned} & \xrightarrow{-1R_1 + R_2}\\ & \xrightarrow{4R_1 + R_2} \end{aligned} \left[\begin{array}{ccc|ccc} 1 & -1 & 2 & 1 & 0 & 0 \\ 0 & 3 & -3 & -1 & 1 & 0 \\ 0 & 0 & 9 & 4 & 0 & 1 \\ \end{array} \right] & \! \begin{aligned} & \xrightarrow{\frac{1}{3} \cdot R_2}\\ & \xrightarrow{\frac{1}{9} \cdot R_3} \end{aligned}\]

    \[\left[\begin{array}{ccc|ccc} 1 & -1 & 2 & 1 & 0 & 0 \\ 0 & 1 & -1 & -\frac{1}{3} & \frac{1}{3} & 0 \\ 0 & 0 & 1 & \frac{4}{9} & 0 & \frac{1}{9} \\ \end{array} \right] & \! \begin{aligned} & \xrightarrow{R_3 + R_2}\\ & \xrightarrow{-2R_3+R_1} \end{aligned} \left[\begin{array}{ccc|ccc} 1 & -1 & 0 & \frac{1}{9} & 0 & -\frac{2}{9} \\ 0 & 1 & 0 & \frac{1}{9} & \frac{1}{3} & \frac{1}{9} \\ 0 & 0 & 1 & \frac{4}{9} & 0 & \frac{1}{9} \\ \end{array} \right]\]

    \[\xrightarrow{R_2+R_1} \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{2}{9} & \frac{1}{3} & -\frac{1}{9} \\ 0 & 1 & 0 & \frac{1}{9} & \frac{1}{3} & \frac{1}{9} \\ 0 & 0 & 1 & \frac{4}{9} & 0 & \frac{1}{9} \\ \end{array} \right].\]

Therefore, an inverse matrix of a matrix \mathbf{A} is

    \[\mathbf{A^{-1}} = \left[\begin{array}{ccc} \frac{2}{9} & \frac{1}{3} & -\frac{1}{9} \\ \frac{1}{9} & \frac{1}{3} & \frac{1}{9} \\ \frac{4}{9} & 0 & \frac{1}{9} \\ \end{array} \right].\]

 

Example 3.

Solve the following system of equations by using the Cramer’s rule:

    \[x_1 + 2x_2 - x_3 + x_4 = -1,\]

    \[2x_1 + 5x_2 - x_3 + 2x_4 = -2,\]

    \[3x_1 - x_2 - 2x_3 + x_4 = 5,\]

    \[x_1 - x_2 + 3x_3 - 5x_4 = 6.\]

Solution:

Firstly, we write the system in a matrix form:

    \[\left[\begin{array}{cccc} 1 & 2 & -1 & 1 \\ 2 & 5  & -1 & 2\\ 3 & -1  & -2 & 1 \\ 1 & -1  & 3 & -5 \\ \end{array} \right] \cdot \left[\begin{array}{c} x_1 \\ x_2\\ x_3 \\ x_4 \\ \end{array} \right]  = \left[\begin{array}{c} -1 \\ -2\\ 5 \\ 6 \\ \end{array} \right].\]

Let a matrix \mathbf{A} be a matrix of the system above, that is

    \[\mathbf{A} =\left[\begin{array}{cccc} 1 & 2 & -1 & 1 \\ 2 & 5  & -1 & 2\\ 3 & -1  & -2 & 1 \\ 1 & -1  & 3 & -5 \\ \end{array} \right].\]

The determinant of a matrix \mathbf{A} is D = det (\mathbf{A}) = -34 (check it!).

Now we need to calculate the determinant D_1 of a matrix \mathbf{A_1} in which the first column is replaced with the column matrix of free coefficients, that is

    \[\mathbf{A_1} = \left[\begin{array}{cccc} -1 & 2 & -1 & 1 \\ -2 & 5  & -1 & 2\\ 5 & -1  & -2 & 1 \\ 6 & -1  & 3 & -5 \\ \end{array} \right].\]

The determinant of a matrix \mathbf{A_1} is D_1 = -68.

Analogously we treat for the remaining columns of the matrix \mathbf{A} – determinants of matrices that we get, in the order amounts D_2 = 34, D_3 = -34, D_4 = 0. According to the Cramer’s rule, we get the final solutions:

    \[x_1 = \frac{D_1}{D} = \frac{-68}{-34} = 2,\]

    \[x_2 = \frac{D_2}{D} = \frac{34}{-34} = -1,\]

    \[x_3 = \frac{D_3}{D} = \frac{-34}{-34} = 1,\]

    \[x_4 = \frac{D_4}{D} = \frac{0}{-34} = 0.\]

 

Example 4.

