Finding the inverse of a 2×2 matrix is a simple process that begins by determining whether the matrix is actually invertible. If the matrix is invertible, we swap the positions of the elements on the main diagonal, change the signs of the off-diagonal elements, and then divide each element by the determinant of the original matrix.
Below, we will explore this through some tangible examples with detailed solutions. We have also included practice problems for you to try finding the inverse of a 2×2 matrix.
How to find the inverse matrix of a 2×2 matrix?
The inverse matrix of a 2×2 matrix is found by dividing each element of the adjoint matrix by the determinant of the original matrix.
Suppose we have the following 2×2 matrix:
$$ A = \begin{pmatrix}a & b \\c & d\end{pmatrix}$$
To find its inverse matrix, we follow the steps below:
Step 1: Calculate the determinant of the matrix (denoted as det(A) or |A|):
$latex \det(A) = ad – bc $
Step 2: Check whether the determinant is non-zero (i.e. det(A) ≠ 0). If the determinant is zero, the matrix A is singular and has no inverse.
If the determinant is non-zero, we proceed to the next step.
Step 3: Create the adjoint matrix by swapping the elements $latex a$ and $latex d$, and changing the signs of the elements $latex b$ and $latex c$:
$$ \operatorname{adj}(A) = \begin{pmatrix}d & -b \\-c & a\end{pmatrix}$$
Step 4: Divide each element of the adjoint matrix by the determinant:
$$A^{-1} = \frac{1}{\det(A)} \cdot \operatorname{adj}(A)$$
$$A^{-1} = \begin{pmatrix}\frac{d}{\det(A)} & -\frac{b}{\det(A)} \\-\frac{c}{\det(A)} & \frac{a}{\det(A)}\end{pmatrix}$$
The result of the inverse matrix of matrix A, denoted as $latex A^{-1}$.
Solved exercises on inverse matrix of 2×2 matrices
EXAMPLE 1
Find the inverse matrix of the given 2×2 matrix:
$$ A = \begin{pmatrix} 3 & 4 \\ 2 & 5 \end{pmatrix} $$
Solution
Step 1: Calculate the determinant of the matrix (det(A) or |A|):
$latex \det(A) = ad – bc $
$latex = (3)(5) – (4)(2) $
$latex = 15 – 8 = 7$
Step 2: Check if the determinant is non-zero (i.e., $latex \det(A) \neq 0$).
$latex \det(A) = 7 \neq 0$
Since the determinant is non-zero, we move on to the next step.
Step 3: Create the adjoint matrix:
$$\operatorname{adj}(A) = \begin{pmatrix}5 & -4 \\-2 & 3\end{pmatrix}$$
Step 4: Divide each element of the adjoint matrix by the determinant:
$$A^{-1} = \frac{1}{\det(A)} \cdot \operatorname{adj}(A)$$
$$A^{-1} = \frac{1}{7} \begin{pmatrix}5 & -4 \\-2 & 3\end{pmatrix}$$
$$A^{-1} = \begin{pmatrix}\frac{5}{7} & -\frac{4}{7} \\-\frac{2}{7} & \frac{3}{7}\end{pmatrix}$$
EXAMPLE 2
Find the inverse matrix of the following matrix:
$$A = \begin{pmatrix} 7 & 2 \\ 3 & 4 \end{pmatrix} $$
Solution
Start by calculating the determinant:
$latex \det(A) = (7)(4) – (2)(3)$
$latex = 28 – 6 = 22$
Since the determinant is non-zero, proceed to find the adjoint matrix:
$$\operatorname{adj}(A) = \begin{pmatrix}4 & -2 \\-3 & 7\end{pmatrix}$$
Divide each element of the adjoint matrix by the determinant:
$$A^{-1} = \frac{1}{\det(A)} \cdot \operatorname{adj}(A)$$
$$A^{-1} = \frac{1}{22} \begin{pmatrix}4 & -2 \\-3 & 7\end{pmatrix}$$
$$A^{-1} = \begin{pmatrix}\frac{4}{22} & -\frac{2}{22} \\ -\frac{3}{22} & \frac{7}{22}\end{pmatrix}$$
$$A^{-1} = \begin{pmatrix}\frac{2}{11} & -\frac{1}{11} \\ -\frac{3}{22} & \frac{7}{22}\end{pmatrix}$$
EXAMPLE 3
What is the inverse matrix of matrix A?
$$A = \begin{pmatrix}6 & -3 \\4 & -2\end{pmatrix}$$
Solution
Calculating the determinant, we have:
$latex \det(A) = (6)(-2) – (-3)(4)$
$latex = -12 + 12 = 0$
Since the determinant is 0, matrix A has no inverse.
