Question

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

Short answer

The inverse matrix is \[\begin{bmatrix} \frac{1}{7} & \frac{1}{7} & \frac{1}{7} \ \frac{3}{7} & \frac{1}{7} & -\frac{4}{7} \ \frac{3}{7} & -\frac{2}{7} & \frac{1}{7} \ \end{bmatrix} \].

Step-by-step solution

Identify the matrix

The matrix given is \[ \begin{bmatrix} 1 & 1 & 1 \ 3 & 2 & -1 \ 3 & 1 & 2 \end{bmatrix} \] . To find the inverse, we need to follow some systematic steps.

Create the augmented matrix

Augment the given matrix with the identity matrix of the same order (3x3). This results in \[\begin{bmatrix} 1 & 1 & 1 & | & 1 & 0 & 0 \ 3 & 2 & -1 & | & 0 & 1 & 0 \ 3 & 1 & 2 & | & 0 & 0 & 1 \ \end{bmatrix} \].

Perform row operations to get the left side into the identity matrix

Apply row operations to convert the matrix on the left side to the identity matrix (row reduction technique).

Row operation 1

Make the first element of the first row a 1 (it already is in this case), but ensure the leading coefficient below it is zero. To make the element in the second row, first column a zero: \( R_2 = R_2 - 3R_1 \). This gives:\[\begin{bmatrix} 1 & 1 & 1 & | & 1 & 0 & 0 \ 0 & -1 & -4 & | & -3 & 1 & 0 \ 3 & 1 & 2 & | & 0 & 0 & 1 \ \end{bmatrix} \].

Row operation 2

Make the element in the third row, first column a zero: \( R_3 = R_3 - 3R_1 \). Now the matrix appears as:\[\begin{bmatrix} 1 & 1 & 1 & | & 1 & 0 & 0 \ 0 & -1 & -4 & | & -3 & 1 & 0 \ 0 & -2 & -1 & | & -3 & 0 & 1 \ \end{bmatrix} \].

Simplify the second row

Make the leading coefficient in the second row a 1 by dividing the whole row by -1:\( R_2 = \frac{R_2}{-1} \). This gives:\[\begin{bmatrix} 1 & 1 & 1 & | & 1 & 0 & 0 \ 0 & 1 & 4 & | & 3 & -1 & 0 \ 0 & -2 & -1 & | & -3 & 0 & 1 \ \end{bmatrix} \].

Row operation 3

Eliminate the second column, third row element by adding twice the second row to the third row:\( R_3 = R_3 + 2R_2 \). Thus, we have:\[\begin{bmatrix} 1 & 1 & 1 & | & 1 & 0 & 0 \ 0 & 1 & 4 & | & 3 & -1 & 0 \ 0 & 0 & 7 & | & 3 & -2 & 1 \ \end{bmatrix} \].

Normalize the third row

Divide the third row by 7 to obtain a leading 1:\( R_3 = \frac{R_3}{7} \). This results in:\[\begin{bmatrix} 1 & 1 & 1 & | & 1 & 0 & 0 \ 0 & 1 & 4 & | & 3 & -1 & 0 \ 0 & 0 & 1 & | & \frac{3}{7} & -\frac{2}{7} & \frac{1}{7} \ \end{bmatrix} \].

Make third column zeros above the leading '1's

Update the first and second rows to make the elements above the leading 1 in the third row equal to zero.First row: \( R_1 = R_1 - R_3 \), so \[\begin{bmatrix} 1 & 1 & 0 & | & \frac{4}{7} & \frac{2}{7} & -\frac{1}{7} \ 0 & 1 & 4 & | & 3 & -1 & 0 \ 0 & 0 & 1 & | & \frac{3}{7} & -\frac{2}{7} & \frac{1}{7} \ \end{bmatrix} \].Second row: \( R_2 = R_2 - 4R_3 \), so \[\begin{bmatrix} 1 & 1 & 0 & | & \frac{4}{7} & \frac{2}{7} & -\frac{1}{7} \ 0 & 1 & 0 & | & \frac{3}{7} & \frac{1}{7} & -\frac{4}{7} \ 0 & 0 & 1 & | & \frac{3}{7} & -\frac{2}{7} & \frac{1}{7} \ \end{bmatrix} \].

