# Video: Solving a System of Three Equation Using Matrices

Find the set of values of 𝑘 for which the simultaneous equations 9𝑥 − 9𝑦 − 5𝑧 = 6, 2𝑥 + 3𝑦 + 7𝑧 = 4, 3𝑥 − 4𝑦 + 𝑘𝑧 = 7, have at least one solution.

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### Video Transcript

Find the set of values of 𝑘 for which the simultaneous equations nine 𝑥 minus nine y minus five 𝑧 equals six, two 𝑥 plus three 𝑦 plus seven 𝑧 equals four, three 𝑥 minus four 𝑦 plus 𝑘𝑧 equals seven, have at least one solution.

So, the first thing we’re gonna do is we’re gonna set up our simultaneous equations as a matrix equation. So, I’ve done that here. So, we’ve got our coefficients matrix, first of all. And make sure you are looking at the signs. So, for instance, we’ve got negative nine and negative four in the second column. Then we’ve got our variables matrix, so we’ve got 𝑥, 𝑦, and 𝑧, and then, finally, our solutions matrix.

So, what we’re looking for in this question is to find the set of values for 𝑘 which the equations have at least one solution. Well, equations will be solvable. So, the set of equations will be solvable. So, they’ll 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 we’re gonna have to do is, first of all, find out the determinant.

So, to work out the determinant of our matrix, what we’re gonna do is multiply each of the values from the first row by the determinants of the submatrices. So, we’ve got two-by-two submatrices here. So, we’ve got nine multiplied by, and then we’ve got three, seven, negative four, 𝑘. And that’s cause these’re 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, 𝑘. And we’re 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, we’ve 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 we’ve worked out the determinant of our submatrices, what we do is we multiply diagonally. So, we have three multiplied by 𝑘, which gives us three 𝑘. 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 𝑘.

So, our first part will be nine multiplied by, then we’ve got three 𝑘 minus then seven multiplied by negative four. And then we’ve got plus. And that’s plus because we had minus negative nine. So, if we minus a negative, it turns into a plus. Then, nine multiplied by two 𝑘 minus seven multiplied by three. That’s cause we used the same method as for the first determinant. Then finally, minus five multiplied by then we’ve got three multiplied by negative four minus three multiplied by three. So, this is gonna give us 27𝑘 plus 252 plus 18𝑘 minus 189 plus 85.

And if we simplify this, we’ve got 27𝑘 plus 18𝑘 which gives us 45𝑘. And then, we’ve got 252 minus 189 plus 85 which gives us plus 148. So, we could say that the determinant is 45𝑘 plus 148.

So, now what we do is because we’ve found out the value of this determinant, we’re gonna set it equal to zero. And the reason we’re going to do that is because we wanna find the value of 𝑘 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 45𝑘 is equal to negative 148. So therefore, to find 𝑘 we divide each side of the equation by 45. When we do that, we get 𝑘 is equal to negative 148 over 45. So therefore, we can say that the equations will be solvable for all real values of 𝑘 apart from negative 148 over 45.

So, how would we write this? So, we’d represent it using this 𝑅. So, this is an 𝑅 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 𝑘 is equal to negative 148 over 45.