### Video Transcript

For the matrix one, π‘, π‘ squared, zero, one, two π‘, π‘, zero, two, does there exist a value of π‘ for which it fails to have an inverse? If so, what is this value?

Well, the first thing weβre gonna have a look at is the fact that what weβre trying to do is find a value of π‘ for which this matrix fails to have an inverse. Well, if it fails to have an inverse, then itβs in fact known as being singular. Well, what we know about a singular matrix is that a matrix is singular if the determinant is equal to zero.

Okay, great. So we can use that to help us solve this problem. So therefore, the first thing we need to remind ourselves is how to find the determinant of the three-by-three matrix. Well, if we have the matrix π΄, which is π, π, π, π, π, π, π, β, π, then the determinant is equal to π multiplied by, then we got the minor π, π, β, π β or this can also be known as the determinant of the two-by-two submatrix π, π, β, π β then minus π multiplied by, then weβve got the minor π, π, π, π plus π multiplied by the minor π, π, π, β. So letβs use this to find out the determinant of our matrix.

Well, weβre gonna call it π΄, our matrix. So therefore, what we want to do is find the determinant of π. Then what weβre gonna have, first of all, is one multiplied by, then weβve got the determinant of the two-by-two submatrix one, two π‘, zero, two. And just to remind ourselves how we got our two-by-two submatrix, well, what we do is we delete the row and column that our first element one is in. And then we look at the other four values or elements that are left. Weβve got one, two π‘, zero, two. And then we have minus π‘ multiplied by the determinant of the two-by-two submatrix zero, two π‘, π‘, two plus π‘ squared multiplied by the determinant of the matrix zero, one, π‘, zero.

Okay, great. So now whatβs the next step? Weβll now we quickly need to remind ourselves how do we find the determinant of the two-by-two submatrix. Well, if we want to find the determinant of a two-by-two matrix like this one here π, π, π, π, then what we do is we cross multiply, and then we subtract. So what we have is π multiplied by π, so ππ, minus π multiplied by π. Well, what weβre gonna have is one multiplied by two. And we get that because what weβve got is one multiplied by the determinant of the two-by-two matrix one, two π‘, zero, two.

Well, if we multiply one by two, thatβs gonna give us two. And then we subtract zero by two π‘. Well, we donβt have to subtract anything cause zero multiplied anything is just zero. Then weβve got minus π‘ multiplied by negative two π‘ squared plus π‘ squared multiplied by negative π‘. Well, this is gonna give us our determinant of π΄ as two plus two π‘ cubed minus π‘ cubed. So what weβre gonna do is just quickly simplify. Well, this is just gonna be two plus π‘ cubed. So okay, great. Well, now whatβs next?

Well, what we remember is that we can see if our matrix is in fact singular by seeing whether our determinant can be equal to zero. So what weβre gonna do is set two plus π‘ cubed equal to zero. So weβre gonna have two plus π‘ cubed equals zero. So then what weβre gonna do is subtract two from each side of the equation, which is gonna give us π‘ cubed equals negative two. And then we take the cube root of both sides. So what we get is π‘ is equal to the negative cube root of two. So therefore, in answer to our question, βDoes there exist a value of π‘ for which our matrix fails to have an inverse?β, the answer is yes. Itβs when π‘ is equal to negative cube root of two.