Mastering matrices is crucial for anyone studying advanced mathematics, computer science, or engineering. This article provides a comprehensive set of practice questions on matrices will guide you through a variety of problems, from basic to advanced levels these problems designed to strengthen your understanding and problem-solving skills in this fundamental area.

Important Formulas and Concepts on Matrices
Some of the important formulas and concepts that will help to solve these practice questions on matrices are mentioned below.
Order of Matrices : Order of a Matrix tells about the number of rows and columns present in a matrix. Let’s say if a matrix has 4 rows and 5 columns then the order of the matrix will be 4⨯5.
Transpose of a Matrix : Transpose of a Matrix is the rearrangement of row elements in column and column elements in a row to yield an equivalent matrix. The transpose matrix is represented as AT. if A = [aij]mxn , then AT = [bij]nxm where bij = aji.
Properties of the transpose of a matrix are mentioned below:
- (AT)T = A
- (A+B)T = AT + BT
- (AB)T = BTAT
Trace of Matrix : Trace of a Matrix is the sum of the principal diagonal elements of a square matrix. Trace of a matrix is only found in the case of a square matrix because diagonal elements exist only in square matrices.
Determinant of a Matrix: Determinant of a matrix is a number associated with that square matrix it is is calculated by adding the product of the elements of a matrix with their cofactors.. The determinant of a matrix can only be calculated for a square matrix. It is represented by |A|.
Minor of a Matrix: Minor of a matrix for an element is given by the determinant of a matrix obtained after deleting the row and column to which the particular element belongs to. Minor of Matrix is represented by Mij.
Cofactor of Matrix : Cofactor of a matrix is found by multiplying the minor of the matrix for a given element by (-1)i+j. Cofactor of a Matrix is represented as Cij. Hence, the relation between the minor and cofactor of a matrix is given as Mij = (-1)i+jMij.
Adjoint of Matrix : Adjoint is calculated for a square matrix. Adjoint of a matrix is the transpose of the cofactor of the matrix. The Adjoint of a Matrix is thus expressed as adj(A) = CT where C is the Cofactor Matrix.
Inverse of a Matrix : For a square matrix of order n its inverse can be defined as a matrix which when multiplied by the original matrix generates an identity matrix I of order n. i.e. A×A-1 = I The inverse is only calculated for a square matrix whose determinant is non-zero.
The formula for the inverse of a matrix is given as:
A-1 = adj(A)/det(A) = (1/|A|)(Adj A), where |A| should not be equal to zero, which means matrix A should be non-singular.
Matrices Formulas
Basic formula for the matrices are discussed below:
- A-1 = adj(A)/|A|
- A(adj A) = (adj A)A = I, where I is an Identity Matrix
- |adj A| = |A|n-1 where n is the order of matrix A
- adj(adj A) = |A|n-2A where n is the order of the matrix
- |adj(adj A)| = |A|(n-1)2
- adj(AB) = (adj B)(adj A)
- adj(Ap) = (adj A)p
- adj(kA) = kn-1(adj A) where k is any real number
- adj(I) = I
- adj 0 = 0
- If A is symmetric then adj(A) is also symmetric
- If A is a diagonal Matrix then adj(A) is also a diagonal matrix
- If A is a triangular matrix then adj(A) is also a triangular matrix
- If A is a singular Matrix then |adj A| = 0
- (AB)-1 = B-1A-1
Practice Questions on Matrices - Solved
Question 1: If (A+B)2= A2 +2AB+ B2 then what can we say about A and B? (Assume AB and BA exists)
Solution:
(A+B)2 = (A+B) (A+B)
⇒ According to question,
⇒ A2 + AB + BA + B2 = A2 + 2AB + B2
⇒ AB+BA = 2AB
⇒ BA = AB⇒ So, we can say that A and B are commutative
Question 2: If A is a n x m matrix such that AB and BA are both defined, then order of B is:
Solution:
If A size is n x m and it is also given that AB is defined then,
⇒ An x m X Bm x ☐ = (AB)n x n
⇒ ☐ = nOR
⇒ B☐ x m X An x m = (AB)n x n
⇒ ☐ = n⇒ So, the size of the matrix B is m x n
Question 3: Under what conditions is the matrix equation A2-B2 = (A-B) (A+B) will be true?
