If and , show that (A + B) (A – B) ≠ A² – B².

We have the matrices A and B, where

We need to show that (A + B) (A – B) ≠ A² – B².

Take L.H.S: (A + B) (A – B)

First, let us compute (A + B).

If two matrices are of same order (say, m × n), then they can be added or subtracted. Example,

If we have matrices and . Then, they can be added as

So,

Now, let us compute (A – B).

Similarly, two matrices having same order can be subtracted in a similar fashion.

So,

Now, let us compute (A + B) (A – B).

In order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(0, 0).(0, 0) = (0 × 0) + (0 × 0)

⇒ (0, 0).(0, 0) = 0 + 0

⇒ (0, 0).(0, 0) = 0

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix B, then sum them up.

(0, 0).(2, 1) = (0 × 2) + (0 × 1)

⇒ (0, 0).(2, 1) = 0 + 0

⇒ (0, 0).(2, 1) = 0

Multiply 2^nd row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(2, 1).(0, 0) = (2 × 0) + (1 × 0)

⇒ (2, 1).(0, 0) = 0 + 0

⇒ (2, 1).(0, 0) = 0

Multiply 2^nd row of matrix A by matching members of 2^nd column of matrix B, then sum them up.

(2, 1).(2, 1) = (2 × 2) + (1 × 1)

⇒ (2, 1).(2, 1) = 4 + 1

⇒ (2, 1).(2, 1) = 5

So, we have

Take R.H.S: A² – B²

Let us compute A² first.

A² = A.A

So, we need to compute A.A.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix A, then sum them up.

(0, 1).(0, 1) = (0 × 0) + (1 × 1)

⇒ (0, 1).(0, 1) = 0 + 1

⇒ (0, 1).(0, 1) = 1

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix A, then sum them up.

(0, 1).(1, 1) = (0 × 1) + (1 × 1)

⇒ (0, 1).(1, 1) = 0 + 1

⇒ (0, 1).(1, 1) = 1

Multiply 2^nd row of matrix A by matching members of 1^st column of matrix A, then sum them up.

(1, 1).(0, 1) = (1 × 0) + (1 × 1)

⇒ (1, 1).(0, 1) = 0 + 1

⇒ (1, 1).(0, 1) = 1

Multiply 2^nd row of matrix A by matching members of 2^nd column of matrix A, then sum them up.

(1, 1).(1, 1) = (1 × 1) + (1 × 1)

⇒ (1, 1).(1, 1) = 1 + 1

⇒ (1, 1).(1, 1) = 2

So,

Now, let us compute B².

B² = B.B

We need to compute B.B.

Multiply 1^st row of matrix B by matching members of 1^st column of matrix B, then sum them up.

(0, -1).(0, 1) = (0 × 0) + (-1 × 1)

⇒ (0, -1).(0, 1) = 0 – 1

⇒ (0, -1).(0, 1) = -1

Multiply 1^st row of matrix B by matching members of 2^nd column of matrix B, then sum them up.

(0, -1).(-1, 0) = (0 × -1) + (-1 × 0)

⇒ (0, -1).(-1, 0) = 0 + 0

⇒ (0, -1).(-1, 0) = 0

Multiply 2^nd row of matrix B by matching members of 1^st column of matrix B, then sum them up.

(1, 0).(0, 1) = (1 × 0) + (0 × 1)

⇒ (1, 0).(0, 1) = 0 + 0

⇒ (1, 0).(0, 1) = 0

Multiply 2^nd row of matrix B by matching members of 2^nd column of matrix B, then sum them up.

(1, 0).(-1, 0) = (1 × -1) + (0 × 0)

⇒ (1, 0).(-1, 0) = -1 + 0

⇒ (1, 0).(-1, 0) = -1

So,

Now, compute A² – B².

Clearly,

and are not equal.

Thus, (A + B)(A – B) ≠ A² – B².