Q. 225.0( 1 Vote )

# If , and , verify:

(i) (AB) C = A (BC)

(ii) A(B + C) = AB + AC

Answer :

We have matrices A, B and C, such that

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.

(i). We need to verify: (AB)C = A(BC)

Take L.H.S = (AB)C

First, compute AB.

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

(1, 2)(2, 3) = (1 × 2) + (2 × 3)

⇒ (1, 2)(2, 3) = 2 + 6

⇒ (1, 2)(2, 3) = 8

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

(1, 2)(3, -4) = (1 × 3) + (2 × -4)

⇒ (1, 2)(3, -4) = 3 – 8

⇒ (1, 2)(3, -4) = -5

Similarly, let us fill for the rest of elements.

Let .

Now, compute for DC. [∵ (AB)C = DC]

Multiply 1^{st} row of matrix D by matching members of 1^{st} column of matrix C, then sum them up.

(8, -5)(1, -1) = (8 × 1) + (-5 × -1)

⇒ (8, -5)(1, -1) = 8 + 5

⇒ (8, -5)(1, -1) = 13

Multiply 1^{st} row of matrix D by matching members of 2^{nd} column of matrix C, then sum them up.

(8, -5)(0, 0) = (8 × 0) + (-5 × 0)

⇒ (8, -5)(0, 0) = 0 + 0

⇒ (8, -5)(0, 0) = 0

Similarly, let us fill for the rest of elements.

So,

Take R.H.S: A(BC)

First, compute BC.

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

(2, 3)(1, -1) = (2 × 1) + (3 × -1)

⇒ (2, 3)(1, -1) = 2 – 3

⇒ (2, 3)(1, -1) = -1

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

(2, 3)(0, 0) = (2 × 0) + (3 × 0)

⇒ (2, 3)(0, 0) = 0 + 0

⇒ (2, 3)(0, 0) = 0

Similarly, let us fill for the rest of the elements.

Let .

Now, compute for AE.

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

(1, 2)(-1, 7) = (1 × -1) + (2 × 7)

⇒ (1, 2)(-1, 7) = -1 + 14

⇒ (1, 2)(-1, 7) = 13

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

(1, 2)(0, 0) = (1 × 0) + (2 × 0)

⇒ (1, 2)(0, 0) = 0 + 0

⇒ (1, 2)(0, 0) = 0

Similarly, repeat the step for the other elements.

So,

Thus, (AB)C = A(BC).

(ii). We need to verify: A(B + C) = AB + AC

Take L.H.S: A(B + C)

Add B + C.

Let B + C = F, such that

Now, multiply A and F.

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

(1, 2)(3, 2) = (1 × 3) + (2 × 2)

⇒ (1, 2)(3, 2) = 3 + 4

⇒ (1, 2)(3, 2) = 7

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

(1, 2)(3, -4) = (1 × 3) + (2 × -4)

⇒ (1, 2)(3, -4) = 3 – 8

⇒ (1, 2)(3, -4) = -5

Similarly, repeat the steps for the other elements.

So,

Now, take R.H.S: AB + AC

Compute AB.

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

(1, 2)(2, 3) = (1 × 2) + (2 × 3)

⇒ (1, 2)(2, 3) = 2 + 6

⇒ (1, 2)(2, 3) = 8

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

(1, 2)(3, -4) = (1 × 3) + (2 × -4)

⇒ (1, 2)(3, -4) = 3 – 8

⇒ (1, 2)(3, -4) = -5

Similarly, repeat the steps for the other elements.

So,

Now, compute AC.

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

(1, 2)(1, -1) = (1 × 1) + (2 × -1)

⇒ (1, 2)(1, -1) = 1 – 2

⇒ (1, 2)(1, -1) = -1

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

(1, 2)(0, 0) = (1 × 0) + (2 × 0)

⇒ (1, 2)(0, 0) = 0 + 0

⇒ (1, 2)(0, 0) = 0

Similarly, repeat the steps for the other elements.

So,

Adding AB + AC.

Matrices of same order can be added or subtracted.

So, clearly L.H.S = R.H.S.

Thus, A(B + C) = AB + AC.

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