Question Video: Using Matrix Inverses to Solve a 2 Γ— 2 Matrix Equation Mathematics

Using matrix inverses, solve the following for 𝑋: 𝑋[βˆ’3, 2 and 4, βˆ’3] = [0, βˆ’2 and 3, 0]

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Video Transcript

Using matrix inverses, solve the following for 𝑋. That’s 𝑋 multiplied by the matrix with elements negative three, two, four, negative three is equal to the matrix with elements zero, negative two, three, zero.

We’re given a matrix equation and asked to solve for 𝑋. We know, in fact, that 𝑋 is a matrix since if 𝑋 were a scalar, we’d be able to find the value for 𝑋 satisfying the equation shown. And by equality of matrices, this would mean that 𝑋 would satisfy all four of the shown equations. 𝑋 is equal to zero satisfies the first and fourth equations. However, for the second equation, the only value of 𝑋 satisfying this is 𝑋 is negative one. And for the third equation, the only value of 𝑋 satisfying this is 𝑋 is three over four. Since a scalar quantity cannot be three things at once, our 𝑋 must be a matrix.

Now we know that for a matrix multiplication to be possible, if a matrix 𝐴 has size π‘š by 𝑛 and a matrix 𝐡 has size 𝑛 by 𝑝, then their product must have the size π‘š by 𝑝, that is, π‘š rows and 𝑝 columns. If we call the resulted matrix 𝐢, in our case, 𝐢 is a two-by-two matrix. This means both π‘š and 𝑝 are two. So if our matrix 𝑋 is labeled 𝐴 and it multiplies in matrix 𝐡, we know 𝐡 is a two-by-two matrix; therefore, 𝑛 is also two in which case our matrix 𝑋, corresponding to 𝐴, is also a two-by-two matrix. So let’s let our matrix 𝑋 have elements 𝑋 one, 𝑋 two, 𝑋 three, and 𝑋 four.

Now we’re asked to use matrix inverses to solve the equation. And we recall that for an 𝑛-by-𝑛 invertible matrix π‘š, its inverse π‘š minus one is an 𝑛-by-𝑛 matrix such that π‘š multiplied by π‘š inverse is equal to π‘š inverse multiplied by π‘š. And that’s equal to the identity, where the identity matrix has all of its elements equal to zero except those on the leading diagonal, which are all equal to one.

Now, remember, for a two-by-two matrix π‘š with elements π‘Ž, 𝑏, 𝑐 and 𝑑, the inverse of π‘š is one over the determinant of π‘š multiplied by the matrix where π‘Ž and 𝑑 have been swapped and we take negative 𝑏 and negative 𝑐. And remember, of course, that the determinant of a two-by-two matrix is π‘Žπ‘‘ minus 𝑏𝑐.

To solve the equation, we’re going to isolate 𝑋 on the left-hand side. If we call our two other matrices 𝐴 and 𝐡, respectively, our equation says 𝑋 multiplied by 𝐴 is equal to 𝐡. We’re going to use the inverse of the matrix 𝐴 to eliminate 𝐴 on the left-hand side. So if we multiply both sides on the right by inverse 𝐴, by the inverse law, we know that 𝐴 multiplied by 𝐴 inverse is equal to the identity matrix. So we now have 𝑋 multiplied by the identity matrix is 𝐡 multiplied by 𝐴 inverse. Any matrix multiplied by the identity matrix is itself. So we have 𝑋 is equal to 𝐡 multiplied by 𝐴 inverse. So now we need to find the inverse of the matrix 𝐴.

Making a little room, our matrix 𝐴 has elements negative three, two, four, and negative three. For the inverse of 𝐴, we’re going to need the determinant. And the determinant is π‘Žπ‘‘ minus 𝑏𝑐. And that’s negative three multiplied by negative three minus two multiplied by four. And that’s nine minus eight, which is equal to one. The inverse is given by swapping 𝑑 and π‘Ž and taking the negatives of 𝑏 and 𝑐 within our matrix and multiplying by the determinant. In our case, π‘Ž and 𝑑 are both negative three. So swapping those leaves them the same. And taking the negative of 𝑏 and 𝑐, where 𝑏 is two and 𝑐 is four, gives us negative two, negative four. The inverse of 𝐴 is therefore the matrix with elements negative three, negative two, negative four, and negative three.

We can easily check that this is actually the inverse of the matrix 𝐴 by multiplying the two together. Our result should be the identity matrix. Using matrix multiplication, we have negative three multiplied by negative three plus two multiplied by negative four, negative three multiplied by negative two plus two multiplied by negative three, and similarly for the remaining two elements. This evaluates to nine minus eight, six minus six, negative 12 plus 12, and negative eight plus nine, which does indeed give us the two-by-two identity matrix. We do therefore have the correct inverse matrix for the matrix 𝐴.

And now inserting this into our equation for the inverse of 𝐴, performing matrix multiplication should then give us the matrix 𝑋. We have zero times negative three plus negative two times negative four, zero times negative two plus negative two times negative three, three times negative three plus zero times negative four, and finally three times negative two plus zero times negative three.

Zero times negative three is zero. Negative two times negative four is positive eight. Zero times negative two is zero. And negative two times negative three is six. Three times negative three is negative nine. Zero times negative four is zero. Three times negative two is negative six. And zero times negative three is zero. Our result is therefore the matrix with elements eight, six, negative nine, and negative six.

Equating this with our matrix 𝑋, our elements are therefore 𝑋 one is equal to eight, 𝑋 two is equal to six, 𝑋 three is equal to negative nine, and 𝑋 four is equal to negative six. And this is our solution 𝑋.

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