Question Video: Finding the Unknown Elements of a Singular Matrix Mathematics

Find the set of real values of (π‘₯) that make the matrix [π‘₯ βˆ’ 3, 8 and 2, π‘₯ + 3] singular.


Video Transcript

Find the set of real values of π‘₯ that make the two-by-two matrix π‘₯ minus three, eight, two, π‘₯ plus three singular.

Let’s begin by defining the word singular when it comes to matrices. We say that a matrix is singular if it’s not invertible; it doesn’t have an inverse. We know that a matrix is invertible if its determinant is not equal to zero, and the converse is also true. So, in other words, a matrix is singular if its determinant is equal to zero. In this case then, we need to find the set of real values of π‘₯ such that the determinant of our matrix is equal to zero. The determinant of a two-by-two matrix π‘Ž, 𝑏, 𝑐, 𝑑 is π‘Žπ‘‘ minus 𝑏𝑐. We subtract the product of the elements in the top right and bottom left from the product of those in the top left and bottom right.

So, in this case, that’s π‘₯ minus three times π‘₯ plus three minus eight times two. If we distribute these parentheses, we get π‘₯ times π‘₯, which is π‘₯ squared, plus three π‘₯ minus three π‘₯ minus three times three, which is nine. That simplifies to π‘₯ squared minus nine. And eight multiplied by two is 16. So the determinant of our matrix is π‘₯ squared minus nine minus 16, which is π‘₯ squared minus 25. We’re trying to find the set of values of π‘₯ that make our matrix singular. In other words, which values of π‘₯ make the determinant zero? So, let’s set our expression for the determinant equal to zero and solve for π‘₯. That is, π‘₯ squared minus 25 equals zero.

Adding 25 to both sides of this equation gives us π‘₯ squared equals 25. And then we’ll take the square root of both sides of our equations, remembering to take the positive and negative square root of 25. That gives us π‘₯ is equal to positive or negative five. We can use these squiggly brackets to help us represent the set of values that make our matrix singular. They are negative five and five. Note that at this stage, we could check our solutions by substituting each value of π‘₯ into our original matrix and then checking that the determinant is indeed equal to zero.

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