Solution:
We are given, A2-B2 = (A-B) (A+B)
⇒ A2 - B2 = A2 + AB - BA + B2
⇒ AB - BA = 0
⇒ AB = BA⇒ So, we can say that A and B should be commutative
Question 4: If AB = A and BA = B, then show that A and B are idempotent matrices.
Solution:
We are given that,
AB = A
⇒ A(BA) = A
⇒ (AB)A = A
⇒ (A)A = A
⇒ A2 = A⇒ So, we can say that A is idempotent matrix
Similarly, we can prove that B is also an idempotent matrix.
Question 5: Show that the sum of two idempotent matrices A and B is idempotent if AB = BA = 0.
Solution:
We have been given that,
AB = BA = 0 and A2 = A and B2 = B
⇒ (A+B)2 = (A+B) (A+B)
⇒ (A+B)2 = A2 + AB + BA + B2
⇒ (A+B)2 = A2 + B2 {since, AB=BA=0}
⇒ (A+B)2 = A + B⇒ Hence, sum of two idempotent matrices A and B is idempotent if AB=BA=0
Question 6: Evaluate
Solution:
Since
\omega is the cube root of unity, we know that1+\omega+\omega^2=0 By applying C1 → C1 + C2 + C3
⇒\begin{vmatrix}1+\omega+\omega^2 & \omega & \omega^2\\\omega+\omega^2+1 & \omega^2 & 1\\\omega^2+1+\omega & 1 & \omega\end{vmatrix}
Now, we know since1+\omega+\omega^2=0 Above determinant is written as,
⇒\begin{vmatrix}0 & \omega & \omega^2\\0 & \omega^2 & 1\\0 & 1 & \omega\end{vmatrix}
⇒0 Hence, the value of the given determinant is 0
Question 7: Evaluate
Solution:
By applying R1 → R1 + R2 + R3
⇒
\Delta = \begin{vmatrix}a-b + (b-c) + (c-a) & m-n + (n-p) + (p-m) & x-y + (y-z) + (z-x)\\b-c & n-p & y-z\\c-a & p-m & z-x\end{vmatrix}
⇒\Delta = \begin{vmatrix}0 & 0 & 0\\b-c & n-p & y-z\\c-a & p-m & z-x\end{vmatrix}
⇒\Delta = 0
Question 8: If A is a symmetric matrix, then prove that adj A (adjoint of A) is also symmetric.
Solution:
Let 'A' is a symmetric matrix, then AT = A
We know that,
⇒ (adj A)T = adj AT
⇒ (adj A)T = adj AHence, adj A is also a symmetric matrix
Question 9: Show that if A is a non-singular matrix, then det(A-1 ) = (det(A))-1
Solution:
We know that, |A-1| = 1 / |A|
⇒ A A-1 = In {where, I is an Identity matrix}
⇒ |A A-1| = |In|
⇒ |A| |A-1| = 1
⇒ |A-1| = 1 / |A|
⇒ |A|-1 = 1 / |A|Hence proved
Question 10: If A and B are n rowed squared matrices and AB = 0 & |B| ≠ 0, then A = 0. State True or False.
Solution:
Since, AB = 0
By multiplying by I on both sides, {where, I is an identity matrix}
⇒ A B B-1 = 0 B-1
⇒ A I = 0
⇒ A = 0Hence the above statement is True.
Practice Questions on Matrices - Unsolved
Question 1: Find x so that
Question 2: Under what conditions is the matrix equation A2- B2 = (A+B).(A+B) true?
Question 3: Evaluate
Question 4: Evaluate
Question 5: Prove that
Question 6: Evaluate
Question 7: If a+b+c = 0, then solve the equation
Question 8: Prove that if A is idempotent and A ≠ I, then A is singular.
Question 9: Only a square, non-singular matrix possesses inverse which is unique. State True or False.
Question 10: If AB = 0, does it imply that it is necessary that BA = 0.