Video: Solving Linear Equations Resulting from Evaluating Second-Order Determinants

Find all values of 𝑥 for which the determinant 𝑥, −2, −2, 𝑥 + 6, 3, 1, 8 = 45.


Video Transcript

Find all values of 𝑥 for which the determinant 𝑥, negative two, negative two, 𝑥 plus the determinant six, three, one, eight equals 45.

So these lines do not mean absolute value, they represent the determinant, and the way to find the determinant of a matrix of numbers is to take 𝑎𝑑 minus 𝑏𝑐. Essentially, you’re kind of cross multiplying and subtracting, 𝑎 times 𝑑 minus 𝑏 times 𝑐. So let’s go ahead and look at our equation.

To take the determinant of the first matrix, 𝑥 times 𝑥 is 𝑥 squared minus negative two times negative two which is positive four. And now we’re adding the determinant of the other matrix, six times eight, which is 48, minus three times one, which is three. And we set it equal to 45. So now we need to combine like terms. 48 minus three is 45. Now there’s two ways we could solve from here. Since there’s 45s in both sides of the equation, we could subtract it. They cancel, and we get 𝑥 squared minus four equal zero. And this is a difference of squares because 𝑥 is being squared and four is a perfect square, it’s two squared. So this was factored to be 𝑥 plus two, 𝑥 minus two, and we set each factor equal to zero. And now we solve.

So we subtract two from the first equation and add two to the second equation. So we get 𝑥 equals negative two or 𝑥 equals positive two. The other way we could’ve solved from this point is to add negative four and 45. And we would’ve had 𝑥 squared plus 41 equals 45. And now we subtract 41 from both sides of the equation. So we get that 𝑥 squared equals four. And now we square root both sides. And we get that 𝑥 equals plus or minus two, which is the exact same thing that we got before.

So our final answer is: 𝑥 equals negative two or 𝑥 equals positive two.

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