Namrata Das Exams Prep Master
Exams Prep Master
A-1 is the inverse of Matrix for a matrix ‘A’. A simple formula can be used to calculate the inverse of a 2x2 matrix. In addition, we must know the determinant and adjoint of a 3x3 matrix to compute its inverse. The inverse of a matrix is another matrix that yields the multiplicative identity when multiplied with the supplied matrix.
The matrix inversion method uses the inverse of a matrix to find the solution to linear equations. Let's look at the inverse of matrix formula, methods, and terminologies along with some solved questions.
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Key Terms: Matrix, square matrix, matrix's cofactor, determinant, matrix formula, adjoint matrix, methods, identity matrix, real number, minor
Inverse Matrix Formula
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The inverse of any real integer is the number a-1, therefore a times a-1 equaled 1. We have already understood that the reciprocal of a real number is the inverse of the number, as long as the number isn't zero. The inverse of a square matrix A, represented by A-1, is the matrix, hence the Identity matrix is the product of A and A-1. The resulting identity matrix will be the same size as matrix A.

The video below explains this:
Matrices Detailed Video Explanation:
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Terms Related to Inverse Matrix
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Each element in a square matrix has its minor. A number produced from the determinant of a square matrix by eliminating a row and a column corresponding to the element of a matrix is known as the minor.
In the case of a square matrix A, the value of the determinant produced by deleting the ith row and jth column of A matrix is called minor of an element [an ij]. M ij is the symbol for it. To discover the minor of a square matrix, we must erase one row and one column at a time and calculate their determinants until all of the minors have been computed. The steps to calculate minor from a matrix are as follows:
- Hide the ith row and jth column from a given matrix one by one, where I denote m and j denotes n, the total number of rows and columns in matrices.
- After hiding a row and a column in Step 1, calculate the value of the determinant of the matrix.
Cofactor: In order representation, the cofactor of an element is computed by multiplying the minor with -1 to the exponent of the sum of the row and column elements.

Matrix determinant: A matrix's determinant is its sole unique value representation. Any row or column of the provided matrix can be used to calculate the determinant of the matrix. The sum of the product of the elements and their cofactors in a given row or column of the matrix is the determinant of the matrix.

Singular Matrix: A singular matrix is defined as a matrix with a determinant value of zero. |A| = 0 for a singular matrix A.

Non-Singular Matrix: A non-singular matrix is one in which the determinant value is not equal to zero. |A| 0 for a non-singular matrix Because its inverse can be found, a non-singular matrix is called an invertible matrix.

Matrix Adjoint: The adjoint of a matrix is the transpose of the provided matrix's cofactor element matrix.
A Determinant's Row and Column Operations Rules
- To conduct row and column operations on determinants, use the following rules.
- If the rows and columns are swapped, the determinant's value remains intact.
- If any two rows or (two columns) are swapped, the determinant's sign changes.
- The determinant has zero value if any two rows or columns of a matrix are equal.
- When each element of a row or column is multiplied by a constant, the determinant's value is likewise multiplied by the constant.
- The determinant can be expressed as a sum of determinants if the elements of a row or column are expressed as a sum of elements.
- The value of the determinant remains intact when the elements of one row or column are added or subtracted with the matching multiples of elements from another row or column.
Read More: Symmetric Matrix
Methods to Find Inverse Matrix
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There are two ways to find the inverse of a matrix. The inverse of a matrix can be determined using basic operations and a matrix's adjoint. Row and column transformations can be used to execute basic operations on a matrix. In addition, using the inverse of matrix formula and the determinant and adjoint of the matrix, the inverse of a matrix can be determined. On the right-hand side of the equation, we use the matrix X and the second matrix B to perform the inverse of the matrix using elementary column operations.
- Elementary Row and column operations
- Matrix formula in reverse (using the adjoint and determinant of the matrix)
Let's have a look at each of the approaches below.
- Elementary Row operation.
