Compute the indicated products: 1 2 3 2 3 4

Given that, A = 1 2 3 , B = 2 3 4 To find : AB Now, AB = 1 2 3 2 3 4 AB = 1 × 2 1 × 3 1 × 4 2 × 2 2 × 3 2 × 4 3 × 2 3 × 3 3 × 4 AB = 2 3 4 4 6 8 6 9 12 Hence, product of given matrix is 2 3 4 4 6 8 6 9 12 .

Compute the indicated products: 1 2 3 2 3 4

Given that, A = 1 2 3 , B = 2 3 4 To find : AB Now, AB = 1 2 3 2 3 4 AB = 1 × 2 1 × 3 1 × 4 2 × 2 2 × 3 2 × 4 3 × 2 3 × 3 3 × 4 AB = 2 3 4 4 6 8 6 9 12 Hence, product of given matrix is 2 3 4 4 6 8 6 9 12 .

Compute the indicated products: 3 - 1 3 - 1 0 2 2 - 3 1 0 3 1

Given that, A = 3 - 1 3 - 1 0 2 , B = 2 - 3 1 0 3 1 To find : AB Now, AB = 3 - 1 3 - 1 0 2 2 - 3 1 0 3 1 AB = 3 × 2 + - 1 × 1 + 3 × 3 3 × - 3 + - 1 × 0 + 3 × 1 - 1 × 2 + 0 × 1 + 2 × 3 - 1 × - 3 + 0 × 0 + 2 × 1 AB = 14 - 6 4 5 Hence, product of given matrix is 14 - 6 4 5 .

If A = 1 0 0 - 1 and B = 0 1 1 0 , find AB and BA .

Given that, A = 1 0 0 - 1 and B = 0 1 1 0 To find : AB and BA Now, AB = 1 0 0 - 1 0 1 1 0 AB = 1 × 0 + 0 × 1 1 × 1 + 0 × 0 0 × 0 - 1 × 1 0 × 1 - 1 × 0 AB = 0 1 - 1 0 Taking, BA = 0 1 1 0 1 0 0 - 1 BA = 0 × 1 + 1 × 0 0 × 0 - 1 × 1 1 × 1 + 0 × 0 1 × 0 + 0 × - 1 BA = 0 - 1 1 0 Hence, AB = 0 1 - 1 0 and BA = 0 - 1 1 0 .

If A = 1 1 1 1 and B = 1 - 1 - 1 1 , find AB and BA .

Given that, A = 1 1 1 1 and B = 1 - 1 - 1 1 To find : AB and BA Now, AB = 1 1 1 1 1 - 1 - 1 1 AB = 1 × 1 + 1 × - 1 1 × - 1 + 1 × 1 1 × 1 + 1 × - 1 1 × - 1 + 1 × 1 AB = 0 0 0 0 Taking, BA = 1 - 1 - 1 1 1 1 1 1 BA = 1 × 1 + - 1 × 1 1 × 1 + - 1 × 1 - 1 × 1 + 1 × 1 - 1 × 1 + 1 × 1 BA = 0 0 0 0 Hence, AB = 0 0 0 0 and BA = 0 0 0 0 .

If A = 1 2 3 4 , B = a 0 0 b and AB = BA , then show that B is a scalar matrix.

Given that, A = 1 2 3 4 , B = a 0 0 b To proof : B is a scalar matrix Now, AB = BA 1 2 3 4 a 0 0 b = 1 2 3 4 a 0 0 b 1 × a + 2 × 0 1 × 0 + 2 × b 3 × a + 4 × 0 3 × 0 + 4 × b = 1 × a + 2 × 0 1 × 0 + 2 × b 3 × 0 + 4 × 0 3 × 0 + 4 × b a + 0 0 + 2 b 3 a + 0 0 + 4 b = a + 0 2 a + 0 0 + 3 b 0 + 4 b a 2 b 3 a 4 b = a 2 a 3 b 4 b Thus their corresponding entries are equal. ∴ 2 b = 2 a ⇒ b = a And 3 a = 3 b ⇒ a = b Hence, B = a 0 0 b is a scalar matrix since all the diagonal elements are equal and all non-diagonal entries are zero.

Give an example of two non-zero 2 × 2 matrices A and B such that AB = O .

To find : 2 × 2 matrices A and B such that AB = O Now, Let A = 1 0 0 0 and B = 0 0 0 1 be two non-zero matrices We will check that AB = O or not AB = 1 0 0 0 0 0 0 1 AB = 1 × 0 + 0 × 0 1 × 0 + 0 × 1 0 × 0 + 0 × 0 0 × 0 + 0 × 1 AB = 0 0 0 0 Hence, A = 1 0 0 0 and B = 0 0 0 1 are two non-zero matrices such that AB = O .

Find the product x y z a h g h b f g f c x y z .

Given that, x y z a h g h b f g f c x y z To find : product of matrices Now, x y z a h g h b f g f c ⇒ x a + y h + z g x h + y b + z f x g + y f + z c ⇒ x a + y h + z g x h + y b + z f x g + y f + z c x y z ⇒ xa + yh + zg x xh + yb + zf y xg + yf + zc z ⇒ a x 2 + x h y + x zg + x h y + y 2 b + z f y + x g z + y f z + z 2 c ⇒ a x 2 + b y 2 + c z 2 + 2 h x y + 2 g x y + 2 y z f Hence, product is a x 2 + b y 2 + c z 2 + 2 h x y + 2 g x y + 2 y z f .

If A = α β γ - α and A 2 = I , find the value of α 2 + βγ .

Given that, A = α β γ - α and A 2 = I To find : value of α 2 + βγ Now, A 2 = I α β γ - α α β γ - α = 1 0 0 1 α × α + β × γ α × β + β × - α γ × α + - α × γ γ × β + - α × - α = 1 0 0 1 α 2 + βγ 0 0 α 2 + βγ = 1 0 0 1 Their corresponding element are equal... ∴ α 2 + βγ = 1 Hence, the required value is 1 .

If A = α 0 1 1 , B = 1 0 2 1 and A 2 = B , then find the value(s) of α .

Given that, A = α 0 1 1 , B = 1 0 2 1 and A 2 = B To find : value(s) of α Now, A 2 = B α 0 1 1 α 0 1 1 = 1 0 2 1 α × α + 0 × 1 α × 0 + 0 × 1 1 × α + 1 × 1 1 × 0 + 1 × 1 = 1 0 2 1 α 2 + 0 0 + 0 α + 1 0 + 1 = 1 0 2 1 α 2 0 α + 1 1 = 1 0 2 1 By equality of matrices... α 2 = 1 and α + 1 = 2 α = ± 1 and α = 1 Hence, value of α = 1 .