Question Video: Finding Unknown Elements of a Matrix Using Equality of Matrices | Nagwa Question Video: Finding Unknown Elements of a Matrix Using Equality of Matrices | Nagwa

Question Video: Finding Unknown Elements of a Matrix Using Equality of Matrices Mathematics • First Year of Secondary School

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Given that [9^(π₯), 9π₯ + 3π¦ and 2π₯ β 6π¦, 9^(π¦)] = [32, π and π, 2], find the value of π/π.

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Video Transcript

Given that the matrix nine to the power of π₯, nine π₯ plus three π¦, two π₯ minus six π¦, nine to the power of π¦ is equal to the matrix 32, π, π, two, find the value of π over π.

We begin by recalling that if two matrices are equal, each of their corresponding terms must be equal. For example, considering the top-left elements, we have nine to the power of π₯ is equal to 32. And considering the bottom-right elements, nine to the power of π¦ is equal to two. We will use these two equations to calculate the values of π₯ and π¦.

We know that two to the power of five is equal to 32. So we can rewrite the first equation as nine to the power of π₯ is equal to two to the power of five. And since nine to the power of π¦ is equal to two, this can be rewritten as nine to the power of π₯ is equal to nine to the power of π¦ to the power of five. Next, we recall the power rule of exponents, which states that π to the power of π raised to the power of π is equal to π to the power of π multiplied by π. This means that our equation simplifies to nine to the power of π₯ is equal to nine to the power of five π¦. And since the base numbers are equal, the exponents themselves must be equal. π₯ is equal to five π¦.

Letβs now consider the terms in the top right and bottom left of our matrices. Equating the elements in the top right, we see that π is equal to nine π₯ plus three π¦. And since π₯ is equal to five π¦, the right-hand side is equal to nine multiplied by five π¦ plus three π¦. This simplifies to 45π¦ plus three π¦, which is equal to 48π¦. Equating the elements in the bottom left of our matrices, we have π is equal to two π₯ minus six π¦. Once again, replacing π₯ with five π¦, we have π is equal to two multiplied by five π¦ minus six π¦, which is equal to four π¦.

We are asked to find the value of π over π, which is therefore equal to 48π¦ divided by four π¦. Since π¦ cannot be equal to zero, we can cancel a common factor of π¦ from the numerator and denominator. We know that π¦ cannot be equal to zero as the element in the bottom right of the first matrix would become nine to the power of zero, which is equal to one. And this is not equal to the corresponding element in the second matrix. π over π is therefore equal to 48 divided by four, which is equal to 12. If the two given matrices are equal, then the value of π over π is 12.

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