Inverse of Matrix
- Introduction
- What Is the Matrix Inverse?
- Inverse Matrix Formula
- Terms Associated With Matrix Inverse
- Methods to Find Matrix Inverse
- Inverse of `2 × 2` Matrix
- Inverse of `3 × 3` Matrix
- Determinant of Inverse Matrix
- Inverse Matrix Properties
- Practice Problems
- Frequently Asked Questions
Introduction
A matrix is like an array of numbers organized into rows and columns. Its size more specifically called its dimension, is written as the number of rows by the number of columns. We can only find the inverse of square matrices, meaning those with an equal number of rows and columns. The inverse of a matrix is another matrix that, when multiplied with the original, gives the identity matrix. To find the inverse of a `2×2` matrix, there's a simple formula. But for larger matrices, like `3×3` or more, we need to calculate the determinant and adjoint of the matrix. The inverse of a matrix is crucial in solving a system of linear equations using the matrix inversion method.
Example of a `2×2` matrix
\( A = \begin{bmatrix} -2 & 4 \\ 5 & -8 \end{bmatrix} \)
Example of a `3×3` matrix
\( A = \begin{bmatrix} 2 & 1 & 3 \\ 0 & -1 & 4 \\ -2 & 1 & 0 \end{bmatrix} \)
What Is the Matrix Inverse?
The inverse of a matrix is another matrix that, when multiplied by the original matrix, gives a multiplicative identity. For a square matrix `A`, its inverse is written as `A^-1`. The multiplication of `A` by `A^-1`, or vice versa, results in the identity matrix, which is denoted by `I`. An invertible matrix is defined as a matrix with a non-zero determinant, for which the inverse matrix can be calculated.
`A · A^-1 = A^-1· A = I`
The identity matrix for the `2 × 2` matrix is given by
\( I = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] \)
Note: For a matrix to have an inverse,
- The given matrix must be non-singular.
- The matrix determinant should not be zero.
Let's get into the formulas and methods for finding inverses.
Inverse Matrix Formula
When you multiply `A` by `A^-1`, you get the identity matrix, a matrix version of the number `1`.
Formula for inverse of matrix:
`A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A)`
where,
- `|A|` represents the determinant of `A`
- `"Adj A"` is the adjoint of `A`
Terms Associated With Matrix Inverse
The terminology listed below are useful for more clear understanding and calculating the inverse of a matrix
Minor:
For each element within a matrix, there exists a corresponding minor. This minor is determined by obtaining the determinant of the matrix resulting from the removal of the row and column containing the specific element.
For example, consider a matrix \( A \).
\(A = \begin{bmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33} \\\end{bmatrix}\)
The minor of the element `a_13` is,
\(M_{13} = \begin{vmatrix}a_{21} & a_{22} \\a_{31} & a_{32} \\\end{vmatrix}\)
Cofactor:
The cofactor of an element is obtained by multiplying the minor of that element by \((-1)^{i+j}\), where `i` and `j` are the row and column indices of the element.
`C_(ij) = (-1)^(i+j) × "Minor of " a_(ij)`
Determinant:
The determinant of a matrix is a single value that represents the entire matrix and can be calculated by summing the products of each element with its corresponding cofactor.
Singular Matrix:
A singular matrix has a determinant value of zero, making it impossible to find an inverse.
`|A| = 0`
Non-Singular Matrix:
A non-singular matrix has a non-zero determinant, allowing for the existence of an inverse.
`|A| ≠ 0`
Adjoint of Matrix:
Adjoint of a matrix is the transpose of the cofactor matrix and plays a significant role in finding the inverse of a matrix. It is also called the adjudicate matrix.
Few Rules for Row and Column Operations of a Determinant:
- Exchanging rows or columns in a matrix does not alter the determinant value.
- If any two rows or columns of a matrix are identical, the determinant becomes zero.
- When elements in a row or column are represented as a sum, the determinant can be expressed as a sum of determinants.
- Exchanging any two rows or columns alters the sign of the determinant.
- Multiplying every element in a given row or column by a constant results in the determinant being multiplied by the same constant.
- Adding or subtracting matching multiples of elements from one row or column to another keeps the determinant value constant.
