WEBVTT
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Find, in its simplest form, an expression for the determinant.
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Hereβs our three-by three-matrix.
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To find the determinate in this three-by-three matrix, weβll use the first row.
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The first row the first column will be positive, first row second column negative, and first row third column positive.
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To find the determinant, weβre gonna break this up into three smaller determinants.
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Starting with nine minus two π, we want to take this value and multiply it by the smaller matrix created by removing both the row and the column that nine minus two π is located in.
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Weβll multiply positive nine minus two π times the determinant of the two-by-two matrix created here.
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And this will be our first term.
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Our second term will be negative negative two π multiplied by the two-by-two matrix we create when we remove the row and the column that negative two π is found in.
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The two-by-two matrix would look like this: negative two π, seven π, negative two π, nine plus seven π.
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This will be our second term.
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Following in the same pattern, weβll take seven π and multiply it by the two-by-two matrix created when you remove the column and the row that seven π is located in, which will look like this and would be our third term.
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Now we have some major algebra to do.
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Iβm going to bring down the part that comprises the first term.
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And for now, weβll ignore the second and third term.
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Nine minus two π is what we start with, but remember we need the determinant of this two-by-two matrix.
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And that means multiplying negative one minus two π times nine plus seven π and then subtracting that value from seven π times negative two π.
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And all of this will solve for our first term.
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Looking inside the brackets, we need to multiply negative one minus two π times nine plus seven π.
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Negative one times nine is negative nine.
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Negative one times seven π, negative seven π.
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Moving to our negative two π, multiply that by nine, and we get negative 18π.
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Negative two π times seven π equals negative 14ππ.
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Still inside the brackets, we need to multiply seven π times negative two π.
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That equals negative 14ππ.
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And weβre dealing with subtraction.
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We know that subtracting a negative can be represented with addition.
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And now inside the brackets, we have negative 14ππ plus 14ππ.
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Those cancel out.
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The remaining portion of our brackets is negative nine minus seven π minus 18π.
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And we need to multiply this by nine minus two π.
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Nine times negative nine equals negative 81.
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Nine times negative seven π equals negative 63π.
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Nine times negative 18π equals negative 162π.
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And with our negative two π, negative two π times negative nine equals 18π.
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Negative two π times negative seven π equals positive 14ππ.
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And negative two π times negative 18π equals positive 36ππ.
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And this value is our first term.
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We need to follow the same process to find terms two and three.
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For term two, weβre dealing with negative negative two π, which we can write as positive two π.
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And we need to multiply two π by the determinant of this two- by-two matrix.
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That determinant is found by multiplying negative two π times nine plus seven π and then subtracting seven π times negative two π.
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Inside the brackets, negative two π times nine is negative 18π.
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Negative two π times seven π equals negative 14ππ.
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Weβre subtracting seven π times negative two π, negative 14ππ.
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Again, weβre subtracting a negative.
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We can make that more clear by making an addition.
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What we see now is negative 14ππ plus 14ππ inside the brackets.
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Adding those together, we get zero.
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Now we need to multiply negative 18π by two π.
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Weβre left with negative 36ππ.
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This is our second term.
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We have the same procedure one more time.
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Weβll put it here: seven π times the determinant of this two-by-two matrix.
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We find that by multiplying negative two π times negative two π and then subtracting negative one minus two π times negative two π.
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Moving inside the brackets, negative two π times negative two π equals four ππ.
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We are subtracting.
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Weβll distribute this negative two π over the negative one minus two π.
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Negative two π times negative one equals two π.
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Negative two π times negative two π equals positive four ππ.
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Now we need to distribute this negative sign to both the two π and the four ππ.
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And inside the brackets, weβll say four ππ minus two π minus four ππ.
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The four ππs cancel out, and we have seven π times negative two π, multiplied together, equals negative 14ππ, which is our third term.
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The next thing we need to do is add our first, second, and third terms together.
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Iβm gonna clear off some space to make room to do that.
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So if you wanna pause the video to copy any of this down, you can do that now.
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So here we go!
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Our first time plus our second term plus our third term.
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Do you notice anything here?
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When weβre adding, we have a positive 14ππ and a negative 14ππ; these will cancel out β as well, positive 36ππ and negative 36ππ β leaving us with negative 81 minus 63π minus 162π and 18π.
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We can simplify this a little bit further by noticing that all the constants are factors of nine.
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If we take out negative nine, divide negative 81 by negative nine, and you get nine, divide negative 63π by negative nine, and you get positive seven π, divide negative 162π by negative nine, and you get positive 18π.
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18π divided by negative nine equals negative two π.
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The simplified expression of this determinant, the simplified form of this determinant, is negative nine times nine plus seven π plus 18π minus two π.