Video: Finding the Inverse of a Two-by-Two Matrix

Consider the matrix equation [3, 1 and 0, 1][π‘Ž, 𝑏 and 𝑐, 𝑑] + [5, 7 and βˆ’10, βˆ’8] = [8, βˆ’2 and 2, 7]. Find the values of π‘Ž, 𝑏, 𝑐, and 𝑑.

04:01

Video Transcript

Consider the matrix equation. Find the values of π‘Ž, 𝑏, 𝑐, and 𝑑.

We can imagine this to be a bit like a linear equation. And to solve, we can start by subtracting the constant matrix from both sides. That’s the matrix five, seven, negative 10, negative eight. Subtracting this matrix from both sides, and we’re left with three, one, zero, one multiplied by π‘Ž, 𝑏, 𝑐, 𝑑 on the left-hand side. For the right-hand side we’ll need to subtract five, seven, negative 10, negative eight from the matrix eight, negative two, two, seven. And we could do this by subtracting each of its individual elements.

Eight minus five is three. Negative two minus seven is negative nine. Two minus negative 10 is 12. And seven minus negative eight is 15. And we’re left with three, one, zero, one multiplied by π‘Ž, 𝑏, 𝑐, 𝑑 being equal to three, negative nine, 12, 15.

Now to solve for the matrix π‘Ž, 𝑏, 𝑐, 𝑑, we can’t divide by the matrix three, one, zero, one. Instead, we’ll need to multiply both sides by its multiplicative inverse. And the reason this works is because when we multiply a matrix by its inverse, we get the identity matrix. So, if we multiply both sides by the inverse of three, one, zero, one, we’ll solve for π‘Ž, 𝑏, 𝑐, 𝑑. For two by two matrix with elements π‘Ž, 𝑏, 𝑐, 𝑑, its inverse is found by one over the determinant of π‘Ž multiplied by 𝑑 negative 𝑏 negative π‘π‘Ž, where the determinant is the difference between the product of the top-left and bottom-right elements and the product of the top-right and bottom-left.

So, let’s begin by finding the determinant of our matrix three, one, zero, one. π‘Ž multiplied by 𝑑 is the top-left element multiplied by the bottom-right element. That’s three multiplied by one. And then we subtract the product of the top-right and bottom-left element. That’s one multiplied by zero. And that gives us a determinant of three.

Let’s use this information to find the inverse of this matrix. One over the determinant is one-third. We then swap the elements on the top left and the bottom right. We swap one and three. And we multiply the elements on the top right and the bottom left by negative one, essentially changing their signs. Then we can multiply each of the elements by one-third and we get the inverse of the matrix to be a third, negative a third, zero and one. And now we have this, we can multiply both sides of our equation by the inverse.

On the left-hand side, we end up with π‘Ž, 𝑏, 𝑐, 𝑑. And on the right-hand side we end up with a third, negative a third, zero, one multiplied by three, negative nine, 12, 15. Remember, order matters, so we must complete this multiplication with the inverse in the front.

To multiply two matrices, we find the dot product of the associated rows and columns. For π‘Ž, we find the dot product of the first row in the first matrix and the first column in the second. That’s negative three. For 𝑏, we find the dot product of the first row in the first matrix and the second column in the second. That’s negative eight. For 𝑐, we find the dot product of the second row in the first matrix and the first column in the second, which is 12. And for 𝑑, we find the dot product of the second row in the first matrix and the second column in the second. That’s 15.

We have found π‘Ž to be negative three, 𝑏 to be negative eight, 𝑐 to be 12. and 𝑑 to be 15. And in matrix form π‘Ž, 𝑏, 𝑐, 𝑑 is negative three, negative eight, 12, 15.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.