Methods to Find Matrix Inverse
There are two methods to find the inverse of a matrix:
`1`. Using Matrix Formula
`2`. Inverse Matrix by Elementary Transformation
`1`. Inverse of a Matrix Formula
The calculation of the inverse matrix \( A^{-1} \) for matrix A involves dividing the adjoint of the matrix by its determinant using the inverse matrix formula.
`A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A)`
where,
`"Adj A = adjoint of the matrix A"`
`"|A| = determinant of the matrix A"`
Here are the steps to find the inverse of a matrix using the inverse matrix formula:
- Step `1`: Calculate the minors for all elements of matrix `A`.
- Step `2`: Compute the cofactors for all elements and construct the cofactor matrix by replacing `A's` elements with their corresponding cofactors.
- Step `3`: Take the transpose of a cofactor of the matrix `A` to find its adjoint (denoted as `"adj A"`).
- Step `4`: Multiply `"adj A"` by the reciprocal of the determinant of `A`.
`2`. Inverse Matrix by Elementary Transformation
Below are the steps to find the inverse matrix using the elementary transformation method:
- Step `1`: Represent the given matrix as \( A = IA \), where \( I \) is the identity matrix of the same order as \( A \).
- Step `2`: Perform a sequence of either row operations or column operations until the identity matrix is achieved on the left-hand side (LHS). Also, use similar elementary operations on the right-hand side (RHS) such that \( I = BA \). Therefore, the matrix \( B \) on the RHS is the inverse of matrix \( A \).
- Step `3`: Ensure to use either row operations or column operations consistently while performing elementary operations.
By following these steps, we can easily determine the inverse of a `2 × 2` matrix using elementary operations.
Inverse of `2 × 2` Matrix
Finding the inverse of a `2×2` matrix is simpler compared to matrices of higher dimensions. We can determine the inverse of a `2×2` matrix by following the general procedure for calculating matrix inverses. Now, let's compute the inverse of the following `2×2` matrix:
\( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)
`A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A)`
\(A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)
Hence, to compute the inverse of a `2 × 2` matrix, we must interchange the positions of elements `a` and `d`, assign negative signs to elements `b` and `c`, and then divide the result by the determinant of the matrix.
Example `1`: Find the inverse of matrix \( A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \)
Solution:
`A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A)`
To find the inverse \( A^{-1} \), we first calculate the determinant of \( A \):
`|A| = (2 \times 4) - (3 \times 1)`
`= 8 - 3`
`= 5`
Now, using the formula for the inverse of a `2×2` matrix, we have:
\( A^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix} \)
Example `2`: Find the inverse of matrix \( A = \begin{bmatrix} -1 & -2 \\ 6 & 3 \end{bmatrix} \)
Solution:
`A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A)`
To find the inverse \( A^{-1} \), we first calculate the determinant of \( A \):
`|A| = (-1 \times 3) - (-2 \times 6)`
`= -3-( - 12)`
`= -3+12`
`= 9`
Now, using the formula for the inverse of a `2×2` matrix, we have:
\( A^{-1} = \frac{1}{9} \begin{bmatrix} 3 & 2 \\ -6 & -1 \end{bmatrix} \)
Inverse of `3 × 3` Matrix
Let's denote the `3 × 3` square matrix as:
\( A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \)
In order to find its inverse, we'll first check if the matrix is invertible, i.e., if its determinant \( |A| \) is non-zero. If it's invertible, we can then calculate the inverse matrix \( A^{-1} \) using the formula:
`A^{-1} = \frac{1}{|A|} \times \text{Adj}(A)`
Where \( \text{Adj}(A) \) denotes the adjugate (or adjoint) matrix of \( A \).
Once we find \( A^{-1} \), we can verify if \( A \times A^{-1} = I \), where \( I \) is the identity matrix.
