WEBVTT
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Consider the determinant π₯, π§, π¦, π¦, π₯, π§, π§, π¦, π₯.
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Given that π₯ cubed add π¦ cubed add π§ cubed equals negative 73 and π₯π¦π§ equals negative eight, determine the determinantβs numerical value.
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Remember that we can choose to calculate this determinant by picking a row or column.
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We often choose the row or column with the most amount of zeros.
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But as we have three unknowns, π₯, π¦, and π§, letβs just choose the top row.
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And here is the general formula to find the determinant of an π-by-π matrix, using a particular row, π.
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As weβre using a three-by-three matrix, π runs from one to three.
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And as weβre using the top row, π is equal to one.
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So here is the particular version of this general formula applied for our question.
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Letβs clear some space before we continue.
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Iβve kept the formula that weβre using on the screen.
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Letβs call the matrix that weβre using π΄.
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The straight lines either side of the matrix denote the determinant.
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Letβs begin by finding the different components in the formula that weβre using.
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The lowercase π one one, π one two, and π one three are the entries in the row that weβre using.
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So π one one is π₯, π one two is π§, and π one three is π¦.
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The uppercase, π΄ one one, π΄ one two, and π΄ one three denote the matrix minors.
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Remember that we get these from removing a particular row and a particular column from the original matrix.
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For example, we get π΄ one one by removing the first row and the first column, like so.
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And weβre left with the two-by-two matrix π₯, π§, π¦, π₯.
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We then do the same for π΄ one two.
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We get the matrix minor by removing the first row and the second column.
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This leaves us with the two-by-two matrix π¦, π§, π§, π₯.
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And finally, we get the matrix minor π΄ one three by removing the first row and the third column.
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This leaves us with the two-by-two matrix π¦, π₯, π§, π¦.
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But what we actually need for our formula is the determinant of each of these matrices.
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Letβs recall a method to find the determinant of a two-by-two matrix.
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So the determinant of the matrix minor π΄ one one is π₯ times π₯ minus π§ times π¦.
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We can equivalently write this as π₯ squared minus π¦π§.
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In the same way, we find the determinant of the matrix minor π΄ one two to be π¦ times π₯ minus π§ times π§, or equivalently π₯π¦ minus π§ squared.
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And we find the determinant of the matrix minor π΄ one three to be π¦ times π¦ minus π₯ times π§, which is equivalently π¦ squared minus π₯π§.
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The final thing we need to calculate for our formula are the values of negative one to the power of one add one, negative one to the power of one add two, and negative one to the power of one add three.
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Remember that raising negative one to an even power gives us one and raising negative one to an odd power gives us negative one.
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So negative one to the power of one add one is negative one squared, which is one.
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Negative one to the power of one add two is negative one cubed, which is negative one.
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And finally, negative one to the power of one add three is negative one to the fourth power, which gives us one.
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So letβs not write out the determinants of this matrix with the components that we found.
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From here, letβs just try and simplify a little bit.
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Letβs begin by multiplying together the brackets with algebraic terms.
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For the first one, we have π₯ multiplied by π₯ squared minus π¦π§.
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So this term becomes one times π₯ cubed minus π₯π¦π§.
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We then do the same with the next term.
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This gives us negative one times π₯π¦π§ minus π§ cubed.
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And repeating the same with the last term, we end up with one times π¦ cubed minus π₯π¦π§.
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So now, weβve multiplied together the brackets containing algebraic terms.
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Letβs multiply each term by the value at the front.
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The first term is one times π₯ cubed minus π₯π¦π§, which is of course going to give us π₯ cubed minus π₯π¦π§.
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The next term is negative one times π₯π¦π§ minus π§ cubed.
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So we can multiply the second bracket through by negative one.
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But weβve got to be careful because the π§ cubed already has a minus at the front of it, so thatβs going to become a positive.
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So weβre gonna end up with minus π₯π¦π§ add π§ cubed.
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Finally, the last term is just one times π¦ cubed minus π₯π¦π§.
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So this gives us π¦ cubed minus π₯π¦π§.
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We can now simplify this a little bit.
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We can gather together the π₯π¦π§ terms.
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Bringing these together gives us minus three π₯π¦π§.
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From here, we notice that we have π₯ cubed add π§ cubed add π¦ cubed.
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Weβre told in the question that π₯ cubed add π¦ cubed add π§ cubed gives us negative 73.
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Additionally, we have negative three π₯π¦π§, and weβre told in the question that π₯π¦π§ is equal to negative eight.
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So substituting in those values gives us negative 73 minus three times negative eight, which is negative 73 add 24.
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And this gives us negative 49.
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So sometimes, we encounter matrices which donβt have numerical values, but we can still find the determinant in exactly the same way.