In Exercises 57–60, solve each equation for x.

|1 x - 2|
|3 1 1...

Question

In Exercises 57–60, solve each equation for x. |1 x - 2| |3 1 1| = - 8 |0 - 2 2|

Matrix equation for exercise 59 in college algebra, solving for x.

Accepted Answer

Welcome back. I am so glad you're here. We are given this matrix that has an X inside of it and we're asked to solve for X. Well, the way we're going to do that is we're actually going to find the determinant of this matrix and then set that determinant equal to negative 22. And then we'll be able to solve for X. So I am going to copy over this matrix starting with the first column negative 32, negative two and then X. The negative 110. Now recalling from previous lessons on how to find the determinant of a three by three matrix, I'm going to copy over the first two columns again. So again negative 32 negative two and 45 X. Now we are going to diagonally multiply, starting in the upper left hand corner and going down negative three times five times zero. Well, anything times zero is zero and then going to the next diagonal four times one times negative two, which will be negative eight. And then the last diagonal is negative one times two times X, which is minus two X. And we are going to combine these three as much as we can. So the like terms here are zero minus eight. Well that's just minus eight and then minus two X. Which is not a like term with the constance That's as simplified as that one gets. So now we'll do the diagonals going in the other direction. Starting in the lower left hand corner and working our way up. So we have negative two times five times negative one which is a positive 10 and then we have X times one times negative three. That will be a minus three X. Then we have zero times two times four. Well anything times zero is zero so plus zero and combining like terms where we can well 10 plus zero are like terms and that's just 10. So this is 10 minus three X. And now we recall from those previous lessons on determinants that we're going to take the first answer and subtract our second answers. So negative eight minus two X minus. And put in parentheses 10 minus three. X. And I put that in parentheses because I want to be sure to distribute that negative to both the 10 and the negative three X. So let's do that. Next copy down negative eight minus two X. And then it'll be minus 10 plus three X. Now combining like terms negative eight minus 10 is negative 18 and our X terms negative two X plus three X. Is plus X. Well those two can't be combined anymore. So now that's our determinant and we're going to set that equal to negative 22 in order to solve for X. So we've got negative 18 plus X equals negative 22 will add 18 to both sides to get X by itself and we get X equals negative 22 plus 18 is negative four. And we've done it we've solved for X. We look at our answer choices and this matches with answer choice D. Well done, we'll catch you on the next one.

In Exercises 57–60, solve each equation for x. |1 x - 2| |3 1 1| = - 8 |0 - 2 2|

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