# Question Video: Finding an Unknown in a System of Three Equations Using a Matrix Inverse Mathematics

Consider the following matrix equation: [−7, −5, −7 and 8, 10, 7 and 9, 7, 7][𝑥 and 𝑦 and 𝑧] = [1 and 2 and 𝑘]. Find the value of 𝑘 that results in 𝑥 = 1/2.

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### Video Transcript

Consider the following matrix equation: the three-by-three matrix with elements negative seven, negative five, negative seven, eight, 10, seven, nine, seven, seven multiplied by the column matrix with elements 𝑥, 𝑦, 𝑧 is equal to the column matrix with elements one, two, 𝑘. Find the value of 𝑘 that results in 𝑥 is equal to one over two.

The given matrix equation represents a system of linear equations which you would normally want to solve for the unknowns 𝑥, 𝑦, and 𝑧. However, in this case, we’re given a value for 𝑥, that’s 𝑥 is equal to one-half, and asked to find the value of 𝑘. Either way, we still have three equations with three unknowns. And to solve, we can use the matrix inverse method.

Beginning by labeling our matrices 𝐴, 𝐮, and 𝐯, we’ll see how this method works. We have the equation 𝐴 multiplied by 𝐮 is equal to 𝐯. If we assume that 𝐴 is invertible, that is, its inverse exists, and multiply through on the left by 𝐴 inverse, we have 𝐴 inverse 𝐴 multiplied by 𝐮 is equal to 𝐴 inverse 𝐯. We know that 𝐴 inverse multiplied by 𝐴 is the identity matrix. So in fact, on our left-hand side, we have the identity matrix multiplied by 𝐮, which is simply 𝐮. This isolates 𝐮 on the left-hand side. And once we’ve found 𝐴 inverse, if we multiply out the right-hand side, we’ll find our solution. So in fact, we need to find 𝐴 inverse.

First, making some room, we know that for an 𝑛-by-𝑛 invertible matrix 𝐴, 𝐴 inverse is equal to one over the determinant of 𝐴 multiplied by the adjoint matrix of 𝐴. And remember that the adjoint matrix of 𝐴 is the transpose of the matrix of cofactors of 𝐴, which we’ll come back to later on. But let’s first find the determinant of 𝐴. Remember that if we expand along row 𝑖 of our matrix, we can find the determinant using the formula the sum from 𝑗 is one to 𝑛 of element 𝑎𝑖𝑗 multiplied by negative one raised to the power 𝑖 plus 𝑗 multiplied by the determinant of the matrix minor capital 𝐴 subscript 𝑖𝑗.

And remember that the matrix minor is the two-by-two matrix resulting from removing row 𝑖 and column 𝑗 from our matrix 𝐴. In our case, if we expand along the first row, then we have the determinant of 𝐴 is equal to the element 𝑎 one one multiplied by the two-by-two determinant of matrix minor 𝐴 sub one one minus the element in row one, column two multiplied by the determinant of its matrix minor plus the element in row one, column three multiplied by the determinant of its matrix minor. And notice that our sign is determined by negative one raised to the power 𝑖 plus 𝑗. If 𝑖 plus 𝑗 is even, then our sign is positive. And if 𝑖 plus 𝑗 is odd, our sign is negative.

Element 𝑎 one one is negative seven. And we obtain our first matrix minor by removing row one and column one from our matrix. And we obtain our next two terms similarly by removing row one, column two and row one, column three. Now recalling that for a two-by-two matrix with elements 𝑎, 𝑏, 𝑐, and 𝑑, the determinant is given by 𝑎𝑑 minus 𝑏𝑐. Our first term is then negative seven multiplied by 10 multiplied by seven minus seven multiplied by seven. That’s negative seven multiplied by 21, which is negative 147.

For our second term, we have plus five multiplied by eight multiplied by seven minus nine multiplied by seven, which is negative seven. That is negative 35. And for our third term, our determinant is eight multiplied by seven minus 10 multiplied by nine, which is negative 34, and multiplied by negative seven, that’s 238. The determinant of our matrix 𝐴 is therefore negative 147 minus 35 plus 238, which is 56.

