Question Video: Using Determinants and Noncollinearity to Determine the value of a Variable Mathematics

If the points (π‘₯, 1), (3, π‘₯), and (0, βˆ’2) are collinear, use determinants to find all the possible values of π‘₯. Round your answers to two decimal places.

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Video Transcript

If the points π‘₯, one; three, π‘₯; and zero, negative two are collinear, use determinants to find all the possible values of π‘₯. Round your answers to two decimal places.

In this question, we’re given three points with an unknown value of π‘₯. We’re told that these three points are collinear, and we need to use determinants to determine all of the possible values of π‘₯. We need to round our answers to two decimal places. To answer this question, let’s start by recalling how we use determinants to determine whether three points are collinear.

We recall if we have three distinct points π‘₯ sub one, 𝑦 sub one; π‘₯ sub two, 𝑦 sub two; and π‘₯ sub three, 𝑦 sub three, then these will be collinear if the determinant of the three-by-three matrix π‘₯ sub one, 𝑦 sub one, one, π‘₯ sub two, 𝑦 sub two, one, π‘₯ sub three, 𝑦 sub three, one is equal to zero. And if this determinant is not equal to zero, then the three points are not collinear.

And it’s worth pointing out this property works in both directions. If we have three points in which are collinear, we know the determinant is zero. And if the determinant is zero, we know the three points are collinear. And we can say something very similar for three distinct noncollinear points. If the determinant of this matrix is nonzero, then the three points are noncollinear. And if we have three noncollinear points, then this determinant must be nonzero.

Therefore, since we’re told the three points given to us in the question are collinear, the determinant of this matrix where we substitute the coordinates of these three points in must be zero. The determinant of the three-by-three matrix π‘₯, one, one, three, π‘₯, one, zero, negative two, one is equal to zero. Now, if we expand this determinant, we’ll have an equation in terms of π‘₯.

And there’s a few different ways we can evaluate this determinant. We’re going to expand over the first column because it includes a value of zero. And remember, when we expand over a row or column, the parity of the sum of the row number and column number will affect the sign of this term. In particular, if the row number plus the column number is even, we’ll have a positive sign, and if it’s odd, we’ll have a negative sign.

We’re now ready to evaluate the determinant of this matrix by expanding over the first column. We get positive π‘₯ multiplied by the matrix minor we get by removing row one, column one. That’s the determinant of the two-by-two matrix π‘₯, one, negative two, one. We then do the same for the second entry in this column. And don’t forget, we need to multiply this by negative one. We get negative three times the determinant of the two-by-two matrix one, one, negative two, one. And finally, we would need to expand over the third entry; however, this is zero. So, the third term will have a factor of zero and is equal to zero. This gives us π‘₯ times the determinant of π‘₯, one, negative two, one minus three times the determinant of one, one, negative two, one.

We now need to evaluate this expression. And to do this, we recall to evaluate the determinant of a two-by-two matrix, we find the difference in the product of the diagonals. The determinant of the first matrix is π‘₯ times one minus negative two times one, which is π‘₯ plus two. And the determinant of the second matrix is one times one minus one times negative two, which is one plus two. This gives us π‘₯ times π‘₯ plus two minus three times one plus two, which if we distribute and simplify we see is equal to π‘₯ squared plus two π‘₯ minus nine.

And remember, because the three points are collinear, we know that this must be equal to zero. Therefore, we have a quadratic equation in π‘₯. We could try solving this by factoring. However, we could see that this is very difficult. So, instead, we’ll solve this by using a quadratic formula. Recall, this tells us if we have a quadratic equation π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 is equal to zero, then the two roots of this equation are given by negative 𝑏 plus or minus the square root of 𝑏 squared minus four π‘Žπ‘ all over two π‘Ž. And there’s a couple of assumptions. First, our value of π‘Ž cannot be equal to zero because this is a quadratic equation. And next, the discriminant 𝑏 squared minus four π‘Žπ‘ cannot be negative; otherwise, we won’t have real solutions.

In our case, our value of π‘Ž is one, 𝑏 is two, and 𝑐 is negative nine. We’ll substitute these into the quadratic formula. Substituting these values in gives us two possible expressions for the roots. We have π‘₯ is equal to negative two plus the square root of two squared minus four times one times negative nine all divided by two times one and π‘₯ is equal to negative two minus the square root of two squared minus four times one multiplied by negative nine all divided by two times one.

We could start simplifying these expressions. However, the question only asks for the answers to two decimal places of accuracy. Although it’s not necessary, we can find exact expressions for the roots. The first root is negative one plus root 10, and the second root is negative one minus root 10. We can then use our calculators to find the decimal expansions of these two roots. The first root we get 2.162, and this continues. And for the second root, we get negative 4.162, and this continues. And in both of these cases, the third decimal digit was two. So, we can just remove the decimal expansion, giving us the roots 2.16 and negative 4.16 to two decimal places.

Therefore, we were able to show if the three points π‘₯, one; three, π‘₯; and zero, negative two are collinear, then there are two possible values for π‘₯. To two decimal places, these are 2.16 and negative 4.16.

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