Example: Find the inverse of matrix \( A = \begin{bmatrix} 2 & 1 & 3 \\ 0 & -1 & 4 \\ -2 & 1 & 0 \end{bmatrix} \)
Solution:
The inverse of matrix `A` is given by the formula `A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A)`
Given:
\( A = \begin{bmatrix} 2 & 1 & 3 \\ 0 & -1 & 4 \\ -2 & 1 & 0 \end{bmatrix} \)
Step `1`: To find the inverse \( A^{-1} \), we can first calculate the determinant of \( A \):
`|A| = (2)((-1)(0) - (4)(1)) - (1)((0)(0) - (4)(-2)) + (3)((0)(1) - (-1)(-2))`
`= (2)(-4) - (1)(8) + (3)(-2)`
`= -8 - 8 - 6`
`= -22`
Step `2`: Finding the determinants of the `2×2` minor matrices:
\(
\begin{align*}
M_{11} &= \begin{vmatrix} -1 & 4 \\ 1 & 0 \end{vmatrix} \\
&= (-1)(0)-(4)(1) \\
&= 0-4 \\
&= -4 \\
M_{12} &= \begin{vmatrix} 0 & 4 \\ -2 & 0 \end{vmatrix} \\
&= (0)(0)-(4)(-2) \\
&= 0+8 \\
&= 8 \\
M_{13} &= \begin{vmatrix} 0 & -1 \\ -2 & 1 \end{vmatrix} \\
&= (0)(1)-(-1)(-2) \\
&= 0-2 \\
&= -2 \\
M_{21} &= \begin{vmatrix} 1 & 3 \\ 1 & 0 \end{vmatrix} \\
&= (1)(0)-(3)(1) \\
&= 0-3 \\
&= -3 \\
M_{22} &= \begin{vmatrix} 2 & 3 \\ -2 & 0 \end{vmatrix} \\
&= (2)(0)-(3)(-2) \\
&= 0+6 \\
&= 6 \\
M_{23} &= \begin{vmatrix} 2 & -1 \\ -2 & 1 \end{vmatrix} \\
&= (2)(1)-(1)(-2) \\
&= 2+2 \\
&= 4 \\
M_{31} &= \begin{vmatrix} 1 & 3 \\ -1 & 4 \end{vmatrix} \\
&= (1)(4)-(3)(-1) \\
&= 4+3 \\
&= 7 \\
M_{32} &= \begin{vmatrix} 2 & 3 \\ 0 & 4 \end{vmatrix} \\
&= (2)(4)-(3)(0) \\
&= 8-0 \\
&= 8 \\
M_{33} &= \begin{vmatrix} 2 & 1 \\ 0 & -1 \end{vmatrix} \\
&= (2)(-1)-(1)(0) \\
&= -2-0 \\
&= -2 \\
\end{align*}
\)
Step `3`: Forming the matrix of cofactors:
Matrix of minors \(=\begin{bmatrix} -4 & 8 & -2 \\ -3 & 6 & 4 \\ 7 & 8 & -2 \end{bmatrix}\)
To generate the matrix of cofactors, multiply each element of the minor matrix by the corresponding sign from the matrix:
\(\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}\)
Matrix of cofactors \(= \begin{bmatrix} -4 & -8 & -2 \\ 3 & 6 & -4 \\ 7 & -8 & -2 \end{bmatrix}\)
Step `4`: Taking the transpose of the cofactor matrix to get the adjugate matrix:
\(Adj(A) = \begin{bmatrix} -4 & 3 & 7 \\ -8 & 6 & -8 \\ -2 & -4 & -2 \end{bmatrix}\)
Step `5`: Finding the Inverse of the `3×3` Matrix:
Substitute the value of `|A|` and the `Adj(A)` in the formula:
`A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A)`
\( A^{-1} = \frac{1}{-22} \cdot \begin{bmatrix} -4 & 3 & 7 \\ -8 & 6 & -8 \\ -2 & -4 & -2 \end{bmatrix}\)
So, the inverse of matrix \( A \) is:
\( A^{-1} = \begin{bmatrix} \frac{2}{11} & -\frac{3}{22} & -\frac{7}{22} \\ \frac{4}{11} & -\frac{3}{11} & \frac{4}{11} \\ \frac{1}{11} & \frac{2}{11} & \frac{1}{11} \end{bmatrix}\)
Determinant of a Inverse Matrix
The determinant of the inverse of a matrix that is invertible is the reciprocal of the determinant of the original matrix.
