### Video Transcript

Find the set of values of ๐ for which the simultaneous equations nine ๐ฅ minus nine y minus five ๐ง equals six, two ๐ฅ plus three ๐ฆ plus seven ๐ง equals four, three ๐ฅ minus four ๐ฆ plus ๐๐ง equals seven, have at least one solution.

So, the first thing weโre gonna do is weโre gonna set up our simultaneous equations as a matrix equation. So, Iโve done that here. So, weโve got our coefficients matrix, first of all. And make sure you are looking at the signs. So, for instance, weโve got negative nine and negative four in the second column. Then weโve got our variables matrix, so weโve got ๐ฅ, ๐ฆ, and ๐ง, and then, finally, our solutions matrix.

So, what weโre looking for in this question is to find the set of values for ๐ which the equations have at least one solution. Well, equations will be solvable. So, the set of equations will be solvable. So, theyโll have at least one solution, if the determinant is not equal to zero. So therefore, to find out where the determinant is not equal to zero what weโre gonna have to do is, first of all, find out the determinant.

So, to work out the determinant of our matrix, what weโre gonna do is multiply each of the values from the first row by the determinants of the submatrices. So, weโve got two-by-two submatrices here. So, weโve got nine multiplied by, and then weโve got three, seven, negative four, ๐. And thatโs cause theseโre the values left if you eliminate the row and column that the nine was in.

And then, this is minus negative nine multiplied by the determinant of the submatrix, which is a two-by-two submatrix, two, seven, three, ๐. And weโre subtracting this because when we look at the pattern, we have add, subtract, add, subtract and it goes on it like this with the columns. So, the first column is positive, second column is gonna be negative, et cetera. And then, finally, weโve got add negative five multiplied by the determinant of the submatrix two, three, three, negative four.

So then, we can calculate these. Just to remind us how weโve worked out the determinant of our submatrices, what we do is we multiply diagonally. So, we have three multiplied by ๐, which gives us three ๐. Then, we have seven multiplied by negative four. And then, what we do is we subtract the seven multiplied by negative four away from the three multiplied by ๐.

So, our first part will be nine multiplied by, then weโve got three ๐ minus then seven multiplied by negative four. And then weโve got plus. And thatโs plus because we had minus negative nine. So, if we minus a negative, it turns into a plus. Then, nine multiplied by two ๐ minus seven multiplied by three. Thatโs cause we used the same method as for the first determinant. Then finally, minus five multiplied by then weโve got three multiplied by negative four minus three multiplied by three. So, this is gonna give us 27๐ plus 252 plus 18๐ minus 189 plus 85.

And if we simplify this, weโve got 27๐ plus 18๐ which gives us 45๐. And then, weโve got 252 minus 189 plus 85 which gives us plus 148. So, we could say that the determinant is 45๐ plus 148.

So, now what we do is because weโve found out the value of this determinant, weโre gonna set it equal to zero. And the reason weโre going to do that is because we wanna find the value of ๐ for which the set of equations is not solvable. Because as we said, the set of equations will be solvable if the determinant is not equal to zero.

So, what we do is we subtract 148 from each side of the equation. And when we do that, we get 45๐ is equal to negative 148. So therefore, to find ๐ we divide each side of the equation by 45. When we do that, we get ๐ is equal to negative 148 over 45. So therefore, we can say that the equations will be solvable for all real values of ๐ apart from negative 148 over 45.

So, how would we write this? So, weโd represent it using this ๐
. So, this is an ๐
which means all real values, then minus. And then, in our brackets, we have negative 148 over 45. So, as we said this means all real values except where ๐ is equal to negative 148 over 45.