Making some room, and now that we have the determinant of 𝐴, let’s find the adjoint matrix of 𝐴. Recall that the adjoint matrix of an 𝑛-by-𝑛 matrix is the transpose of the matrix of cofactors. That’s where the cofactor of element 𝑖𝑗 is negative one raised to the power 𝑖 plus 𝑗 multiplied by the determinant of the matrix minor 𝐴 𝑖𝑗. The negative one raised to the power 𝑖 plus 𝑗 gives us the sign or parity. So for a three-by-three matrix, our signs are positive, negative, positive; negative, positive, negative; and positive, negative, positive.

To find the adjoint matrix of 𝐴 then, we need to find the cofactors. In fact, we’ve already seen three of these cofactors, those for the first row. The cofactor 𝑐 one one is the positive determinant of the matrix minor 𝐴 subscript one one. That’s the matrix with elements 10, seven, seven, seven, and similarly for cofactors 𝑐 one two and 𝑐 one three. And we know that our signs are positive, negative, and positive.

Our first determinant, as we saw earlier, is 10 multiplied by seven minus seven multiplied by seven, and that’s 21. Our second cofactor is negative eight multiplied by seven minus seven multiplied by nine. And that’s negative negative seven, which is positive seven. Our third cofactor 𝑐 one three is eight multiplied by seven minus 10 times nine, which is negative 34.

So now we do the same thing for the other two rows in our matrix. So by exactly the same process, but noticing our signs again negative, positive, and negative, we have the cofactor 𝑐 two one is negative 14, the cofactor 𝑐 two two is positive 14, and the cofactor 𝑐 two three is four. And finally, for our third row, we have the cofactor 𝑐 three one is 35, 𝑐 three two is negative seven, and 𝑐 three three is negative 30 and where our signs were positive, negative, and positive.

Our matrix of cofactors, therefore, has the elements 21, seven, negative 34; negative 14, 14, and four; and 35, negative seven, and negative 30. And recall that the adjoint of our matrix 𝐴 is the transpose of the matrix of cofactors. The transpose is the matrix where we swap the rows and columns. In our case, this means, for example, that the row with elements 21, seven, and negative 34 becomes the column with elements 21, seven, and negative 34. Similarly, for our second row, which becomes our second column, and our third row becomes our third column. And this is the adjoint matrix of our matrix 𝐴.

Now remember, it’s the inverse of our matrix 𝐴 that we’re trying to find because it’s the inverse that will help us to solve our equation. And the inverse of the matrix 𝐴 is one over the determinant of the matrix 𝐴 multiplied by the adjoint. Our determinant, we found before, is 56, so the inverse of matrix 𝐴 is one over 56 multiplied by the adjoint of our matrix 𝐴. So now we can use this to solve our system of linear equations.

We have 𝐮, which is the column matrix with elements 𝑥, 𝑦, and 𝑧, is equal to 𝐴 inverse multiplied by the column matrix 𝐯. Now recall going back to our original question, we have a value for 𝑥. That’s 𝑥 is equal to one over two. And now is as good a time as any to substitute this into our equation. On our left-hand side then, we have the column matrix with elements one-half, 𝑦, and 𝑧. Now, in fact, to find the value of 𝑘, we need only multiply out the first row with the column matrix on the right-hand side, since in the resulting equation there’s only one unknown, and that’s 𝑘.

For the sake of completion, however, let’s multiply out the whole of our right-hand side. We can then also find the values of 𝑦 and 𝑧. Our first row is 21 times one plus negative 14 times two plus 35 times 𝑘, and similarly for our second and third rows. Now collecting like terms, we have one over 56 multiplied by the matrix with row one negative seven plus 35𝑘, row two 35 minus seven 𝑘, and row three negative 26 minus 30𝑘.

So now, by equality of matrices, we have one-half is equal to one over 56 multiplied by negative seven plus 35𝑘. Multiplying both sides by 56, we have 56 over two is negative seven plus 35𝑘. We have 28 is negative seven plus 35𝑘. Adding seven to both sides and dividing through by 35 then gives us 𝑘 is equal to one. Hence, the value of 𝑘 that results in 𝑥 is equal to a half in our matrix equation is 𝑘 is equal to one.

And now for completion’s sake, we can use this value for 𝑘 to solve for 𝑦 and 𝑧. And making some room, we then have 𝑦 is equal to one over 56 multiplied by 35 minus seven and 𝑧 is equal to one over 56 multiplied by negative 26 minus 30. These then evaluate to one over two and negative one. Hence, we have 𝑘 is equal to one, and 𝑦 is equal to a half, 𝑧 is equal to negative one.