`\det(A^{-1}) = \frac{1}{\det(A)}`
The proof of the above statement is given below:
We know that, `det(A × B) = det (A) × det(B)`
Also, `A × A^-1 = I`
`det(A × A^-1) = det(I)`
or, `det(A) × det(A^-1) = det(I)`
Since, `det(I) = 1`
`det(A) × det(A^-1) = 1`
or, `det(A^-1) = 1 / det(A)`
Hence, proved.
Inverse Matrix Properties
Properties of inverse matrix are as follows:
- For any non-singular matrix \(A\), \((A^{-1})^{-1} = A\)
- Inverse of a non-singular matrix exists, while for a singular matrix, the inverse does not exist.
- For any non-singular matrix \(A\), \((A^T)^{-1} = (A^{-1})^T\)
- If \( A \) and \( B \) are two nonsingular matrices, then \( AB \) is nonsingular. Thus, \( (AB)^{-1} = B^{-1}A^{-1} \)
- If \( A \) is any matrix and \( A^{-1} \) is its inverse, then \( AA^{-1} = A^{-1}A = I_n \), where \( n \) is the order of matrices.
Solved Examples
Example `1`: Find the inverse of matrix \( A = \begin{bmatrix} 2 & 7 & 3 \\ 5 & 1 & 8 \\ 4 & 6 & 9 \end{bmatrix} \)
Solution:
The inverse of matrix A is given by the formula `A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A)`
Given:
\( A = \begin{bmatrix} 2 & 7 & 3 \\ 5 & 1 & 8 \\ 4 & 6 & 9 \end{bmatrix} \)
Step `1`: To find the inverse \( A^{-1} \), we can first calculate the determinant of \( A \):
`|A| = (2)((1)(9) - (8)(6)) - (7)((5)(9) - (8)(4)) + (3)((5)(6) - (1)(4))`
`= (2)(-39) - (7)(13) + (3)(26)`
`= -78 - 91 + 78`
`= -91`
Step `2`: Finding the determinants of the `2×2` minor matrices:
\(\begin{align*}
M_{11} &= \begin{vmatrix} 1 & 8 \\ 6 & 9 \end{vmatrix} \\
&= (1)(9)-(8)(6) \\
&= 9-48 \\
&= -39 \\
M_{12} &= \begin{vmatrix} 5 & 8 \\ 4 & 9 \end{vmatrix} \\
&= (5)(9)-(8)(4) \\
&= 45-32 \\
&= 13 \\
M_{13} &= \begin{vmatrix} 5 & 1 \\ 4 & 6 \end{vmatrix} \\
&= (5)(6)-(1)(4) \\
&= 30-4 \\
&= 26 \\
M_{21} &= \begin{vmatrix} 7 & 3 \\ 6 & 9 \end{vmatrix} \\
&= (7)(9)-(3)(6) \\
&= 63-18 \\
&= 45 \\
M_{22} &= \begin{vmatrix} 2 & 3 \\ 4 & 9 \end{vmatrix} \\
&= (2)(9)-(3)(4) \\
&= 18-12 \\
&= 6 \\
M_{23} &= \begin{vmatrix} 2 & 7 \\ 4 & 6 \end{vmatrix} \\
&= (2)(6)-(7)(4) \\
&= 12-28 \\
&= -16 \\
M_{31} &= \begin{vmatrix} 7 & 3 \\ 1 & 8 \end{vmatrix} \\
&= (7)(8)-(3)(1) \\
&= 56-3 \\
&= 53 \\
M_{32} &= \begin{vmatrix} 2 & 3 \\ 5 & 8 \end{vmatrix} \\
&= (2)(8)-(3)(5) \\
&= 16-15 \\
&= 1 \\
M_{33} &= \begin{vmatrix} 2 & 7 \\ 5 & 1 \end{vmatrix} \\
&= (2)(1)-(7)(5) \\
&= 2-35 \\
&= -33 \\
\end{align*}\)
Step `3`: Forming the matrix of cofactors:
Matrix of minors \( = \begin{bmatrix} -39 & 13 & 26 \\ 45 & 6 & -16 \\ 53 & 1 & -33 \end{bmatrix} \)
To generate the matrix of cofactors, multiply each element of the minor matrix by the corresponding sign from the matrix:
\(\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}\)
Matrix of cofactors \(= \begin{bmatrix} -39 & -13 & 26 \\ -45 & 6 & 16 \\ 53 & -1 & -33 \end{bmatrix}\)
Step `4`: Taking the transpose of the cofactor matrix to get the adjugate matrix:
\(Adj(A) = \begin{bmatrix} -39 & -45 & 53 \\ -13 & 6 & -1 \\ 26 & 16 & -33 \end{bmatrix}\).
