How to Find the Inverse of a Matrix Using Elementary Row Operations with Steps and Examples
An arrangement of numbers, expressions, or symbols in a rectangular array is called a matrix (plural: matrices). With an order of the number of rows × the number of columns, this layout comprises horizontal and vertical columns. In three-dimensional space, every pair of points represents a different equation with one or more solutions.
Matrix algebra involves matrix operations like addition, subtraction, multiplication, etc.
Inverse of Matrix
There exists an inverse matrix A-1 if A is a non-singular square matrix, which satisfies the below condition:
AA-1 = A-1A = I, where I’ is an Identity matrix.
Elementary operations
A matrix is subjected to six operations or transformations, three caused by the rows and three by the columns. These actions are referred to as elementary actions. Only square matrices are used for these operations.
These elementary operations are:
Interchanging any two rows or columns ($R_i \leftrightarrow R_j$ or $C_i \leftrightarrow C_j$)
Multiplying the elements of any row or column by a positive integer ($R_i \rightarrow kR_i$ or $C_i \rightarrow kC_i$)
Addition or subtraction of multiples of one row to another ($R_i \rightarrow R_i + kR_j$ or $C_i$ → $C_i+ kC_j$)
How to find the inverse of the 3×3 matrix?
In order to determine a matrix's inverse, we must do the following steps:
Finding the minors matrix is the first step.
Change that matrix into a matrix of cofactors
Find the adjoint of the matrix
Multiply by \[\dfrac{1}{\text{determinant}}\]
Finding Inverse of Matrix Using Elementary Operations
If the three matrices, X, A, and B, are all of the same order, then X = AB is an equation of matrices. The basic row operations will be applied simultaneously to the matrices X and A, which is on the RHS side of the matrix equation of the product of AB for the given matrix equation.
Similarly, we can apply the basic column operations simultaneously on the matrices X and B (on the RHS side of the matrix equation) of the product of AB.
Thus, in general, if we calculate the inverse of a matrix A by doing simple row operations on A = IA in a certain order until we get: I = BA.
Also, the value of the inverse of matrix A must be obtained by performing basic column operations on the value of A = AI sequentially until we reach I = AB.
Inverse of Matrix Using Elementary Row transformation
As the names suggest, only the rows of the matrices are changed; the columns remain unchanged. A certain set of guidelines is followed while performing these row operations to ensure that the converted matrix is identical to the original matrix. Let’s understand how to find the inverse of a matrix using elementary row operations with the help of an example:
Example:
A = \[\left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&1&1\end{array}} \right]\]
Applying $R_1 \rightarrow R_2$
\[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\3&1&1\end{array}} \right]\]
Inverse of Matrix Using Elementary Column Transformation
The basic column operation is obtained by similarly applying those three-row operations to columns.
Solved Examples:
1. Find the inverse of the matrix using elementary operations.
A = \[\left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&1&1\end{array}} \right]\]
Solution:
A = \[\left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&1&1\end{array}} \right]\]
Using the elementary row operation, we know;
$A = IA$
\[\left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&1&1\end{array}} \right]\] = \[\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\]. A
Applying $R_1 \rightarrow R_2$
\[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\3&1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0&1&0\\1&0&0\\0&0&1\end{array}} \right].A\]
Applying $R_3 \rightarrow R_3 - 3R_1$
\[\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&{ - 5}&{ - 8}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0&1&0\\1&0&0\\0&{ - 3}&1\end{array}} \right].A\]
Applying $R_1 \rightarrow R_1-2R_2$ and $R_3\rightarrow R-3 + 5R_2$
$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{bmatrix}= \begin{bmatrix} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \end{bmatrix} \cdot A$
Applying $ R_3 \rightarrow \frac{R_3}{2}$
\[\left[ {\begin{array}{*{20}{c}}1&0&{ - 1}\\0&1&2\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2}&1&0\\1&0&0\\{\dfrac{5}{2}}&{\dfrac{{ - 3}}{2}}&{\dfrac{1}{2}}\end{array}} \right].A\]
Applying $R_1 \rightarrow R_1+R_3, R_2 \rightarrow R_2 – 2R_3$
\[\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\dfrac{1}{2}}&{\dfrac{{ - 1}}{2}}&{\dfrac{1}{2}}\\{ - 4}&3&1\\{\dfrac{5}{2}}&{\dfrac{{ - 3}}{2}}&{\dfrac{1}{2}}\end{array}} \right].A\]
Therefore,
$A^{-1}=\begin{bmatrix}{\dfrac{1}{2}}&{\dfrac{{ - 1}}{2}}&{\dfrac{1}{2}}\\{ - 4}&3&1\\{\dfrac{5}{2}}&{\dfrac{{ - 3}}{2}}&{\dfrac{1}{2}}\end{bmatrix}$
