Examples of Matrices problems with Solution

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 nxm 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,

⇒ A_{n x m} X B_{m x ☐} = (AB)_{n x n}

⇒ ☐ = n

OR

⇒ B_{☐ x m} X A_{n 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 [Tex]\Delta = \begin{vmatrix} 1 & \omega & \omega^2\\ \omega & \omega^2 & 1\\ \omega^2 & 1 & \omega \end{vmatrix}[/Tex] where [Tex]\omega[/Tex] is one of the cube roots of the unity.

Solution:

Since [Tex]\omega[/Tex] is the cube root of unity, we know that [Tex]1+\omega+\omega^2=0[/Tex]

By applying C₁ → C₁ + C₂ + C₃

⇒[Tex]\begin{vmatrix} 1+\omega+\omega^2 & \omega & \omega^2\\ \omega+\omega^2+1 & \omega^2 & 1\\ \omega^2+1+\omega & 1 & \omega \end{vmatrix}[/Tex]

Now, we know since [Tex]1+\omega+\omega^2=0[/Tex]

Above determinant is written as,

⇒ [Tex]\begin{vmatrix} 0 & \omega & \omega^2\\ 0 & \omega^2 & 1\\ 0 & 1 & \omega \end{vmatrix}[/Tex]

⇒ [Tex]0[/Tex]

Hence, the value of the given determinant is 0

Question 7: Evaluate [Tex]\Delta = \begin{vmatrix} a-b & m-n & x-y\\ b-c & n-p & y-z\\ c-a & p-m & z-x \end{vmatrix}[/Tex]

Solution:

By applying R₁ → R₁ + R₂ + R₃

⇒[Tex]\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}[/Tex]

⇒[Tex]\Delta = \begin{vmatrix} 0 & 0 & 0\\ b-c & n-p & y-z\\ c-a & p-m & z-x \end{vmatrix}[/Tex]

⇒[Tex]\Delta = 0[/Tex]

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 A^T = A

We know that,

⇒ (adj A)^T = adj A^T

⇒ (adj A)^T = adj A

Hence, 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 = I_n {where, I is an Identity matrix}

⇒ |A A^-1| = |I_n|

⇒ |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 = 0

Hence the above statement is True.

Matrices are rectangular arrays of numbers, symbols, or characters where all of these elements are arranged in each row and column. An array is a collection of items arranged at different locations.