Description of matrices and simultaneous equations

There is a quick method for obtaining an inverse for a 2 x 2 matrix. This is a special case of Cramer’s rule used to solve sets of equations.

Die Inverse von \(\displaystyle \begin{bmatrix}a & b \\ c & d \end{bmatrix}\) ist \(\displaystyle \begin{bmatrix}a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad-bc}\begin{bmatrix}d & -b\\-c & a\end{bmatrix}\)

There are three steps to invert a 2x2 matrix

Swap the diagonal elements

Change the sign of the other elements

Divide each element by \(ad-bc\)

If for example in the formula above \(ad = bc\), then we would be attempting to divide by zero and consequently there is no inverse. In this case we call the original matrix \(A\) a singular matrix. If the matrix has an inverse, we say that the matrix is non-singular.

Another way to can get \(ad = bc\)is, if the second row of the matrix is a multiple of the first.

Without actually calculating the inverse matrix, we can make a decision whether an inverse exists by just calculating a single number, the denominator in the formula. This denominator is called the determinant.

If the determinate is zero, it is a singular matrix, which cannot be inverted.

Cramer’s rule also exists for larger matrices but is computationally very inefficient. Therefore, it is helpful especially for large matrices if we can determine before starting the calculation, whether the inverse exists. This can be done by defining also for large matrices a single number that characterizes the matrix – again, it is called the determinant of the matrix.

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The matrix multiplication can be used to calculate simultaneous equations.

The equations

\(x+y=3\)

\(2x+3y=1\)

is conformed to the following matrix expression by applying the multiplication rule

\(\displaystyle \begin{bmatrix}1 & 1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix} =\begin{bmatrix}3 \\ 1 \end{bmatrix}\)

The way that multiplying matrices allows a whole set of linear equations to be written as a single equation, containing matrices instead of numbers.

If the symbols in this equation are numbers, it is easily solve it by division. Even though we have symbols, we can avoid using division, when we multiply with reciprocal value. Instead of dividing by \(x\), we can just as well multiply by \(x^{-1}\).

The inverse of a matrix is not easy to calculate, and you can only find an inverse if the matrix is a square matrix and even then, not always.

In this case above the inverse of \(\displaystyle \begin{bmatrix}1 & 1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix}3 & -1 \\ -2 & 1 \end{bmatrix}\)

Now you can use the inverse above to calculate the values of \(x\) and \(y\).

\(\displaystyle \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix}3 & -1 \\ -2 & 1 \end{bmatrix} · \begin{bmatrix} 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 8 \\ -5 \end{bmatrix}\)

The result is \(x = 8\) and \(y = -5\)

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