Answer :

We have,

We need to verify (BA)^{2} ≠ B^{2}A^{2}.

Take L.H.S: (BA)^{2}

First, compute BA.

We know what order of matrix is,

If a matrix has M rows and N columns, the order of matrix is M × N.

Order of matrix B:

Number of rows = 2

⇒ M = 2

Number of columns = 3

⇒ N = 3

Then, order of matrix = M × N

⇒ Order of matrix B = 2 × 3

Order of matrix A:

Number of rows = 3

⇒ M = 3

Number of columns = 2

⇒ N = 2

Then, order of matrix = M × N

⇒ Order of matrix A = 3 × 2

Since, 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.

So, A and B can be multiplied.

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

(2, 1, 2)(3, 1, 2) = (2 × 3) + (1 × 1) + (2 × 2)

⇒ (2, 1, 2)(3, 1, 2) = 6 + 1 + 4

⇒ (2, 1, 2)(3, 1, 2) = 11

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

(2, 1, 2)(-4, 1, 0) = (2 × -4) + (1 × 1) + (2 × 0)

⇒ (2, 1, 2)(-4, 1, 0) = -8 + 1 + 0

⇒ (2, 1, 2)(-4, 1, 0) = -7

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

(1, 2, 4)(3, 1, 2) = (1 × 3) + (2 × 1) + (4 × 2)

⇒ (1, 2, 4)(3, 1, 2) = 3 + 2 + 8

⇒ (1, 2, 4)(3, 1, 2) = 13

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

(1, 2, 4)(-4, 1, 0) = (1 × -4) + (2 × 1) + (4 × 0)

⇒ (1, 2, 4)(-4, 1, 0) = -4 + 2 + 0

⇒ (1, 2, 4)(-4, 1, 0) = -2

So,

(BA)^{2} = (BA).(BA)

Similarly,

Take R.H.S: B^{2}A^{2}

Let us first compute B^{2}.

B^{2} = B.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.

Note that in matrix B, number of columns ≠ number of rows.

This means, we can’t find B^{2}.

⇒ L.H.S ≠ R.H.S

Thus, (BA)^{2} ≠ B^{2}A^{2}.

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