Question Video: Using the Inverse Matrix to Solve a System of Linear Equations Mathematics • 10th Grade

Video Transcript

Use the inverse matrix to solve the matrix equation, where the three-by-three matrix with elements two, three, four; negative five, five, six; and seven, eight, nine is multiplied by the column matrix with elements 𝑥, 𝑦, and 𝑧 is equal to the column matrix with elements zero, 45, and negative 45, giving your answer as an appropriate matrix.

We’re given a system of linear equations in the form 𝐴 multiplied by 𝐮 is equal to 𝐯, where 𝐴 is a three-by-three matrix, 𝐮 is a column matrix, and 𝐯 is a column matrix. The column matrix 𝐮 contains the three unknowns 𝑥, 𝑦, and 𝑧. And we’re asked to solve this system using the matrix inverse. What this means is that assuming 𝐴 is nonsingular, that is, an inverse exists for 𝐴. If we find the inverse of 𝐴, we can multiply our equation through on the left by 𝐴 inverse. And we can use the fact that 𝐴 inverse multiplied by 𝐴 is equal to 𝐴𝐴 inverse, which is equal to the identity for an 𝑛-by-𝑛 nonsingular matrix.

Recalling that the 𝑛-by-𝑛 identity matrix is the matrix whose elements are all zero except those on the leading diagonal which are all equal to one, which for a three-by-three matrix is as shown. On the left-hand side then, we have 𝐴 inverse 𝐴, which is equal to 𝐼, multiplied by 𝐮. And 𝐼 multiplied by 𝐮 is simply 𝐮. And so we’ve isolated 𝐮 on the left-hand side, remembering that 𝐮 is our column matrix with elements 𝑥, 𝑦, 𝑧. And these are our unknowns. And on our right-hand side, we have 𝐴 inverse multiplied by 𝐯. So if we find 𝐴 inverse, we can solve for 𝑥, 𝑦, and 𝑧.

To find the inverse of the matrix 𝐴, we’re going to use the adjoint method. That is, for an 𝑛-by-𝑛 nonsingular matrix 𝐴, the inverse of 𝐴 is given by one over the determinant of 𝐴 multiplied by the adjoint of 𝐴. This means we’ll have to find the determinant of the matrix 𝐴 and the adjoint of matrix 𝐴. For the determinant, we can expand along the first row of our matrix 𝐴 so that we have the determinant of 𝐴 is equal to the element 𝑎 one one, which is two, multiplied by the determinant of the two-by-two matrix minor 𝐴 one one minus the element 𝑎 one two multiplied by the determinant of its matrix minor plus the element 𝑎 one three, which is four, multiplied by the determinant of its matrix minor. Recalling that the matrix minor 𝐴 𝑖𝑗 is the matrix 𝐴 with row 𝑖 and column 𝑗 removed.

So, for example, in our case, the matrix minor 𝐴 one one is the two-by-two matrix resulting in the removal of row one and column one from our matrix 𝐴. Matrix minor 𝐴 one one is therefore the two-by-two matrix with elements five, six, eight, and nine. Similarly, for our second term, the matrix minor 𝐴 one two is the two-by-two matrix with elements negative five, six, seven, and nine. And for our third term, the matrix minor 𝐴 one three is the two-by-two matrix with elements negative five, five, seven, and eight. To evaluate this, we’re going to need to work out our two-by-two determinants. And we recall that for a two-by-two matrix 𝑀 with elements 𝑎, 𝑏, 𝑐, 𝑑, the determinant of 𝑀 is equal to 𝑎𝑑 minus 𝑏𝑐.

So, for example, in our first term, the determinant is five multiplied by nine minus six multiplied by eight. That is 45 minus 48, which is negative three, so that our first term evaluates to two times negative three, which is negative six. For our second term, we’ll have negative three multiplied by negative five times nine minus six times seven, which is negative three times negative 87. And for our third term, we have four multiplied by negative five times eight minus five times seven, which is four times negative 75. Our determinant is then negative six plus 261 minus 300, which is negative 45. And now that we have the determinant of matrix 𝐴, we need to find its adjoint matrix.

Now making some room and making a note of our determinant, we recall that the adjoint of a matrix is the transpose of the matrix of its cofactors, which for a three-by-three matrix is as shown. For element 𝑎 𝑖𝑗, the cofactor 𝑐 𝑖𝑗 is negative one raised to the power 𝑖 plus 𝑗 multiplied by the determinant of the matrix minor 𝐴 𝑖𝑗. And note that negative one raised to the power 𝑖 plus 𝑗 gives us the parity or sign of the cofactor. For a three-by-three matrix, that’s positive, negative, positive; negative, positive, negative; and positive, negative, positive.

We’ve actually already seen three of the cofactors when calculating the determinant of our matrix. 𝑐 one one is the positive determinant of matrix minor 𝐴 one one. And that is five multiplied by nine minus six multiplied by eight, which is 45 minus 48, which, as we saw before, is negative three. Similarly, our second cofactor 𝑐 one two is negative negative 87. That’s positive 87. And our third cofactor 𝑐 one three is negative 75. So now let’s write out the second two rows of cofactors. Our cofactor 𝑐 two one is five, 𝑐 two two is negative 10, and 𝑐 two three is five. And our final row of cofactors are 𝑐 three one is negative two, 𝑐 three two is negative 32, and 𝑐 three three is 25. And note that it’s very important that we get our parity correct, which is positive, negative, positive, and so on.

So now we can write out our adjoint matrix. That’s the transpose of the matrix with elements negative three, 87, negative 75; five, negative 10, and five; negative two, negative 32, and 25. Now making some room, we can write out our adjoint matrix, which is the transpose of the matrix of cofactors, where to obtain the transpose, we interchange the rows and columns of the matrix. This means that our first row becomes our first column, our second row becomes our second column, and our third row becomes our third column. Now remember, it’s actually the inverse of our matrix 𝐴 that we’re looking for at the moment. And the inverse is one over the determinant multiplied by the adjoint matrix, which in our case is one over negative 45 times our adjoint matrix. And of course we can move our negative sign to the numerator.

Now remember, our original equation is 𝐮 is equal to 𝐴 inverse multiplied by 𝐯. And writing this out with our matrices, we have the equation as shown. And to solve this, we simply multiply out the right-hand side and use equality of matrices. Before we do this, however, we can simplify things a little if we multiply our matrix 𝐯 by negative one over 45. This will give us zero, negative 45 over 45, which is negative one, and negative negative 45 over 45, which is positive one. This then eliminates our scalar factor of negative one over 45. So on the right, we replace negative one over 45 times 𝐯 with the column matrix with elements zero, negative one, one.

Using matrix multiplication then, we have negative three multiplied by zero plus five multiplied by negative one plus negative two multiplied by one. And similarly, for our second and third rows, we have 87 multiplied by zero plus negative 10 times negative one plus negative 32 times one and in the third row negative 75 times zero plus five times negative one plus 25 times one. These three rows then evaluate to negative seven, negative 22, and 20. An appropriate matrix for the solution of our matrix equation is therefore the column matrix with elements negative seven, negative 22, and 20. And these are our values for 𝑥, 𝑦, and 𝑧.