Let us examine three square matrices, X, A, and B, for determining the inverse of a matrix using simple row operations. AX = B is the matrix equation. X = A-1B is the matrix equation. This idea is used to conduct the basic row operations. To begin, we write A = IA for the supplied matrix A. To obtain an identity matrix, perform the elementary row operations on the L.H.S. matrix and apply the same operations to the R.H.S. matrix "I." After transformations, the final matrix obtained in R.H.S. with "A" is the inverse of the given matrix.
- Elementary Column operation.
Consider three square matrices, X, A, and B, for finding the inverse of a matrix using elementary column operations. AX = B is the matrix equation. X = A-1B is the matrix equation. Begin by expressing the supplied matrix A as A = IA to conduct the basic row operations. To obtain an identity matrix, do the elementary column operations on the L.H.S. matrix and the same operations on the R.H.S. matrix "I." After transformations, the final matrix obtained in R.H.S. with "A" is the inverse of the given matrix.
Read More: Square matrix
How to Calculate Inverse Matrix?
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By dividing the adjoint of a matrix by the determinant of the matrix, the inverse of a matrix may be obtained using the inverse of the matrix formula. The following steps can be used to calculate the inverse of a matrix:
Calculate the minor for the supplied matrix in step one.
Step 2: Convert the acquired matrix to a co-factors matrix.
Step 3: Next, adjugate, and multiply by the reciprocal of the determinant.
Find the inverse of the following matrix
\(\begin{bmatrix}211 \\[0.3em]111 \\[0.3em] 1^-_12 \\[0.3em] \end{bmatrix} \)
Solution: Let A = \(\begin{bmatrix}2 & 1 & 1 \\[0.3em]1 & 1 & 1 \\[0.3em] 1 & -1 & 2 \\[0.3em]\end{bmatrix} \)
= 2\(\begin{bmatrix}1 & 1 \\[0.3em]-1 & 2 \\[0.3em] \end{bmatrix} - 1 \begin{bmatrix}1 & 1 \\[0.3em]1 & 2 \\[0.3em] \end{bmatrix}+ 1 \begin{bmatrix}1 & 1 \\[0.3em]1 & -1 \\[0.3em] \end{bmatrix}\)
|A| = 2(2+1) - 1(2-1) + 1(-1-1)
|A| = 2(3) - 1(1) + 1(-2)
|A| = 6-1-2
|A| = 3 ≠ 0
Since A is a non-singular matrix. A-1 exists.

A-1 = \(\frac{1}{3} \begin{bmatrix}3 & -3 & 0 \\[0.3em]-1 & 3 &-1 \\[0.3em]-2 & 3 & 1 \\[0.3em] \end{bmatrix}\)
Read More: Scalar Matrix
Things to Remember
- Matrixes, like numbers, have reciprocals. This reciprocal is known as an inverse matrix in the case of matrices. The product of two matrices equals the unit matrix if A is a square matrix and B is its inverse.
- The terms adjoint matrix and inverse matrix are not interchangeable. The inverse matrix is obtained by dividing each term of the adjugate matrix by the determinant of the original matrix.
- If the determinant of a matrix is greater than zero, the matrix is invertible, and we can compute the inverse of it. This implies that the supplied matrix is non-singular.
- We may apply row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse, by augmenting a 3 3 matrix with the identity on the right.
Previous Year Questions
- If the three linear equations x+4ay+az=0x+4ay+az=0 x+3by+bz=0x+3by+bz=0 and x+2cy+cz=0x+2cy+cz=0 have a non-trivial solution, then a, b, c are in
- The matrix 'X' in the equation AX=B, such that A=[1301] and B=[1−101] is given
- If ω≠1 is the complex cube root of unity and matrix H=[ω00ω], then H70 is equal to -
- The number of 3×3 non-singular matrices, with four entries as 1 and all other entries as 0, is
- IfBistheinverseofA,then.IfBistheinverseofA,then\alpha$ is
- A square matrix A=[aij]n×n is called a diagonal matrix if aij=0 for
- A square matrix A is said to be singular if
- If R(t)=[costsint−sintcost], then R(s) R(t) equals...[BITSAT 2017]
- If for a matrix A, |A| = 6 and adj A=⎡⎢⎣1−24411−1k0⎤⎥⎦ , then k is equal to :...[JEE Main 2018]
- If X+Y=[7025] and X−Y=[3003], then X is equal to…..[JEE Main 2013]
- If ∣∣∣x+75x+−33∣∣∣=26, then x is equal to….[JKCET 2013]
- If $2A+3B =[2−14325]andA+2B[503162]thenB =$….[JEE Advanced 2006]
- The only integral root of the equation ∣∣∣∣|2−y2325−y63410−y|=0 is
- The value of ∣∣ ∣∣b+caabc+abcca+b∣∣ ∣∣|b+caabc+abcca+b| is
- A square matrix A=[aij]n×n is called a lower triangular matrix if aij=0 for
Sample Questions
Ques 1: Given C = \(\begin{bmatrix}-10 & -5 \\[0.3em]6 & -\frac{2}{5} \\[0.3em] \end{bmatrix}\), find C-1. (3 marks)
Ans: We will use the formula for the inverse of a 2x2 matrix in order to find the inverse of matrix C.