Make first and second column zeros above leading '1's

Finally update the first row to make the element in the second column, first row equal to zero: \( R_1 = R_1 - R_2 \), resulting in:\[\begin{bmatrix} 1 & 0 & 0 & | & \frac{1}{7} & \frac{1}{7} & \frac{1}{7} \ 0 & 1 & 0 & | & \frac{3}{7} & \frac{1}{7} & -\frac{4}{7} \ 0 & 0 & 1 & | & \frac{3}{7} & -\frac{2}{7} & \frac{1}{7} \ \end{bmatrix} \].

Write the inverse matrix

The inverse of the matrix is the right-hand side of this augmented matrix:\[\begin{bmatrix} \frac{1}{7} & \frac{1}{7} & \frac{1}{7} \ \frac{3}{7} & \frac{1}{7} & -\frac{4}{7} \ \frac{3}{7} & -\frac{2}{7} & \frac{1}{7} \ \end{bmatrix} \].

Key concepts

Nonsingular Matrix

A nonsingular matrix, also called an invertible or non-degenerate matrix, is a square matrix that has an inverse. This means the matrix has a full rank, or in simpler terms, there is a unique solution to the matrix equation \( AX = I \). For a matrix to be nonsingular, its determinant must be non-zero. When you have a nonsingular matrix, you can find another matrix such that when you multiply them, you get the identity matrix. This process is critical in solving systems of linear equations and various mathematical applications. Every step in finding the inverse of a matrix assumes that the matrix is nonsingular.

In our exercise, we begin with the given matrix and identify it as the subject for which we wish to find the inverse. The property that it's nonsingular guarantees that an inverse matrix exists.

Augmented Matrix

An augmented matrix is formed by appending the identity matrix to the original matrix. This is a crucial step in finding the inverse of a matrix using row operations.
In the given solution, the augmented matrix was created from the original matrix and the 3x3 identity matrix:
\[ \begin{bmatrix} 1 & 1 & 1 & | & 1 & 0 & 0 \ 3 & 2 & -1 & | & 0 & 1 & 0 \ 3 & 1 & 2 & | & 0 & 0 & 1 \end{bmatrix} \]
This augmented matrix is used to carry out row operations and transform the left side into the identity matrix. As we do this, the right side transforms into the inverse of the original matrix.

The process of working with an augmented matrix simplifies the complex operations needed to find an inverse by keeping both the system and its identity transformation side-by-side.

Row Operations

Row operations are essential to convert the given matrix into its reduced row echelon form, ultimately aiming to achieve the identity matrix on the left side of the augmented matrix. These operations include:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting the multiple of one row to another row

Let’s break down the use of row operations in our solution:

1. We made elements below the main diagonal zeros (
\[ R_2 = R_2 - 3R_1 \]
and
\[ R_3 = R_3 - 3R_1 \]).
2. We simplified the second row by dividing by -1, making the leading coefficient a 1 (
\[ R_2 = \frac{R_2}{-1} \]).
3. We eliminated the second column element in the third row (
\[ R_3 = R_3 + 2R_2 \]).
4. We normalized the third row by dividing by 7 to get a leading 1 (
\[ R_3 = \frac{R_3}{7} \]).
5. We made elements above leading 1's zeros.

By executing these operations systematically, the matrix gradually transforms, making the inverse computation straightforward and logical.

Identity Matrix

The identity matrix is a square matrix with ones on the diagonal and zeros elsewhere. For any matrix \( A \), multiplying it by the identity matrix \( I \) yields \( A \) itself (
\( AI = IA = A \)).

In our step-by-step solution, transforming the left side of the augmented matrix to the identity matrix is our goal. Look at our identity matrix:
\[ \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \].
The right side of our augmented matrix eventually reveals the inverse of the original matrix.

By performing row operations systematically, we ensure that the left side of the matrix turns into the identity matrix. Once that is achieved, we know the right side of the augmented matrix is the inverse of the original matrix.
This concept underpins how we find the matrix inverse and why reaching the identity matrix through row operations is critical.