Step `5`: Finding the Inverse of the `3×3` Matrix:
Substitute the value of |A| and the Adj(A) in the formula:
\( A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A)\)
\( A^{-1} = \frac{1}{-91} \cdot \begin{bmatrix} -39 & -45 & 53 \\ -13 & 6 & -1 \\ 26 & 16 & -33 \end{bmatrix}\)
So, the inverse of matrix \( A \) is:
\( A^{-1} = \begin{bmatrix} \frac{3}{7} & \frac{45}{91} & -\frac{53}{91} \\ \frac{1}{7} & -\frac{6}{91} & \frac{1}{91} \\ -\frac{2}{7} & -\frac{16}{91} & \frac{33}{91} \end{bmatrix}\)
Example `2`: Find the inverse of matrix \( A = \begin{bmatrix} 4 & 7 \\ 2 & 3 \end{bmatrix} \)
Solution:
`A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A)`
To find the inverse \( A^{-1} \), we first calculate the determinant of \( A \):
`|A| = (4 \times 3) - (7 \times 2)`
`= 12 - 14`
`= -2`
Now, using the formula for the inverse of a `2×2` matrix, we have:
\( A^{-1} = \frac{1}{-2} \begin{bmatrix} 3 & -7 \\ -2 & 4 \end{bmatrix} \)
So, the inverse of matrix \( A \) is:
\( A^{-1} = \begin{bmatrix} \frac{-3}{2} & \frac{7}{2} \\ 1 & -2 \end{bmatrix} \)
Practice Problems
Q`1`. Find the inverse of matrix \( A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \)
- \(A^{-1} = \begin{bmatrix} \frac{4}{5} & -\frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5} \end{bmatrix} \)
- \(A^{-1} = \begin{bmatrix} -\frac{4}{5} & \frac{3}{5} \\ \frac{1}{5} & -\frac{2}{5} \end{bmatrix} \)
- \(A^{-1} = \begin{bmatrix} \frac{3}{5} & -\frac{4}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{bmatrix} \)
- \(A^{-1} = \begin{bmatrix} -\frac{3}{5} & \frac{4}{5} \\ \frac{2}{5} & -\frac{1}{5} \end{bmatrix} \)
Answer: a
Q`2`. Find the inverse of matrix \( C = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 3 & 2 \\ 0 & 2 & -1 \end{bmatrix} \)
- \( C^{-1} = \begin{bmatrix} -1 & 2 & 1 \\ 2 & -4 & -2 \\ -1 & 2 & 1 \end{bmatrix} \)
- \( C^{-1} = \begin{bmatrix} \frac{7}{15} & \frac{1}{15} & \frac{2}{15} \\ -\frac{1}{15} & \frac{2}{15} & \frac{4}{15} \\ -\frac{2}{15} & \frac{4}{15} & -\frac{7}{15} \end{bmatrix} \)
- \(C^{-1} = \begin{bmatrix} \frac{7}{15} & \frac{1}{15} & \frac{2}{15} \\ -\frac{1}{15} & \frac{2}{15} & \frac{4}{15} \\ -\frac{4}{15} & \frac{4}{15} & -\frac{7}{15} \end{bmatrix} \)
- \(C^{-1} = \begin{bmatrix} \frac{7}{15} & \frac{1}{15} & \frac{2}{15} \\ -\frac{1}{15} & \frac{2}{15} & \frac{4}{15} \\ -\frac{2}{15} & \frac{4}{15} & -\frac{4}{15} \end{bmatrix} \)
Answer: b
Q`3`. Find the inverse of matrix \( C = \begin{bmatrix} 4 & 7 \\ -2 & -3 \end{bmatrix} \)
- \(C^{-1} = \begin{bmatrix} \frac{3}{2} & \frac{7}{2} \\ 1 & 2 \end{bmatrix} \)
- \(C^{-1} = \begin{bmatrix} -\frac{3}{2} & -\frac{7}{2} \\ 1 & 2 \end{bmatrix} \)
- \(C^{-1} = \begin{bmatrix} -\frac{3}{2} & \frac{7}{2} \\ -1 & 2 \end{bmatrix} \)
- \(C^{-1} = \begin{bmatrix} \frac{3}{2} & -\frac{7}{2} \\ -1 & 2 \end{bmatrix} \)
Answer: b