EXAMPLE 4
If we have the following matrix M, find its inverse matrix:
$$M = \begin{pmatrix}5 & 1 \\2 & 3\end{pmatrix}$$
Solution
Start by calculating the determinant of the matrix:
$latex \det(M) = (5)(3) – (1)(2)$
$latex = 15 – 2 = 13$
Now, find the adjoint matrix:
$$\operatorname{adj}(M) = \begin{pmatrix}3 & -1 \\-2 & 5\end{pmatrix}$$
Dividing each element of the attached matrix by the determinant, we have:
$$M^{-1} = \frac{1}{\det(M)} \cdot \operatorname{adj}(M)$$
$$M^{-1} = \frac{1}{13} \begin{pmatrix}3 & -1 \\-2 & 5\end{pmatrix}$$
$$M^{-1} = \begin{pmatrix}\frac{3}{13} & -\frac{1}{13} \\-\frac{2}{13} & \frac{5}{13}\end{pmatrix}$$
EXAMPLE 5
Determine the inverse matrix of matrix B:
$$B = \begin{pmatrix}8 & 5 \\-6 & -3\end{pmatrix}$$
Solution
The determinant of matrix B is:
$latex \det(B) = (8)(-3) – (5)(-6)$
$latex = -24 + 30 = 6$
The adjoint matrix of B is:
$$\operatorname{adj}(B) = \begin{pmatrix}-3 & -5 \\6 & 8\end{pmatrix}$$
Dividing each element of the adjoint matrix of B by the determinant, we have:
$$B^{-1} = \frac{1}{\det(B)} \cdot \operatorname{adj}(B)$$
$$B^{-1} = \frac{1}{6} \begin{pmatrix}-3 & -5 \\6 & 8\end{pmatrix}$$
$$B^{-1} = \begin{pmatrix}-\frac{1}{2} & -\frac{5}{6} \\1 & \frac{4}{3}\end{pmatrix}$$
EXAMPLE 6
What is the inverse of the given matrix?
$$A = \begin{pmatrix}1 & 3 \\4 & 2\end{pmatrix}$$
Solution
Start by calculating the determinant of the matrix A:
$latex \det(A) = (1)(2) – (3)(4)$
$latex = 2 – 12 = -10$
The adjoint matrix of A is:
$$\operatorname{adj}(A) = \begin{pmatrix}2 & -3 \\-4 & 1\end{pmatrix}$$
Dividing each element of the adjoint matrix by the determinant, we have:
$$A^{-1} = \frac{1}{\det(A)} \cdot \operatorname{adj}(A)$$
$$A^{-1} = \frac{1}{-10} \begin{pmatrix}2 & -3 \\-4 & 1\end{pmatrix}$$
$$A^{-1} = \begin{pmatrix}-\frac{1}{5} & \frac{3}{10} \\ \frac{2}{5} & -\frac{1}{10}\end{pmatrix}$$
EXAMPLE 7
Find the inverse matrix of matrix C:
$$C = \begin{pmatrix}3 & -7 \\-1 & 2\end{pmatrix}$$
Solution
Calculate the determinant of the matrix C:
$latex \det(C) = (3)(2) – (-7)(-1) $
$latex = 6 – 7 = -1$
The adjoint matrix of C is:
$$\operatorname{adj}(C) = \begin{pmatrix}2 & 7 \\1 & 3\end{pmatrix}$$
When we divide each element of the adjoint matrix of C by the determinant, we have:
$$C^{-1} = \frac{1}{\det(C)} \cdot \operatorname{adj}(C)$$
$$C^{-1} = \frac{1}{-1} \begin{pmatrix}2 & 7 \\1 & 3\end{pmatrix}$$
$$C^{-1} = \begin{pmatrix}-2 & -7 \\-1 & -3\end{pmatrix}$$
EXAMPLE 8
If we have the following matrix, what is its inverse matrix?
