4.7 Inverse Matrix – user's Blog!

4.7 Inverse Matrix

1. If A is a square matrix, B is another square matrix and A × B = B × A = I, then matrix A is the inverse matrix of matrix B and vice versa. Matrix A is called the inverse matrix of B for multiplication and vice versa.

2. The symbol A^-1 denotes the inverse matrix of A.

3. Inverse matrices can only exist for square matrices but not all square matrices have inverse matrices.

4. If AB ≠ I or BA ≠ I, then A is not the inverse of B and B is not the inverse of A.

Example 1:

Determine whether matrix

$A = (\begin{matrix} 2 & 9 \\ 1 & 5 \end{matrix})$ is an inverse matrix of matrix

$B = (\begin{matrix} 5 & - 9 \\ - 1 & 2 \end{matrix}) .$

Solution:

$\begin{array}{l} A B = (\begin{matrix} 2 & 9 \\ 1 & 5 \end{matrix}) (\begin{matrix} 5 & - 9 \\ - 1 & 2 \end{matrix}) \\ = (\begin{matrix} 2 \times 5 + 9 \times - 1 & 2 \times - 9 + 9 \times 2 \\ 1 \times 5 + 5 \times - 1 & 1 \times - 9 + 5 \times 2 \end{matrix}) \\ = (\begin{matrix} 10 + (- 9) & - 18 + 18 \\ 5 + (- 5) & - 9 + 10 \end{matrix}) \\ = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) = I \\ A B = (\begin{matrix} 5 & - 9 \\ - 1 & 2 \end{matrix}) (\begin{matrix} 2 & 9 \\ 1 & 5 \end{matrix}) \\ = (\begin{matrix} 5 \times 2 + (- 9) \times 1 & 5 \times 9 + (- 9) \times 5 \\ - 1 \times 2 + 2 \times 1 & - 1 \times 9 + 2 \times 5 \end{matrix}) \\ = (\begin{matrix} 10 + (- 9) & 18 - 18 \\ - 2 + 2 & - 9 + 10 \end{matrix}) \\ = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) = I \\ A B = B A = I \\ ∴ A is the inverse matrix of B and vice versa . \end{array}$

5. The inverse of a matrix may also be found using a formula.

If

$A = (\begin{matrix} a & b \\ c & d \end{matrix})$ , then the inverse matrix of A, A^-1, is given by the formula below.

$A^{- 1} = \frac{1}{a d - b c} (\begin{matrix} d & - b \\ - c & a \end{matrix}), where a d - b c \neq 0$

6. ad – bc is known as the determinant of matrix A.

7. If the determinant, ad – bc = 0, then the inverse matrix of A does not exist.

Example 2:

Find the inverse matrix of

$A = (\begin{matrix} 6 & 1 \\ - 9 & - 1 \end{matrix})$ using the formula.

Solution:

$\begin{array}{l} A = (\begin{matrix} 6 & 1 \\ - 9 & - 1 \end{matrix}) \\ a = 6, b = 1, c = - 9, d = - 1 \\ A^{- 1} = \frac{1}{a d - b c} (\begin{matrix} d & - b \\ - c & a \end{matrix}) \\ A^{- 1} = \frac{1}{6 \times - 1 - (1 \times - 9)} (\begin{matrix} - 1 & - 1 \\ 9 & 6 \end{matrix}) \\ A^{- 1} = \frac{1}{- 6 + 9} (\begin{matrix} - 1 & - 1 \\ 9 & 6 \end{matrix}) \\ A^{- 1} = \frac{1}{3} (\begin{matrix} - 1 & - 1 \\ 9 & 6 \end{matrix}) = (\begin{matrix} - \frac{1}{3} & - \frac{1}{3} \\ 3 & 2 \end{matrix}) \end{array}$

Example 3:

The inverse matrix of

$(\begin{matrix} - 7 & 2 \\ - 9 & 2 \end{matrix}) is r (\begin{matrix} 2 & s \\ 9 & t \end{matrix}) .$ Find the value of r, of s and of t.

Solution:

$\begin{array}{l} Let A = (\begin{matrix} - 7 & 2 \\ - 9 & 2 \end{matrix}) \\ A^{- 1} = \frac{1}{- 7 \times 2 - (- 9) \times 2} (\begin{matrix} 2 & - 2 \\ 9 & - 7 \end{matrix}) \\ A^{- 1} = \frac{1}{4} (\begin{matrix} 2 & - 2 \\ 9 & - 7 \end{matrix}) \\ ∴ r (\begin{matrix} 2 & s \\ 9 & t \end{matrix}) = \frac{1}{4} (\begin{matrix} 2 & - 2 \\ 9 & - 7 \end{matrix}) \\ By comparison, \\ r = \frac{1}{4}, s = - 2, t = - 7. \end{array}$