If A =[(2,6)(3,9)] , B =[(3,x)(y,2)] and AB = 0, find the values of x and y.

Given, A = , B = AB = 0 Solve for x and y: ∴ 6 + 6y = 0 ⇒ 6y = -6 ⇒ y = ⇒ y = -1. ∴ 2x + 12 = 0 ⇒ 2x = -12 ⇒ x = ⇒ x = -6. Hence, x = -6 and y = -1.

Mathematics

If A = $\begin{bmatrix} 2 & 6 \ 3 & 9 \end{bmatrix}$ , B = $\begin{bmatrix} 3 & x \ y & 2 \end{bmatrix}$ and AB = 0, find the values of x and y.

Matrices

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Answer

Given,

A = $\begin{bmatrix} 2 & 6 \ 3 & 9 \end{bmatrix}$ , B = $\begin{bmatrix} 3 & x \ y & 2 \end{bmatrix}$

AB = 0

$\Rightarrow \begin{bmatrix} 2 & 6 \ 3 & 9 \end{bmatrix} \times \begin{bmatrix} 3 & x \ y & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (2)(3) + 6y & 2x + (6)(2) \ (3)(3) + 9y & 3x + (9)(2) \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 + 6y & 2x + 12 \ 9 + 9y & 3x + 18 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$

Solve for x and y:

∴ 6 + 6y = 0

⇒ 6y = -6

⇒ y = $\dfrac{-6}{6}$

⇒ y = -1.

∴ 2x + 12 = 0

⇒ 2x = -12

⇒ x = $\dfrac{-12}{2}$

⇒ x = -6.

Hence, x = -6 and y = -1.

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Mathematics

If A = $\begin{bmatrix} 2 & 6 \ 3 & 9 \end{bmatrix}$ , B = $\begin{bmatrix} 3 & x \ y & 2 \end{bmatrix}$ and AB = 0, find the values of x and y.

Matrices

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