Question

Each matrix is nonsingular. Find the inverse of each matrix. $$ \left[\begin{array}{rr} 3 & -1 \\ -2 & 1 \end{array}\right] $$

Short answer

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

Step-by-step solution

- Determine the Determinant

Calculate the determinant of the matrix to ensure it is nonsingular. The determinant \text{det}(A) of matrix \( A = \left[\begin{array}{rr} 3 & -1 \ -2 & 1 \end{array}\right] \) is given by \[ \text{det}(A) = (3)(1) - (-1)(-2) \] Simplifying this, we get: \[ \text{det}(A) = 3 - 2 = 1 \]

- Swap the Diagonal Elements

For a 2x2 matrix, the inverse can be found by swapping the elements on the main diagonal. For our matrix, the main diagonal elements are 3 and 1. After swapping, the new main diagonal of the matrix becomes: \[ \left[ \begin{array}{rr} 1 & -1 \ -2 & 3 \end{array} \right] \]

- Change the Sign of Non-Diagonal Elements

Next, change the signs of the off-diagonal elements of the matrix. The non-diagonal elements are -1 and -2, so we get: \[ \left[ \begin{array}{rr} 1 & 1 \ 2 & 3 \end{array} \right] \]

- Divide by the Determinant

Finally, we need to divide every element of the matrix by the determinant. Since the determinant is 1, the matrix remains the same: \( A^{-1} = \frac{1}{1} \left[ \begin{array}{rr} 1 & 1 \ 2 & 3 \end{array} \right] \). Hence, \( A^{-1} = \left[ \begin{array}{rr} 1 & 1 \ 2 & 3 \end{array} \right] \)

Key concepts

nonsingular matrix

A nonsingular matrix, also known as an invertible or non-degenerate matrix, is a crucial concept in linear algebra. It refers to a square matrix that has an inverse. This means there exists another matrix that, when multiplied with the original, results in the identity matrix.
The importance of a nonsingular matrix lies in its determinant. The determinant of a nonsingular matrix is always nonzero. This property distinguishes it from singular matrices, which do not have an inverse.
In solving matrix equations, confirming that a matrix is nonsingular guarantees that a unique solution exists. For example, in our exercise, the given matrix must be nonsingular to proceed with finding its inverse.

determinant

The determinant is a special number associated with a square matrix. It's a scalar value that provides essential information about the matrix, such as whether it is nonsingular or singular. For a 2x2 matrix, the determinant can be calculated using the formula:
\[ \text{det}(A) = ad - bc \] where \ a, b, c, \ and \ d \ are the elements of the 2x2 matrix. In the given exercise, the matrix is: \[ \begin{pmatrix} 3 & -1 \-2 & 1 \ \end{pmatrix} \] The determinant is calculated as follows: \[ \text{det}(A) = (3)(1) - (-1)(-2) = 3 - 2 = 1 \] A nonzero determinant confirms the matrix is nonsingular, allowing us to find its inverse.

2x2 matrix inverse

The inverse of a 2x2 matrix is a process that follows a specific set of steps. Here is a detailed breakdown with the matrix \ \ \[ \begin{pmatrix} 3 & -1 \-2 & 1 \end{pmatrix} \]:

• Step 1 - Find the Determinant: As shown earlier, \ \[ \text{det}(A) = 1 \]
• Step 2 - Swap the Diagonal Elements: Swap elements \( a \) and \( d \). This modifies our matrix to: \ \[ \begin{pmatrix} 1 & -1 \-2 & 3 \end{pmatrix} \]
• Step 3 - Change the Sign of Non-Diagonal Elements: Change the signs of off-diagonal elements \( b \) and \( c \). Our matrix now becomes: \ \[ \begin{pmatrix} 1 & 1 \ 2 & 3 \end{pmatrix} \]
• Step 4 - Divide by the Determinant: Finally, divide each element by the determinant 1, so our inverse matrix remains: \ \[ A^{-1} = \begin{pmatrix} 1 & 1 \ 2 & 3 \end{pmatrix} \]

The resulting matrix is indeed the inverse, illustrating the step-by-step process clearly.