Video: Finding the Unknown Elements of a Singular Matrix

Find the set of real values of π‘₯ that make the matrix [π‘₯ βˆ’ 4, 3, 1 and 3, 3, 1 and βˆ’1, π‘₯ βˆ’5, βˆ’4] singular.

04:24

Video Transcript

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.

Well, the key word here is this word here, singular. Well, what does singular mean? Well, when we’re looking at matrices, a matrix is singular if the determinant is equal to zero. Okay, great. So what we’re gonna do is find the determinant of our matrix. Well, before we can find the determinant of our matrix, what we need to do is remind ourselves how we do that.

Well, if we have the three-by-three matrix π‘Ž, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓, 𝑔, β„Ž, 𝑖, then the determinant of this matrix is gonna be π‘Ž, so our first element in the first row and first column, multiplied by. Then we’ve got a submatrix, a two-by-two submatrix, which you get if you cross out the column and row that π‘Ž is in. So that’s the two-by-two submatrix 𝑒, 𝑓, β„Ž, 𝑖. And what we’ve got here is the determinant of that. Then we subtract 𝑏 multiplied by the determinant of the two-by-two submatrix 𝑑, 𝑓, 𝑔, 𝑖. And then finally, we add on 𝑐 multiplied by the determinant of the two-by-two submatrix 𝑑, 𝑒, 𝑔, β„Ž. And it’s also worth noting at this point that these submatrices are also known as minors.

Okay, so now we know how to find the determinant of our matrix. Let’s go ahead and work it out. So if we’re going to call our matrix 𝐴, 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.

Okay, great, but what do we do to find out the determinant of our submatrices? Well, if we just remind ourselves how we find the determinant of a two-by-two matrix, then if we’ve got the matrix π‘Ž, 𝑏, 𝑐, 𝑑 and we want to find the determinant of it, then this is just gonna be equal to π‘Žπ‘‘ minus 𝑏𝑐. So, we multiply diagonally and then subtract. So first of all, we’re gonna have π‘₯ minus four multiplied by. Then we’ve got negative 12, so that’s three multiplied by negative four, then minus π‘₯ minus five. Then we’ve got minus three multiplied by negative 12 minus negative one. And then finally, we add on, and it’s just one multiplied by, so let’s not worry about that. We’ve got three multiplied by π‘₯ minus five minus negative three.

So now what we can do is start to tidy up and simplify this. So when we do that, what we’re gonna have is π‘₯ minus four multiplied by negative π‘₯ minus seven plus 33 plus three π‘₯ minus 15 plus three. So then, if we distribute across our parentheses, we get negative π‘₯ squared minus seven π‘₯ plus four π‘₯ plus 28. And then we add on 21, because we had 33 minus 15 plus three, then add on three π‘₯. So now all that we need to do is tidy this up. And when we do that, we’re gonna have the determinant of our matrix because it’s gonna be negative π‘₯ squared plus 49.

Okay, great, have we finished there? Well, no, because what we want to do now is set this equal to zero. 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. And what we want to do is find the set of real values of π‘₯ that make our matrix singular. So to show this working out, what we’re gonna do is clear a bit of space on the right-hand side.

So what we’re gonna have is negative π‘₯ squared plus 49 equals zero. Well then, if we divide both sides of the equation by negative one, well we’re gonna have π‘₯ squared minus 49 equals zero. Well, what we can see is that on that left-hand side now we have the difference of two squares. So if we factor this, what we’re gonna get is π‘₯ plus seven multiplied by π‘₯ minus seven equals zero. So therefore, the solutions to this equation are π‘₯ equals negative seven or seven. 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.

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