Question Video: Finding an Unknown in a System of Three Equations Using a Matrix Inverse Mathematics • 10th Grade

Consider the following matrix equation: [−6, 3, −1 and 4, 0, −6 and 2, 2, 1][𝑥 and 𝑦 and 𝑧] = [3 and 𝑘 and 5]. Find the value of 𝑘 that results in 𝑦 = 31/16.

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Video Transcript

Consider the following matrix equation: the three-by-three matrix with elements negative six, three, negative one, four, zero, negative six, two, two, one multiplied by the column matrix with elements 𝑥, 𝑦, and 𝑧 is equal to the column matrix with elements three, 𝑘, and five. Find the value of 𝑘 that results in 𝑦 is equal to 31 over 16.

The given matrix equation represents a system of three linear equations, which we would normally want to solve for the unknowns 𝑥, 𝑦, and 𝑧. However, in this case, we’re given a value for 𝑦; that’s 𝑦 is equal to 31 over 16. And we’re asked to find the value of 𝑘. We still therefore have three equations and three unknowns. And to solve these, we can use the matrix inverse method.

If we label our three matrices 𝐴, 𝐮, and 𝐯 to begin with, we’ll see how this method works on our equation 𝐴 multiplied by 𝐮 is equal to 𝐯. Assuming that 𝐴 is invertible, that is, its inverse exists, then multiplying our equation through on the left by 𝐴 inverse, we have 𝐴 inverse 𝐴 multiplied by 𝐮 is equal to 𝐴 inverse multiplied by 𝐯. Now we know that for an 𝑛-by-𝑛 invertible matrix 𝐴, 𝐴 multiplied by 𝐴 inverse is 𝐴 inverse multiplied by 𝐴 is the identity matrix. That’s the 𝑛-by-𝑛 matrix with all elements equal to zero except the leading diagonal with elements all equal to one and which for a three-by-three matrix is as shown.

Then, on the left-hand side of our equation, we have 𝐼 multiplied by 𝐮. And we know that 𝐼 multiplied by 𝐮 is simply 𝐮. So, this isolates the matrix 𝐮 with elements 𝑥, 𝑦, and 𝑧 on the left-hand side. And if we then multiply out the right-hand side, we’ll find our solution. And to multiply out the right-hand side, we need to find the inverse of the matrix 𝐴. Now making some room, we recall that for an 𝑛-by-𝑛 invertible matrix 𝐴, 𝐴 inverse is given by one over the determinant of 𝐴 multiplied by the adjoint matrix of 𝐴. So, we’ll need to find both the determinant of 𝐴 and the adjoint matrix of 𝐴, remembering that the adjoint matrix of 𝐴 is the transpose of the matrix of cofactors. So, let’s begin by finding the determinant of the matrix 𝐴.

Since the second row of our matrix 𝐴 has a zero element, let’s expand along this row. And remember, to find the determinant of a matrix by expanding along the 𝐼th row, we use the formula the sum from 𝑗 is one to 𝑛 multiplied by the element 𝑎 𝑖𝑗 multiplied by negative one raised to the power 𝑖 plus 𝑗 multiplied by the determinant of the matrix minor 𝐴 𝑖𝑗. Remember, the matrix minor 𝐴 𝑖𝑗 of a three-by-three matrix is the two-by-two matrix obtained by removing row 𝑖 and column 𝑗. In our case then, we have the determinant of 𝐴 is equal to negative the element 𝑎 two one multiplied by the determinant of the matrix minor 𝐴 two one and so on.

Note that it’s very important to get the signs correct at this point. And these are given by negative one raised to the power 𝑖 plus 𝑗. So, for example, in our first term, we have negative one raised to the power two plus one, which is negative one raised to the power three, and that’s negative one. Our first term is then four multiplied by negative one multiplied by the determinant of the two-by-two matrix obtained by removing row two and column one. That is the two-by-two matrix with elements three, negative one, two, and one. Similarly, we obtain our second term and our third term. And we can evaluate our determinants by recalling that for a two-by-two matrix 𝑀 with elements 𝑎, 𝑏, 𝑐, 𝑑, its determinant is 𝑎𝑑 minus 𝑏𝑐.

For our first term, this then gives us negative four multiplied by three multiplied by one minus negative one multiplied by two and similarly for our second and third terms. Evaluating then gives us negative four times five plus zero times negative four plus six times negative 18. That is negative 20 minus 108, which is negative 128. The determinant of our matrix 𝐴 is then negative 128. So, we have our determinant. Let’s now find the adjoint matrix of 𝐴.

Recalling that the adjoint of a matrix 𝐴 is the transpose of the matrix of cofactors and a cofactor 𝐶 𝑖𝑗 is negative one raised to the power 𝑖 plus 𝑗 multiplied by the determinant of the matrix minor 𝐴 𝑖𝑗. The term negative one raised to the power 𝑖 plus 𝑗 gives us the sign or parity of the cofactor. And for a three-by-three matrix, this is as shown. In fact, we’ve already worked out three of the cofactors, those of the second row. But let’s write them all out in full. Our nine cofactors are as shown.

And so, now, we work out the two-by-two determinants of the matrix minors. So, for example, for our cofactor 𝐶 one one, we have the determinant zero multiplied by one minus negative six multiplied by two, and that’s equal to 12. Our cofactors are then 12, negative 16, eight, negative five, negative four, 18, and negative 18, negative 40, negative 12.

Now, writing out our matrix of cofactors, the adjoint of our matrix 𝐴 is then the transpose of the matrix of cofactors. That is the matrix where the rows and columns are interchanged. Now, remember, we’re trying to find the inverse of the matrix 𝐴. And that’s given by one over the determinant times the adjoint of matrix 𝐴. Our determinant is negative 128 and our adjoint is as shown. So, the inverse of our matrix 𝐴 is negative one over 128 multiplied by the adjoint of matrix 𝐴. So, now, we can substitute this into our original equation. That is 𝐮 is equal to 𝐴 inverse multiplied by 𝐯.

Now, remember, we’re to find the value of 𝑘 that results in 𝑦 is equal to 31 over 16. So, before solving our equation, let’s substitute this value in for 𝑦. We can solve this by multiplying out the right-hand side. And using matrix multiplication, for example, in our first row, we have 12 times three plus negative five times 𝑘 plus negative 18 times five. And evaluating each of our rows, we have a first row of negative 54 minus five 𝑘, a second row of negative 248 minus four 𝑘, and a third row of negative 36 plus 18𝑘 all multiplied by negative one over 128. Making some room and rewriting, we can now equate our right- and left-hand sides. And so, by equality of matrices, we have the three equations shown.

Since we want to find the value of 𝑘, we use the second equation because 𝑘 is the only variable in this equation. So, let’s begin by multiplying both sides by 128. Dividing numerator and denominator by 128 on the right-hand side gives us one. And dividing numerator and denominator on the left-hand side by 16 gives us eight multiplied by 31 divided by one. So, on the left, we have 248 and, on the right, 248 plus four 𝑘. Subtracting 248 from both sides gives us zero is equal to four 𝑘. Hence, 𝑘 is equal to zero.

This means that the value of 𝑘 that results in 𝑦 is equal to 31 over 16 in the given matrix equation is 𝑘 is equal to zero. And now, for completion’s sake, we could use this value to solve for 𝑥 and for 𝑧. 𝑥 is then 54 divided by 128, and 𝑧 is then 36 divided by 128. And evaluating these gives us 𝑥 is equal to 27 over 64 and 𝑧 is nine over 32.