Ideas
Inverse notation
In mathematics, the inverse of a number is the number needed so that when you multiply them together the result is 1. In matrices this would be equivalent to the identity matrix, I.
For instance, 6\times \dfrac{1}{6}=1, where \dfrac16and 6 are inverses of one another. The concept of an inverse is often used when solving equations, like solving 5x=20.
| \displaystyle 5x | \displaystyle = | \displaystyle 20 | Write the equation |
| \displaystyle 5x \times \dfrac15 | \displaystyle = | \displaystyle 20 \times \dfrac15 | Multiply both sides by the inverse of 5 |
| \displaystyle x | \displaystyle = | \displaystyle 4 | Evaluate both sides |
The inverse of 6 can also be written as 6^{-1}. As with this example or when using \sin^{-1} to represent inverse sine, we have that:
A^{-1} represents the inverse of matrix A.
The inverse of a matrix A is represented by A^{-1}.
Determinant of a 2 x 2 matrix
The determinant is a number that is used to find the inverse of a matrix (if it exists). If the determinant is zero then the inverse is said to be undefined, this can be useful when solving systems of linear equations later.
Generally, the determinant of a matrix A is written as det(A) and is calculated as follows:
For a matrix A= \begin{bmatrix} a & b \\ c & d \end{bmatrix} the determinant is det(A)= \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad- cb .
Examples
Example 1
Evaluate the determinant \begin{vmatrix} -4 & -6 \\ 3 & 1 \end{vmatrix} .
For a matrix A= \begin{bmatrix} a & b \\ c & d \end{bmatrix} the determinant is det(A)= \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad- cb .
Inverse of a 2 x 2 matrix
To find the inverse of a 2 \times 2 matrix, say A, we swap the entries along the main-diagonal, and multiply the entries in the off-diagonal by -1. Then we multiply the result by \dfrac{1}{det(A)}.
For a matrix A= \begin{bmatrix} a & b \\ c & d \end{bmatrix} , the inverse is A^{-1} = \dfrac{1}{det(A)} \begin{vmatrix} d & -b \\ -c & a \end{vmatrix} .
The inverse matrix A^{-1} has the property that when we multiply it by A, we get the identity matrix.
AA^{-1}=A^{-1}A=I
Examples
Example 2
Consider the matrix A = \begin{bmatrix} -8 & 2 \\ -1 & 9 \end{bmatrix} .
Find the determinant of A.
Find the inverse A^{-1}.
The inverse of a matrix A= \begin{bmatrix} a & b \\ c & d \end{bmatrix} is given by A^{-1} = \dfrac{1}{det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} .
Matrix equations
Just as we have equations that we can solve using algebraic manipulation, we also have matrix equations we can solve using manipulation according to the laws of matrices.
We know that AA^{-1}=A^{-1}A=I where A^{-1} is the inverse of A and I is the identity matrix.
Examples
Example 3
Consider the following matrix equation.
X + \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix} = 3X - \begin{bmatrix} 0 & 4 \\ 3 & 3 \end{bmatrix}
Solve the equation for X.
Solving a matrix equation is like solving an ordinary equation.
Division of matrices
Division for matrices is actually not defined, but this doesn't mean we can't manipulate equations. Anything that we would normally need to do using division, we instead do multiplication by using the inverse, provided the inverse of a matrix exists.
Because AA^{-1}=A^{-1}A we can use this property to simplify by carry out pre-multiplication and post-multiplication in equations instead of division. This is because we are aiming to multiply a matrix and its inverse to form the identity matrix.
For instance consider the matrix equation AX=B. To solve for X we need to pre-multiply. This means placing A^{-1} out the front of both sides of the equation.
| \displaystyle AX | \displaystyle = | \displaystyle B | Write down the equation |
| \displaystyle A^{-1}AX | \displaystyle = | \displaystyle A^{-1}B | Pre-multiply both sides by A^{-1} |
| \displaystyle IX | \displaystyle = | \displaystyle A^{-1}B | Substitute A^{-1}A=I |
| \displaystyle X | \displaystyle = | \displaystyle A^{-1}B | Substitute IX=X |
For the matrix equation XA=B, to solve for X we need to post-multiply. This means placing A^{-1} at the end of both sides of the equation.
XAA^{-1}=BA^{-1}. Here A^{-1} is post-multiplied on both sides.
| \displaystyle XA | \displaystyle = | \displaystyle B | Write down the equation |
| \displaystyle XAA^{-1} | \displaystyle = | \displaystyle BA^{-1} | Post-multiply both sides by A^{-1} |
| \displaystyle XI | \displaystyle = | \displaystyle BA^{-1} | Substitute AA^{-1}=I |
| \displaystyle X | \displaystyle = | \displaystyle BA^{-1} | Substitute XI=X |
Examples
Example 4
A,B, and C are matrices such that AB=C. Using matrix algebra, fill in the gaps to solve for matrix B.
Multiply both sides of the equation by the inverse of ⬚:
(⬚)^{-1} ⬚B = (⬚)^{-1}C
The product of any matrix and its inverse results in the identity matrix:
⬚B = (⬚)^{-1}C
The product of any matrix and the identity matrix is the matrix itself:
⬚ = (⬚)^{-1}C
Since division of matrices is not defined, we can pre-multiply or post-multiply the inverse of the matrix that is multiplied to the matrix that needs to be solved.