Question

If $$ A=\left|\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right| $$ then \(\mathrm{A}^{-1}=\ldots\) (a) \(\mathrm{A}\) (b) \(\mathrm{A}^{2}\) (c) \(\mathrm{A}^{3}\) (d) \(\mathrm{A}^{4}\)

Short answer

\(-\frac{1}{9} \begin{bmatrix} -1 & -3 & 4 \\ 2 & -3 & 0 \\ -2 & 3 & 3 \end{bmatrix}\)

Step-by-step solution

Write Down the Given Matrix A and Identity Matrix

First, write down the given matrix A and the 3x3 identity matrix (I). Matrix A: \[ \left[\begin{array}{lll} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right] \] Identity Matrix (I): \[ \left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \]

Perform Row Operations

Now, we will perform row operations on both A and I simultaneously. The goal is to convert A into the identity matrix (I). 1. Subtract \(\frac{2}{3}\) times row 1 from row 2 (R2 - \(\frac{2}{3}\) * R1): New A: \[ \left[\begin{array}{lll} 3 & -3 & 4 \\ 0 & -1 & 0 \\ 0 & -1 & 1 \end{array}\right] \] New I: \[ \left[\begin{array}{lll} 1 & 0 & 0 \\ -\frac{2}{3} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \] 2. Add row 2 to row 3 (R3 + R2): New A: \[ \left[\begin{array}{lll} 3 & -3 & 4 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right] \] New I: \[ \left[\begin{array}{lll} 1 & 0 & 0 \\ -\frac{2}{3} & 1 & 0 \\ -\frac{2}{3} & 1 & 1 \end{array}\right] \] 3. Divide row 2 by -1 (R2 = -R2): New A: \[ \left[\begin{array}{lll} 3 & -3 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \] New I: \[ \left[\begin{array}{lll} 1 & 0 & 0 \\ \frac{2}{3} & -1 & 0 \\ -\frac{2}{3} & 1 & 1 \end{array}\right] \] 4. Add row 1 to row 2 (R1 + R2): New A: \[ \left[\begin{array}{lll} 3 & 0 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \] New I: \[ \left[\begin{array}{lll} -\frac{1}{3} & -1 & 0 \\ \frac{2}{3} & -1 & 0 \\ -\frac{2}{3} & 1 & 1 \end{array}\right] \] 5. Divide row 1 by 3 (R1 = \(\frac{1}{3}\) * R1): New A: \[ \left[\begin{array}{lll} 1 & 0 & \frac{4}{3} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \] New I: \[ \left[\begin{array}{lll} -\frac{1}{9} & -\frac{1}{3} & 0 \\ \frac{2}{3} & -1 & 0 \\ -\frac{2}{3} & 1 & 1 \end{array}\right] \] 6. Subtract \(\frac{4}{3}\) times row 3 from row 1 (R1 - \(\frac{4}{3}\) * R3): New A: \[ \left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \] New I: \[ \left[\begin{array}{lll} -\frac{1}{9} & -\frac{1}{3} & \frac{4}{9} \\ \frac{2}{3} & -1 & 0 \\ -\frac{2}{3} & 1 & 1 \end{array}\right] \]

Check the Result and Select the Correct Option

Matrix A has now been converted into the identity matrix (I), and New I is the inverse of the original matrix A. So, A^{-1} = \[ \left[\begin{array}{lll} -\frac{1}{9} & -\frac{1}{3} & \frac{4}{9} \\ \frac{2}{3} & -1 & 0 \\ -\frac{2}{3} & 1 & 1 \end{array}\right] \] This does not match with any of the given options (a), (b), (c), or (d).

Key concepts

3x3 matrix

A 3x3 matrix is simply a square matrix that contains 3 rows and 3 columns. For students learning linear algebra, the concept of a 3x3 matrix is one of the building blocks. In mathematics, a matrix is used to represent a collection of numbers in an organized way in order to perform various operations, such as addition, subtraction, and multiplication.
A 3x3 matrix is often used to solve systems of equations and to represent transformations in three-dimensional space.
Many problems in physics and engineering involve 3x3 matrices, especially when dealing with vectors or transformations of 3D objects. This type of matrix is easy to manipulate in terms of its shape, and it allows for a variety of computations and operations which are essential for higher-level mathematics.

row operations

Row operations are methods used in linear algebra to simplify matrices and solve systems of equations. The idea is to manipulate the rows of a matrix with the goal of transforming one matrix into another that is easier to work with, typically reducing it into its row echelon form or even into the identity matrix.
Common row operations include:

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

These operations help in solving equations more efficiently and are key steps in methods like Gaussian elimination.
In the context of finding the inverse of a matrix, row operations are used to simplify the given matrix into the identity matrix, while performing the same operations on the identity matrix to find the inverse.

identity matrix

An identity matrix is a special type of square matrix, often represented as "I". It has 1's on the diagonal from the top left to the bottom right and 0's in all other positions. For any square matrix form, such as a 3x3 matrix, there is a corresponding identity matrix of the same size. Its standard form looks like this for a 3x3 matrix:

• 1 0 0
  0 1 0
  0 0 1

The identity matrix acts like the number 1 in matrix algebra. When any matrix is multiplied by the identity matrix of compatible dimensions, it remains unchanged.
This property is crucial when finding the inverse of a matrix, as the goal of row operations is to transform the original matrix into the identity matrix, while also transforming the initial identity matrix into the inverse of the original matrix.

inverse of a matrix

The inverse of a matrix is essentially the matrix that, when multiplied by the original matrix, yields the identity matrix. Not all matrices have inverses, but when they do, they are a crucial tool for solving systems of linear equations.
To find the inverse of a 3x3 matrix, you can use methods such as row operations (also known as Gaussian elimination) or the determinant-adjoint method.
During row operations, you perform a series of operations to transform the original matrix (A) into the identity matrix (I), while simultaneously transforming the identity matrix (of the same size) into the inverse matrix ( A^{-1} ).
It's important to note that a matrix must be square (n x n) and have a non-zero determinant to possess an inverse. In cases where the matrix does have an inverse, it provides a powerful method for solving linear systems, making it invaluable in areas such as physics and engineering.