C-1 = \(\frac{1}{ad-bc}\begin{bmatrix}d & -b \\[0.3em]-c & a \\[0.3em] \end{bmatrix}\)
C-1 = \(\frac{1}{(-10)(-\frac{2}{5})-(-5)(6)}\begin{bmatrix}-1 & 2 \\[0.3em]3 & 1 \\[0.3em] \end{bmatrix}\)
C-1 = \(\frac{1}{(4+30)}\begin{bmatrix}-\frac{2}{5} & 5 \\[0.3em]-6 & -10 \\[0.3em] \end{bmatrix}\)
C-1 = \(\frac{1}{(34)}\begin{bmatrix}-\frac{2}{5} & 5 \\[0.3em]-6 & -10 \\[0.3em] \end{bmatrix}\)
C-1 = \(\begin{bmatrix}-\frac{1}{85} & \frac{5}{34} \\[0.3em]-\frac{3}{17} & -\frac{5}{17} \\[0.3em] \end{bmatrix}\)
Ques 2: Given A = and B = , confirm if matrix B is the inverse of matrix A. (3 marks)
Ans: For Matrix B to be the inverse of Matrix, A, the matrix multiplication between these two matrices should result in an identity matrix (2×2 identity matrix). If so, B is the inverse of A.
A x B = \(\begin{bmatrix}0 & -4 \\[0.3em]-1 & 1 \\[0.3em] \end{bmatrix} \times \begin{bmatrix}-\frac{1}{4} & -1 \\[0.3em]-\frac{1}{4} & 0 \\[0.3em] \end{bmatrix}\)
= \(\begin{bmatrix}(0)(-\frac{1}{4}) + (-4)(-\frac{1}{4}) & (0)(-1) + (-4)(0) \\[0.3em](-1)(-\frac{1}{4}) + (1)(-\frac{1}{4}) & (-1)(-1)+(1)(0) \\[0.3em] \end{bmatrix}\)
= \(\begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em] \end{bmatrix}\)
This is the 2x2 identity matrix.
Therefore, Matrix B is the inverse of Matrix A.
Ques 3: Using the matrices XX and YY given below: X = \(\begin{bmatrix}1 & -3 \\[0.3em]5 & -6 \\[0.3em] \end{bmatrix}\) Y = \(\begin{bmatrix}-\frac{2}{3} & \frac{1}{3} \\[0.3em]-\frac{5}{9} & \frac{1}{9}\\[0.3em] \end{bmatrix}\)Determine whether the two matrices are inverses. (3 marks)
Ans: We approach this problem by noting that if two matrices are inverses of one other, multiplying them produces an identity matrix of the same dimensions, as described in equation 2 for 2x2 matrices. So, let's put that expression to work and multiply the matrices XX and YY to see what we get:
\(\begin{bmatrix}1 & -3 \\[0.3em]5 & -6 \\[0.3em] \end{bmatrix}\). \(\begin{bmatrix}-\frac{2}{3} & \frac{1}{3} \\[0.3em]-\frac{5}{9} & \frac{1}{9}\\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix}(1)(-\frac{2}{3})+(-3)(-\frac{5}{9}) & (1)(\frac{1}{3})+(-3)(\frac{1}{9}) \\[0.3em](5)(-\frac{2}{3})+(-6)(-\frac{5}{9}) & (5)(\frac{1}{3})+(-6)(\frac{1}{9}) \\[0.3em] \end{bmatrix}\)
= \(\begin{bmatrix}-\frac{2}{3} + \frac{5}{3} & \frac{1}{3}- \frac{1}{3} \\[0.3em]-\frac{10}{3} + \frac{10}{3} & \frac{5}{3} - \frac{2}{3} \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix}\frac{3}{3} & 0 \\[0.3em]0 & \frac{3}{3} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em] \end{bmatrix}\) = I2
As you can see, the product of matrices XX and YY is the second-order identity matrix, hence these two matrices are inverses of each other.