2. Find the inverse of the following matrix by elementary operations?
$\begin{bmatrix} -2 & 1 & 3 \\ 0 & -1 & 1 \\ 1 & 2 & 0 \end{bmatrix}$
Solution: First rewrite the matrix as:
$[A|I]=\left[\begin{array}{ccc:ccc} 1 & 0 & 0 & -2 & 1 & 3 \\ 0 & 1 & 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 1 & 2 & 0 \end{array}\right]$
Applying $R_3 \rightarrow 2R_3+R_1$
$\left[\begin{array}{ccc:ccc} 1 & 0 & 0 & -2 & 1 & 3 \\ 0 & 1 & 0 & 0 & -1 & 1 \\ 1 & 0 & 2 & 0 & 5 & 3 \end{array}\right]$
Applying $R_1$ → $R_1 + R_2$ and $R_3$ → $R_3+ 5R_2$
$\left[\begin{array}{ccc:ccc} 1 & 1 & 0 & -2 & 0 & 4 \\ 0 & 1 & 0 & 0 & -1 & 1 \\ 1 & 5 & 2 & 0 & 0 & 8 \end{array}\right]$
Applying $R_1$ → $2R_1 - R_3$ and $R_2$ → $8R_2 - R_3$
$\left[\begin{array}{ccc:ccc} 1 & -3 & -2 & -4 & 0 & 0 \\ -1 & 3 & -2 & 0 & -8 & 0 \\ 1 & 5 & 2 & 0 & 0 & 8 \end{array}\right]$
Applying $R_1$ → $\frac{1}{-4}, R_2$ → $\frac{1}{-8}$, and $R_3$ → $\frac{1}{8} $
$\left[\begin{array}{ccc:ccc} -1 / 4 & +3 / 4 & +2 / 4 & 1 &0 &0 \\ +1 / 8 & -3 / 8 & 2 / 8 & 0 &1 & 0 \\ 1 / 8 & 5 / 8 & 2 / 8 & 0 & 1 & 0 \end{array}\right]$
Hence $A^{-1}=\left[\begin{array}{ccc:ccc} -1 / 4 & 3 / 4 & 2 / 4 \\ 1 / 8 & -3 / 8 & 2 / 8 \\ 1 / 8 & 5 / 8 & 2 / 8 \end{array}\right]$
Important formulas
$AA^{-1} = A^{-1} A = I$, where I’ is an Identity matrix.
$A^{-1} = \dfrac{(adjA)}{(\det A)}$
Interesting facts
If A and B are square matrices such that AB = BA = I, then B is the inverse matrix of A and is denoted by A-1 and A is the inverse of B.
The inverse of the inverse matrix is equal to the original matrix.
If A and B are invertible matrices, then AB is also invertible.
Summary
We learn about the Inverse of a matrix and how to find the inverse of a matrix using elementary operations. An arrangement of numbers, expressions, or symbols in a rectangular array is called a matrix and there exists an inverse matrix $A^{-1}$ if A is a non-singular square matrix, which satisfies the $AA^{-1} = A^{-1} A = I$ condition. A matrix is subjected to six elementary operations.
Problems for solving
Find the inverse of matrix A using elementary column operation.