$$A = \begin{pmatrix}-1 & 4 \\5 & -2\end{pmatrix}$$
Solution
Find the determinant of the matrix:
$latex \det(A) = (-1)(-2) – (4)(5)$
$latex = 2 – 20 = -18$
Now, find the adjoint matrix of A:
$$\operatorname{adj}(A) = \begin{pmatrix}-2 & -4 \\-5 & -1\end{pmatrix}$$
Finally, divide each element of the adjoint matrix by the determinant:
$$A^{-1} = \frac{1}{\det(A)} \cdot \operatorname{adj}(A)$$
$$A^{-1} = \frac{1}{-18} \begin{pmatrix}-2 & -4 \\-5 & -1\end{pmatrix}$$
$$A^{-1} = \begin{pmatrix}\frac{1}{9} & \frac{2}{9} \\ \frac{5}{18} & \frac{1}{18}\end{pmatrix}$$
EXAMPLE 9
Find the inverse matrix of the following matrix:
$$ A = \begin{pmatrix} \frac{1}{2} & \frac{5}{4} \\ \frac{7}{4} & -\frac{3}{2} \end{pmatrix} $$
Solution
Start by finding the determinant of the matrix:
$$\det(A) = \left(\frac{1}{2}\right)\left(-\frac{3}{2}\right) – \left(\frac{5}{4}\right)\left(\frac{7}{4}\right)$$
$$ = -\frac{3}{4} – \frac{35}{16} = -\frac{47}{16}$$
Now, find the adjoint matrix:
$$\operatorname{adj}(A) = \begin{pmatrix} -\frac{3}{2} & -\frac{5}{4} \\ -\frac{7}{4} & \frac{1}{2}\end{pmatrix}$$
Divide each element of the attached matrix by the determinant:
$$A^{-1} = \frac{1}{\det(A)} \cdot \operatorname{adj}(A)$$
$$A^{-1} = \frac{1}{-\frac{47}{16}} \begin{pmatrix} -\frac{3}{2} & -\frac{5}{4} \\-\frac{7}{4} & \frac{1}{2}\end{pmatrix}$$
$$ A^{-1} = -\frac{16}{47} \begin{pmatrix} -\frac{3}{2} & -\frac{5}{4} \\ -\frac{7}{4} & \frac{1}{2} \end{pmatrix} $$
$$ A^{-1} = \begin{pmatrix} \frac{48}{94} & \frac{80}{94} \\ \frac{112}{94} & \frac{16}{94} \end{pmatrix} $$
$$ A^{-1} = \begin{pmatrix} \frac{24}{47} & \frac{40}{47} \\ \frac{56}{47} & \frac{8}{47} \end{pmatrix} $$
EXAMPLE 10
Given the following matrix, find its inverse matrix:
$$M = \begin{pmatrix}\sqrt{2} & -\sqrt{3} \\2\sqrt{3} & -\sqrt{6}\end{pmatrix}$$
Solution
Calculamos el determinante:
$$\det(A) = (\sqrt{2})(-\sqrt{6}) – (-\sqrt{3})(2\sqrt{3})$$
$$ = -2\sqrt{3} + 6 = 6 – 2\sqrt{3}$$
Encontramos la matriz adjunta:
$$\operatorname{adj}(A) = \begin{pmatrix}-\sqrt{6} & \sqrt{3} \\ -2\sqrt{3} & \sqrt{2}\end{pmatrix}$$
Dividimos a cada elemento de la matriz adjunta por el determinante:
$$A^{-1} = \frac{1}{\det(A)} \cdot \operatorname{adj}(A)$$
$$A^{-1} = \frac{1}{6 – 2\sqrt{3}} \begin{pmatrix}-\sqrt{6} & \sqrt{3} \\-2\sqrt{3} & \sqrt{2}\end{pmatrix}$$
Racionalizamos el denominador:
$$A^{-1} = \frac{6 + 2\sqrt{3}}{(6 – 2\sqrt{3})(6 + 2\sqrt{3})} \begin{pmatrix}-\sqrt{6} & \sqrt{3} \\-2\sqrt{3} & \sqrt{2}\end{pmatrix}$$
$$A^{-1} = \frac{6 + 2\sqrt{3}}{24} \begin{pmatrix}-\sqrt{6} & \sqrt{3} \\-2\sqrt{3} & \sqrt{2}\end{pmatrix}$$
$$A^{-1} = \begin{pmatrix} \frac{6 + 2\sqrt{3}}{24} (-\sqrt{6}) & \frac{6 + 2\sqrt{3}}{24} (\sqrt{3}) \\ \frac{6 + 2\sqrt{3}}{24} (-2\sqrt{3}) & \frac{6 + 2\sqrt{3}}{24} (\sqrt{2}) \end{pmatrix} $$
$$A^{-1} = \begin{pmatrix} \frac{-6\sqrt{6} – 2\sqrt{18}}{24} & \frac{6\sqrt{3} + 2\sqrt{9}}{24} \\ \frac{-12\sqrt{3} – 4\sqrt{9}}{24} & \frac{6\sqrt{2} + 2\sqrt{6}}{24} \end{pmatrix}$$
$$A^{-1} = \begin{pmatrix} \frac{-6\sqrt{6} – 6\sqrt{2}}{24} & \frac{6\sqrt{3} + 6}{24} \\ \frac{-12\sqrt{3} – 12}{24} & \frac{6\sqrt{2} + 2\sqrt{6}}{24} \end{pmatrix} $$
$$A^{-1}= \begin{pmatrix} \frac{-\sqrt{6}-\sqrt{2}}{4}& \frac{\sqrt{3}+1}{4} \\ \frac{-\sqrt{3}-1}{2} & \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{12} \end{pmatrix} $$
Inverse of 2×2 matrices – Practice problems
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See also
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–
Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.