Ques 4: We'll use the matrices AA and BB presented below:A = \(\begin{bmatrix}-3 & -4 \\[0.3em]-6 & -5 \\[0.3em] \end{bmatrix}\) B = \(\begin{bmatrix}\frac{5}{9} & -\frac{4}{9} \\[0.3em]-\frac{2}{3} & \frac{1}{3} \\[0.3em] \end{bmatrix}\)(4 marks)
Ans: By multiplying the two matrices, you may see if they are inverses:
\(\begin{bmatrix}-3 & -4 \\[0.3em]-6 & -5 \\[0.3em] \end{bmatrix}\) \(\begin{bmatrix}\frac{5}{9} & -\frac{4}{9} \\[0.3em]-\frac{2}{3} & \frac{1}{3} \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix}(-3)(-\frac{5}{9})+(-4)(-\frac{2}{3}) & (-3)(-\frac{4}{9})+(-4)(\frac{1}{3}) \\[0.3em](-6)(\frac{5}{9})+(-5)(-\frac{2}{3}) & (-6)(-\frac{1}{9})+(-5)(\frac{1}{3}) \\[0.3em] \end{bmatrix}\)
\(\begin{bmatrix}-\frac{5}{3} + \frac{8}{3} & \frac{4}{3}- \frac{4}{3} \\[0.3em]-\frac{10}{3} + \frac{10}{3} & \frac{8}{3} - \frac{5}{3} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix}\frac{3}{3} & 0 \\[0.3em]0 & \frac{3}{3} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em] \end{bmatrix}\) = I2
We've shown that AA and BB are inverses of each other once more since the identity matrix is the outcome of their multiplication.
Before we go on to the next set of tasks, it's crucial to remember that the order in which you multiply inverted matrices has no bearing on their result; their multiplication will still produce the identity matrix. As a result, if a square matrix is invertible, its matrix multiplication and inverse are commutative (no matter the order in which they are multiplied, they always produce the same result: the identity). You can check this on your own, and we suggest you do it for practice.
Ques 5: Find the inverse of 2x2 matrix XX defined below: (3 marks)
\(X = \begin{bmatrix}3 & 1 \\[0.3em]5 & 2 \\[0.3em] \end{bmatrix}\)
Ans: We utilize equation 5 (inverse of 2x2 matrix formula) for this if the matrix X matches the element notation from equation 3. As a result, the 2x2 inverse matrix is computed as follows:
X-1 = \(\frac{1}{(3)(2) - (1)(5)} \begin{bmatrix}2 & -1 \\[0.3em]-5 & 3 \\[0.3em] \end{bmatrix} = \frac{1}{6-5} \begin{bmatrix}2 & -1 \\[0.3em]-5 & 3 \\[0.3em] \end{bmatrix}\) = \(\frac{1}{1} \begin{bmatrix}2 & -1 \\[0.3em]-5 & 3 \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix}2 & -1 \\[0.3em]-5 & 3 \\[0.3em] \end{bmatrix}\)= X-1
Ques 6: If CC is defined as a second-order identity matrix (just as shown below). In this example, what is the inverse of matrix 2x2? (3 marks)
\(C = \begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em] \end{bmatrix}\)
Ans: We don't need to conduct any computations to answer this question because we learned that the identity matrix is an involutory matrix during our session on the 2x2 invertible matrix (or just an involution for simplicity). Remember that an involutory matrix yields the identity matrix when multiplied by itself (squaring the matrix), therefore an involutory matrix is its inverse. As a result, we can immediately deduce that the inverse of CC in this situation is CC itself.