Using the elementary transformations, solve the following system of equations:

    \[x_1 + 2x_2 + 2x_3 + 3x_4 + x_5 = 3,\]

    \[2x_1 - x_3 - x_4 + 5x_5 = 2,\]

    \[x_1 + 2x_2 + 6x_3 - x_4 + 5x_5 = 3,\]

    \[x_1 -  2x_2 + 5x_3 - 12x_4 + 12x_5 = -1.\]

Solution:

    \[\left[\begin{array}{ccccc|c} 1 & 2 & 2 & 3 & 1 & 3 \\ 2 & 0 & -1 & -1 & 5 & 2 \\ 1 & 2 & 6 &  -1 & 5 & 3 \\ 1 & -2 & 5 &  -12 & 12 & -1 \\ \end{array} \right] & \! \begin{aligned} & \xrightarrow{-2 \cdot R_1 + R_2}\\ & \xrightarrow{-1 \cdot R_1+R_3} \\ & \xrightarrow{-1 \cdot R_1+R_4} \end{aligned} \left[\begin{array}{ccccc|c} 1 & 2 & 2 & 3 & 1 & 3 \\ 0 & -4 & -5 & -7 & 3 & -4 \\ 0 & 0 & 4 &  -4 & 4 & 0 \\ 0 & -4 & 3 &  -15 & 11 & -4 \\ \end{array} \right]\]

    \[& \! \begin{aligned} & \xrightarrow{1/4 \cdot R_3}\\ \end{aligned} \left[\begin{array}{ccccc|c} 1 & 2 & 2 & 3 & 1 & 3 \\ 0 & -4 & -5 & -7 & 3 & -4 \\ 0 & 0 & 1 &  -1 & 1 & 0 \\ 0 & -4 & 3 &  -15 & 11 & -4 \\ \end{array} \right] & \! \begin{aligned} & \xrightarrow{-1 \cdot R_3 + R_1}\\ & \xrightarrow{5 \cdot R_3 + R_2}\\ & \xrightarrow{-3 \cdot R_3 + R_4}\\ \end{aligned} \left[\begin{array}{ccccc|c} 1 & 2 & 0 & 5 & -1 & 3 \\ 0 & -4 & 0 & -12 & 8 & -4 \\ 0 & 0 & 1 &  -1 & 1 & 0 \\ 0 & -4 & 0 &  -12 & 8 & -4 \\ \end{array} \right]\]

    \[& \! \begin{aligned} & \xrightarrow{-1/4 \cdot R_2}\\ & \xrightarrow{-1/4 \cdot R_4}\\ & \xrightarrow{R_4 + R_2}\\ & \xrightarrow{-2 \cdot R_4 + R_1}\\ \end{aligned} \left[\begin{array}{ccccc|c} 1 & 0 & 0 & -1 & 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 &  -1 & 1 & 0 \\ 0 & 1 & 0 &  3 & -2 & 1 \\ \end{array} \right] & \! \begin{aligned} & \xrightarrow{R_2 \leftrightarrow R_4} \end{aligned} \left[\begin{array}{ccccc|c} 1 & 0 & 0 & -1 & 3 & 1 \\ 0 & 1 & 0 & 3 & -2 & 1 \\ 0 & 0 & 1 &  -1 & 1 & 0 \\ 0 & 0 & 0 &  0 & 0 & 0 \\ \end{array} \right]\]

The rank of a matrix \mathbf{A} (matrix of the system) and augmented matrix is 3. Therefore, the solution depends on the two free parameters. We have (from the system above):

    \[x_1 - x_4 + 3x_5 = 1,\]

    \[x_2 + 3x_4 - 2x_5 = 1,\]

    \[x_3 - x_4 + x_5 = 0.\]

If x_4 = t and x_5 = s; t, s \in \mathbb{R}, then

    \[x_1 =1 + t - 3s,\]

    \[x_2 = 1 -3 t + s,\]

    \[x_3 = t - s.\]

 

The solution we can write in a matrix form:

 

    \[\left[\begin{array}{c} x_1  \\ x_2  \\ x_3  \\ x_4 \\ x_5 \\ \end{array} \right] = \left[\begin{array}{c} 1 + t - s  \\ 1 - 3t + 2s  \\ t - s  \\ t \\ s \\ \end{array} \right] = t \left[\begin{array}{c} 1  \\ -3  \\ 1  \\ 1 \\ 0 \\ \end{array} \right] + s  \left[\begin{array}{c} -3  \\ 2  \\ -1  \\ 0 \\ 1 \\ \end{array}\right] + \left[\begin{array}{c} 1  \\ 1  \\ 0  \\ 0\\ 0\\ \end{array}\right].\]

 

Example 5.

Solve the following system of equations:

    \[15 x_1 + 2x_2 + 7x_4 = -1,\]

    \[x_2 + 2x_3 + 5x_4 = -1,\]

    \[2x_1 + 4x_2 + 7x_3 - 2x_4 = 0,\]

    \[x_2 + 2x_3 + 5x_4 = 3.\]

 

Solution:

 

    \[\left[\begin{array}{cccc|c} 15 & 2 & 0 & 7  & -1 \\ 0 & 1 & 2 & 5  & -1 \\ 2 & 4 & 7 &  -2  & 0 \\ 0 & 1 & 2 &  5 &  3 \\ \end{array} \right] & \! \begin{aligned} & \xrightarrow{R_2 - R_4} \end{aligned} \left[\begin{array}{cccc|c} 15 & 2 & 0 & 7  & -1 \\ 0 & 1 & 2 & 5  & -1 \\ 2 & 4 & 7 &  -2  & 0 \\ 0 & 0 & 0 &  0 &  -4 \\ \end{array} \right]\]

 

The 4-th row gives the equation 0=-4, which is not possible, therefore, the given system of equations does not have solutions.

 

 

 

 

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