Question Video: Finding the Values of Scalars from a Vector Equation Mathematics

Find the values of the scalars π‘Ÿ and 𝑠 if ⟨0, 3⟩ = π‘ŸβŸ¨2, 1⟩ + π‘ βŸ¨βˆ’3, βˆ’1⟩.

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Video Transcript

Find the values of the scalars π‘Ÿ and 𝑠 if the vector zero, three is equal to π‘Ÿ times the vector two, one plus 𝑠 times the vector negative three, negative one.

In this question, we’re given a vector equation involving two scalars π‘Ÿ and 𝑠. We need to use this to determine the values of these scalars. To answer this question, we need to recall to multiply a vector by a scalar, we need to multiply all of the components of the vector by the scalar. For example, to multiply the vector two, one by the scalar π‘Ÿ, we need to multiply each component by π‘Ÿ. We get the vector two π‘Ÿ, one π‘Ÿ.

We can apply the same process to the second vector on the right-hand side of this equation. We multiply each component of the vector by 𝑠. We get the vector negative three times 𝑠, negative one times 𝑠. And we know that this is equal to the vector zero, three. We can now simplify the right-hand side of this equation. In our first vector, we have one π‘Ÿ, which is equal to π‘Ÿ. And in our second vector, we have negative one 𝑠, which is just negative 𝑠. So we have that the vector zero, three is equal to the vector two π‘Ÿ, π‘Ÿ plus the vector negative three 𝑠, negative 𝑠.

To solve this equation, we want to add the two vectors on the right-hand side of the equation together. And we can do this by recalling to add two vectors of the same dimension together, we just add the corresponding components. Adding the first component of each vector on the right-hand side of our equation gives us two π‘Ÿ plus negative three 𝑠, which is two π‘Ÿ minus three 𝑠. And adding the second component of each vector on the right-hand side of the equation gives us π‘Ÿ plus negative 𝑠, which is π‘Ÿ minus 𝑠. Therefore, we have the vector zero, three is equal to the vector two π‘Ÿ minus three 𝑠, π‘Ÿ minus 𝑠.

So now we have an equation involving two vectors being equal to each other. And we know for two vectors to be equal, they must be of the same dimension and all of their corresponding components must be equal. Therefore, the first components of these two vectors must be equal. So zero must be equal to two π‘Ÿ minus three 𝑠. And the second components of these vectors must also be equal. Three must be equal to π‘Ÿ minus 𝑠.

We have two linear equations and two unknowns. We need to solve a pair of simultaneous equations. And we can solve this in many different ways. We’re going to use elimination. And we’re going to do this by multiplying the second equation through by two. We have three times two is equal to six and two times π‘Ÿ minus 𝑠 is two π‘Ÿ minus two 𝑠. So our second equation now becomes six is equal to two π‘Ÿ minus two 𝑠.

We can now eliminate the variable π‘Ÿ by finding the difference between the two equations. Finding the difference in the left-hand sides of the equations, we get zero minus six, which is equal to negative six. And finding the difference in the π‘Ÿ-terms on the right-hand sides of these equations, we have two π‘Ÿ minus two π‘Ÿ, which is equal to zero. Similarly, finding the difference in the 𝑠-terms on the right-hand sides of these equations, we get negative three 𝑠 minus negative two 𝑠. That’s negative three 𝑠 plus two 𝑠, which is negative 𝑠.

Therefore, we’ve shown negative six is equal to zero minus 𝑠. We can solve this equation for 𝑠. We just multiply the equation through by negative one and remove the zero term. We get that 𝑠 is equal to six. We now need to determine the value of π‘Ÿ. And we can use this by using any of the two equations. Let’s use the fact that three is equal to π‘Ÿ minus 𝑠 and 𝑠 is equal to six. Substituting 𝑠 is equal to six into the equation gives us that three is equal to π‘Ÿ minus six. And then we can add six to both sides of the equation to find the value of π‘Ÿ. We get that π‘Ÿ is equal to nine. And we could check that these values are correct by substituting them into the other of our simultaneous equations. We get two times nine minus three times six is 18 minus 18, which is equal to zero. So they do satisfy the other equation. Or we could substitute them back into our vector equation. Either method would work.

Either way, we were able to show if the vector zero, three is equal to π‘Ÿ times the vector two, one plus 𝑠 times the vector negative three, negative one, then the value of π‘Ÿ is nine and the value of 𝑠 is six.

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