A =\[\left[ {\begin{array}{*{20}{c}}3&5&1\\4&3&1\\9&8&1\end{array}} \right]\]
Find the inverse of matrix A using elementary column operation.
A = \[\left[ {\begin{array}{*{20}{c}}1&0&0\\0&0&1\\0&1&1\end{array}} \right]\]
Answer:
\[\left[{\begin{array}{*{20}{c}}{-\frac{1}{3}}&{\frac{1}{5}}&{\frac{2}{{15}}}\\{\frac{1}{3}}&{ - \frac{2}{5}}&{\frac{1}{{15}}}\\{\frac{1}{3}}&{\frac{7}{5}}&{ - \frac{{11}}{{15}}}\end{array}} \right]\]
\[\left[ {\begin{array}{*{20}{c}}1&0&0\\0&{ - 1}&1\\0&1&0\end{array}} \right]\]
FAQs on Inverse of Matrix by Elementary Operations Explained
1. What is the inverse of a matrix by elementary operations?
The inverse of a matrix by elementary operations is found by applying row operations to convert the matrix into an identity matrix while performing the same operations on an identity matrix to obtain the inverse. For a square matrix A, we form the augmented matrix [A | I] and apply elementary row operations until it becomes [I | A-1]. This method is also called the Gauss–Jordan method and works only if the matrix is non-singular (determinant ≠ 0).
2. How do you find the inverse of a matrix using elementary row operations?
To find the inverse of a matrix using elementary row operations, follow these steps:
- Write the augmented matrix [A | I].
- Apply row operations (Ri ↔ Rj, kRi, Ri + kRj) to transform A into I.
- When the left side becomes the identity matrix, the right side becomes A-1.
3. What are the elementary row operations used to find the inverse of a matrix?
The elementary row operations used to find a matrix inverse are three standard operations:
- Interchanging two rows (Ri ↔ Rj).
- Multiplying a row by a non-zero constant (kRi).
- Adding a multiple of one row to another row (Ri + kRj).
4. What is the condition for a matrix to have an inverse using elementary operations?
A matrix has an inverse using elementary operations only if its determinant is non-zero. In other words, the matrix must be non-singular and square (same number of rows and columns). If det(A) = 0, the matrix is singular and no inverse exists.
5. Can you show an example of finding the inverse of a 2×2 matrix by elementary operations?
Yes, the inverse of a 2×2 matrix can be found using elementary operations by reducing it to the identity matrix. For example, let A = [[1, 2], [3, 4]].
- Form the augmented matrix: [1 2 | 1 0; 3 4 | 0 1].
- Apply row operations to convert the left side to identity.
- The result becomes: [1 0 | -2 1; 0 1 | 3/2 -1/2].
6. Why do we use the identity matrix when finding the inverse by elementary operations?
We use the identity matrix because it acts as the multiplicative identity in matrix multiplication. When we transform matrix A into I using row operations, the same operations convert I into A-1. This works because A × A-1 = I.
7. What is the difference between finding the inverse by formula and by elementary operations?
The difference is that the formula method uses a direct formula (mainly for 2×2 matrices), while the elementary operations method works for matrices of any order. For a 2×2 matrix A = [[a, b], [c, d]], the formula is A-1 = (1/(ad − bc)) [[d, −b], [−c, a]]. However, for 3×3 or higher matrices, Gauss–Jordan elimination is more practical.
8. What happens if a matrix cannot be reduced to the identity matrix?
If a matrix cannot be reduced to the identity matrix, then it does not have an inverse. This situation occurs when the matrix is singular, meaning its determinant is zero and at least one row becomes entirely zero during row reduction.
9. How do you verify that the inverse found by elementary operations is correct?
To verify the inverse, multiply the original matrix A by the computed matrix A-1 and check if the result is the identity matrix. If A × A-1 = I and A-1 × A = I, then the inverse is correct.
10. Is the inverse of a matrix unique when found by elementary operations?
Yes, the inverse of a matrix is unique if it exists. No matter which valid sequence of elementary row operations is used, the final result will always be the same matrix A-1, provided the matrix is non-singular.
