WEBVTT
00:00:00.040 --> 00:00:13.320
Find the set of real values of π₯ that make the matrix π₯ minus four, three, one, three, three, one, negative one, π₯ minus five, negative four singular.
00:00:14.440 --> 00:00:17.360
Well, the key word here is this word here, singular.
00:00:17.900 --> 00:00:19.310
Well, what does singular mean?
00:00:20.290 --> 00:00:25.960
Well, when weβre looking at matrices, a matrix is singular if the determinant is equal to zero.
00:00:26.250 --> 00:00:26.960
Okay, great.
00:00:27.530 --> 00:00:30.610
So what weβre gonna do is find the determinant of our matrix.
00:00:31.810 --> 00:00:36.250
Well, before we can find the determinant of our matrix, what we need to do is remind ourselves how we do that.
00:00:36.840 --> 00:00:49.770
Well, if we have the three-by-three matrix π, π, π, π, π, π, π, β, π, then the determinant of this matrix is gonna be π, so our first element in the first row and first column, multiplied by.
00:00:50.050 --> 00:00:56.980
Then weβve got a submatrix, a two-by-two submatrix, which you get if you cross out the column and row that π is in.
00:00:57.210 --> 00:01:00.590
So thatβs the two-by-two submatrix π, π, β, π.
00:01:00.740 --> 00:01:02.550
And what weβve got here is the determinant of that.
00:01:02.920 --> 00:01:09.020
Then we subtract π multiplied by the determinant of the two-by-two submatrix π, π, π, π.
00:01:09.020 --> 00:01:16.810
And then finally, we add on π multiplied by the determinant of the two-by-two submatrix π, π, π, β.
00:01:18.260 --> 00:01:22.470
And itβs also worth noting at this point that these submatrices are also known as minors.
00:01:22.990 --> 00:01:26.630
Okay, so now we know how to find the determinant of our matrix.
00:01:26.770 --> 00:01:28.190
Letβs go ahead and work it out.
00:01:29.460 --> 00:01:54.020
So if weβre going to call our matrix π΄, then the determinant of our matrix is π₯ minus four multiplied by the determinant of the submatrix three, one, π₯ minus four, negative four minus three multiplied by the determinant of the submatrix three, one, negative one, negative four plus one multiplied by the determinant of the submatrix three, three, negative one, π₯ minus five.
00:01:54.660 --> 00:01:59.010
Okay, great, but what do we do to find out the determinant of our submatrices?
00:02:00.380 --> 00:02:12.340
Well, if we just remind ourselves how we find the determinant of a two-by-two matrix, then if weβve got the matrix π, π, π, π and we want to find the determinant of it, then this is just gonna be equal to ππ minus ππ.
00:02:12.640 --> 00:02:15.560
So, we multiply diagonally and then subtract.
00:02:17.370 --> 00:02:20.920
So first of all, weβre gonna have π₯ minus four multiplied by.
00:02:20.920 --> 00:02:25.990
Then weβve got negative 12, so thatβs three multiplied by negative four, then minus π₯ minus five.
00:02:27.400 --> 00:02:32.010
Then weβve got minus three multiplied by negative 12 minus negative one.
00:02:32.010 --> 00:02:37.350
And then finally, we add on, and itβs just one multiplied by, so letβs not worry about that.
00:02:37.830 --> 00:02:42.020
Weβve got three multiplied by π₯ minus five minus negative three.
00:02:43.640 --> 00:02:46.580
So now what we can do is start to tidy up and simplify this.
00:02:47.880 --> 00:02:56.560
So when we do that, what weβre gonna have is π₯ minus four multiplied by negative π₯ minus seven plus 33 plus three π₯ minus 15 plus three.
00:02:57.910 --> 00:03:03.380
So then, if we distribute across our parentheses, we get negative π₯ squared minus seven π₯ plus four π₯ plus 28.
00:03:04.690 --> 00:03:10.240
And then we add on 21, because we had 33 minus 15 plus three, then add on three π₯.
00:03:11.530 --> 00:03:13.410
So now all that we need to do is tidy this up.
00:03:14.510 --> 00:03:19.630
And when we do that, weβre gonna have the determinant of our matrix because itβs gonna be negative π₯ squared plus 49.
00:03:19.730 --> 00:03:21.530
Okay, great, have we finished there?
00:03:21.560 --> 00:03:24.500
Well, no, because what we want to do now is set this equal to zero.
00:03:24.650 --> 00:03:33.040
And the reason we want to set it equal to zero is because if we remind ourselves of what we looked at at the beginning, we know that a matrix is singular if the determinant is equal to zero.
00:03:33.270 --> 00:03:37.680
And what we want to do is find the set of real values of π₯ that make our matrix singular.
00:03:39.170 --> 00:03:42.340
So to show this working out, what weβre gonna do is clear a bit of space on the right-hand side.
00:03:43.650 --> 00:03:47.110
So what weβre gonna have is negative π₯ squared plus 49 equals zero.
00:03:48.360 --> 00:03:55.250
Well then, if we divide both sides of the equation by negative one, well weβre gonna have π₯ squared minus 49 equals zero.
00:03:55.750 --> 00:03:59.350
Well, what we can see is that on that left-hand side now we have the difference of two squares.
00:04:00.570 --> 00:04:05.470
So if we factor this, what weβre gonna get is π₯ plus seven multiplied by π₯ minus seven equals zero.
00:04:06.880 --> 00:04:10.430
So therefore, the solutions to this equation are π₯ equals negative seven or seven.
00:04:11.620 --> 00:04:23.040
So therefore, we can say that the set of real values of π₯ that make the matrix π₯ minus four, three, one, three, three, one, negative one, π₯ minus five, negative four singular are negative seven, seven.