Q`4`. Determine the inverse of matrix \( A = \begin{bmatrix} 2 & 1 & 3 \\ 0 & -1 & 4 \\ -2 & 1 & 0 \end{bmatrix} \)
- \( A^{-1} = \begin{bmatrix} \frac{1}{11} & -\frac{2}{11} & -\frac{3}{11} \\ \frac{4}{11} & -\frac{3}{11} & \frac{2}{11} \\ \frac{1}{11} & \frac{4}{11} & \frac{3}{11} \end{bmatrix} \)
- \( A^{-1} = \begin{bmatrix} \frac{2}{11} & -\frac{1}{11} & -\frac{1}{11} \\ \frac{4}{11} & -\frac{3}{11} & \frac{4}{11} \\ \frac{1}{11} & \frac{1}{11} & \frac{2}{11} \end{bmatrix} \)
- \( A^{-1} = \begin{bmatrix} \frac{2}{11} & -\frac{3}{22} & -\frac{7}{22} \\ \frac{4}{11} & -\frac{3}{11} & \frac{4}{11} \\ \frac{1}{11} & \frac{2}{11} & \frac{1}{11} \end{bmatrix} \)
- \( A^{-1} = \begin{bmatrix} \frac{1}{11} & \frac{2}{11} & \frac{4}{11} \\ \frac{3}{11} & -\frac{3}{11} & \frac{4}{11} \\ \frac{1}{11} & -\frac{1}{11} & \frac{2}{11} \end{bmatrix} \)
Answer: c
Q`5`. Determine the inverse of matrix \( E = \begin{bmatrix} 1 & -2 \\ -3 & 4 \end{bmatrix} \)
- \(E^{-1} = \begin{bmatrix} -\frac{1}{2} & -1 \\ \frac{3}{2} & -2 \end{bmatrix} \)
- \(E^{-1} = \begin{bmatrix} -\frac{1}{2} & -2 \\ \frac{3}{2} & -1 \end{bmatrix} \)
- \(E^{-1} = \begin{bmatrix} -\frac{1}{2} & 2 \\ -\frac{3}{2} & -1 \end{bmatrix} \)
- \(E^{-1} = \begin{bmatrix} -2 & -1 \\ -\frac{3}{2} & -\frac{1}{2} \end{bmatrix} \)
Answer: d
Frequently Asked Questions
Q`1`. What is an inverse matrix?
Answer: An inverse matrix is a matrix that, when multiplied by the original matrix, yields the identity matrix. For a square matrix \( A \), denoted as \( A^{-1} \), if \( A \times A^{-1} = A^{-1} \times A = I \), where \( I \) is the identity matrix, then \( A^{-1} \) is the inverse of \( A \).
Q`2`. When does a matrix have an inverse?
Answer: A square matrix has an inverse if and only if its determinant is nonzero. If the determinant of a matrix is zero, it is said to be singular, and it does not have an inverse.
Q`3`. How do you find the inverse of a matrix?
Answer: To find the inverse of a matrix, you typically use methods such as Gaussian elimination, cofactor expansion, or row reduction. These methods involve manipulating the original matrix to transform it into its inverse.
Q`4`. What is the significance of the inverse matrix?
Answer: The inverse matrix is crucial in solving systems of linear equations, as it allows you to solve for the unknown variables by multiplying both sides of the equation by the inverse matrix. It also plays a vital role in various mathematical and engineering applications, including solving optimization problems, computing algorithms, and performing transformations.
Q`5`. Can all square matrices be inverted?
Answer: No, not all square matrices can be inverted. A square matrix must be non-singular, meaning it must have a nonzero determinant, to have an inverse. If the determinant is zero, the matrix is singular and does not have an inverse. Therefore, only non-singular matrices can be inverted.