\(\begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em] \end{bmatrix}\) x \(\begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix}(1)(1)+(0)(0)& (1)(0) + (0)(1) \\[0.3em](0)(1)+(1)(0)&0)(0) + (1)(1) \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em] \end{bmatrix}\)
Let us not work the inverse using the formula for the inverse of a matrix 2x2 stated in equation 5 for the sake of clarity, so that we can see that the same result will be obtained:
C-1 = \(\frac{1}{(1)(1)-0}\)\(\begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em] \end{bmatrix}\)
Due to involution, the inverse of matrix 2x2 named CC is CC itself, as anticipated previously.
Ques 7: Find out the inverse of \(\begin{bmatrix}1 & -1 & 2 \\[0.3em]4 & 0 & 6 \\[0.3em]0 & 1 & -1 \\[0.3em] \end{bmatrix}\). (4 marks)
Ans: Let A = \(\begin{bmatrix}1 & -1 & 2 \\[0.3em]4 & 0 & 6 \\[0.3em]0 & 1 & -1 \\[0.3em] \end{bmatrix}\) be the given matrix.
A-1 = \(\frac{adj (A)}{|A|}\)
To find out the adj(A), forst we have to find out cofactor (A).
a11 = -6, a12 = 4, a13 = 4
a21 = 1, a22 = -1, a23 = -1
a31 = -6, a32 = 2, a23 = 4 so, cofactor (A) = \(\begin{bmatrix}-6 & 4 & 4 \\[0.3em]1 & -1 & -1 \\[0.3em]-6 & 2 & 4 \\[0.3em] \end{bmatrix}\)
adj(A) = $ [cofactor(A)] * (T) $
adj(A) = [cofactor(A)] T = \(\begin{bmatrix}-6 & 4 & 4 \\[0.3em]1 & -1 & -1 \\[0.3em]-6 & 2 & 4 \\[0.3em] \end{bmatrix}\)T
adj(A) = \(\begin{bmatrix}-6 & 1 & -6 \\[0.3em]4 & -1 & 2 \\[0.3em]4 & -1 & 4 \\[0.3em] \end{bmatrix}\)
Then, |A| = 1(0-6) + 1(-4-0) + 2(4-0) = -6-4+8 = -2
A-1 = \(\frac{adj (A)}{|A|}\) = \(\frac{\begin{bmatrix}-6 & 1 & -6 \\[0.3em]4 & -1 & 2 \\[0.3em]4 & -1 & 4 \\[0.3em] \end{bmatrix}}{2}\)
A-1 = \(\begin{bmatrix}3 & -\frac{1}{2} & 3 \\[0.3em]-2 & \frac{1}{2} & -1 \\[0.3em]-2 & \frac{1}{2} & -2\\[0.3em] \end{bmatrix}\)
Ques 8: Find the inverse, A-1, of A = (\(\begin{smallmatrix}2&-3 \\ 4&-7 \end{smallmatrix}\)) (4 marks)
Ans: [1] Interchange leading diagonal elements: \((\begin{smallmatrix}-7 & -3 \\ 4 & 2 \end{smallmatrix})\)
- 7 → 2; 2 → - 7
[2] Change signs of the other two elements
- 3 → 3; 4 → - 4
\((\begin{smallmatrix}-7 & -3 \\ 4 & 2 \end{smallmatrix})\)
[3] Find the deteminant |A|
| \(\begin{smallmatrix}2&-3 \\ 4&-7 \end{smallmatrix}\) | = – 14 + 12 = – 2
[4] Multiply result of [2] by \(\frac{1}{|A|}\)
A-1 = \(\frac{1}{|A|}\)\((\begin{smallmatrix}-7 & 3 \\ -4 & 2 \end{smallmatrix})\)
= \(\frac{1}{-2}\)\((\begin{smallmatrix}-7 & 3 \\ -4 & 2 \end{smallmatrix})\)
= \((\begin{smallmatrix}3.5 & -1.5 \\ 2 & -1 \end{smallmatrix})\)
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