Question

Each matrix is nonsingular. Find the inverse of each matrix. $$ \left[\begin{array}{rrr} 1 & 0 & 2 \\ -1 & 2 & 3 \\ 1 & -1 & 0 \end{array}\right] $$

Short answer

The inverse matrix is \( \left[\begin{array}{ccc} 3 & -3 & 3 \ -3 & -2 & -1 \ -2 & -3 & 0 \end{array}\right] \).

Step-by-step solution

- Identify the Matrix

The given matrix is \[ A = \left[\begin{array}{rrr} 1 & 0 & 2 \ -1 & 2 & 3 \ 1 & -1 & 0 \end{array}\right] \]

- Compute the Determinant

Calculate the determinant of matrix \( A \) by expanding along the first row:\[ \text{det}(A) = 1 \cdot \left(2 \cdot 0 - 3 \cdot (-1)\right) - 0 \cdot \left(-1 \cdot 0 - 3 \cdot 1\right) + 2 \cdot \left(-1 \cdot (-1) - 2 \cdot 1\right) \]This simplifies to:\[ \text{det}(A) = 1 \cdot 3 - 0 \cdot (-3) + 2 \cdot (-1) = 3 + 0 - 2 = 1 \]

- Compute the Adjugate Matrix

Find the cofactor matrix of \( A \) and take its transpose to get the adjugate matrix. The cofactor matrix is calculated as follows:\[ \text{Cof}(A) = \left[\begin{array}{ccc} (2 \cdot 0 - 3 \cdot (-1)) & -(0 \cdot 0 - 3 \cdot 1) & (0 \cdot (-1) - 2 \cdot 1) \ -(1 \cdot 0 - 3 \cdot 1) & (1 \cdot 0 - 2 \cdot 1) & -(1 \cdot (-1) - (-1 \cdot 1)) \ (1 \cdot (-1) - (-1\cdot 2)) & -(1 \cdot 1 - (-1 \cdot -1)) & (1 \cdot 2 - (-1 \cdot -1))\end{array}\right] \]Simplifying further, we get:\[ \text{Cof}(A) = \left[\begin{array}{ccc} 3 & -3 & -2 \ -3 & -2 & -3 \ 3 & -1 & 0 \end{array}\right] \]Taking the transpose of this cofactor matrix to get the adjugate matrix, we obtain:\[ \text{adj}(A) = \left[\begin{array}{ccc} 3 & -3 & 3 \ -3 & -2 & -1 \ -2 & -3 & 0 \end{array}\right] \]

- Calculate the Inverse Matrix

To find the inverse of \( A \), use the formula: \[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \]Given \( \text{det}(A) = 1 \), the inverse of \( A \) is:\[ A^{-1} = 1 \cdot \left[\begin{array}{ccc} 3 & -3 & 3 \ -3 & -2 & -1 \ -2 & -3 & 0 \end{array}\right] = \left[\begin{array}{ccc} 3 & -3 & 3 \ -3 & -2 & -1 \ -2 & -3 & 0 \end{array}\right] \]

Key concepts

Determinant Calculation

To find the inverse of a matrix, the first step is determinant calculation. The determinant is a special number that can be calculated from a square matrix. For a 3x3 matrix, the determinant can be found by expanding along any row or column. Here’s how it’s done:

For the matrix \(\begin{bmatrix} 1 & 0 & 2 \ -1 & 2 & 3 \ 1 & -1 & 0 \end{bmatrix}\), we can expand along the first row:
\[\text{det}(A) = 1 \times (2 \times 0 - 3 \times (-1)) - 0 \times (-1 \times 0 - 3 \times 1) + 2 \times (-1 \times (-1) - 2 \times 1)\text{.}\]

Simplifying the equation, we get:
\[\text{det}(A) = 1 \times 3 - 0 \times (-3) + 2 \times (-1) = 3 + 0 - 2 = 1\text{.}\]

When det(A) ≠ 0, the matrix is nonsingular and has an inverse.

Adjugate Matrix

The adjugate matrix is an essential step in finding the inverse of a matrix. It involves creating a cofactor matrix and then taking its transpose.

For the matrix \(\begin{bmatrix} 1 & 0 & 2 \ -1 & 2 & 3 \ 1 & -1 & 0 \end{bmatrix}\), we first find the cofactor matrix. The cofactor matrix is computed by taking the determinant of the 2x2 minor matrices obtained by removing the row and column of each element, and then applying a sign based on the element's position.

Once we have the cofactor matrix \(\text{Cof}(A) = \begin{bmatrix} 3 & -3 & -2 \ -3 & -2 & -3 \ 3 & -1 & 0 \end{bmatrix}\), we take its transpose to get the adjugate matrix. So,

\(\text{adj}(A) = \begin{bmatrix} 3 & -3 & 3 \ -3 & -2 & -1 \ -2 & -3 & 0 \end{bmatrix}\).

Cofactor Matrix

The cofactor matrix is a matrix where each element is replaced by its respective cofactor. The cofactor of an element is the determinant of the 2x2 matrix that remains after removing the element's row and column, multiplied by a sign depending on its position.

Signs are determined by the position (i, j) of each element, following the pattern:
\[ (-1)^{i+j} \]

For example, in matrix \(\begin{bmatrix} 1 & 0 & 2 \ -1 & 2 & 3 \ 1 & -1 & 0 \end{bmatrix}\), the cofactor for the element in the first row and first column is calculated using:

The minor matrix for element (1,1) is \(\begin{bmatrix} 2 & 3 \ -1 & 0 \end{bmatrix}\)

Its determinant is:
\(\text{det} = 2 \times 0 - 3 \times (-1) = 3 \)

With the sign, the cofactor becomes \(3\text{.}\) Repeat this process for every element to fill the cofactor matrix.

Nonsingular Matrix

For matrix operations, especially finding the inverse, it is crucial that the matrix is nonsingular. A nonsingular matrix is one that has a non-zero determinant. This ensures that the matrix has an inverse.

For a 3x3 matrix like \(\begin{bmatrix} 1 & 0 & 2 \ -1 & 2 & 3 \ 1 & -1 & 0 \end{bmatrix}\), if the determinant is non-zero, the matrix is nonsingular and invertible.

This property assures:

• The matrix represents a system of linear equations with a unique solution.
• The rows (or columns) of the matrix are linearly independent.

A nonsingular matrix is also called an invertible matrix. If a matrix is singular (its determinant is zero), it does not have an inverse, making solving certain equations impossible.