Question Video: Solving a Matrix Equation Mathematics

Solve for matrix 𝑋 in the matrix equation 𝑇𝑋 + 𝐡 = 𝐢, where 𝑇 = [3, 0 and 0, 3], 𝐡 = [5, 7 and βˆ’10, βˆ’8], 𝐢 = [8, βˆ’2 and 2, 7].

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Video Transcript

Solve for matrix 𝑋 in the matrix equation 𝑇𝑋 plus 𝐡 equals 𝐢, where 𝑇 is equal to three, zero, zero, three; 𝐡 is equal to five, seven, negative 10, negative eight; 𝐢 is equal to eight, negative two, two, seven.

Let’s begin by imagining this was a standard linear equation and solving for 𝑋. If this was the case, the first thing we would do is subtract 𝐡 from both sides. And that tells us that 𝑇𝑋 is equal to 𝐢 minus 𝐡. Next, we recall that the inverse of a matrix multiplied by that matrix is simply the identity matrix. So to solve this equation, we’ll multiply both sides by the inverse of 𝑇.

Remember, order does matter. So we’ll work out the inverse of 𝑇 multiplied by the matrix found from subtracting 𝐡 from 𝐢. And of course, since the inverse of a matrix multiplied by that matrix is just the identity matrix, that tells us that 𝑋 is equal to the inverse of 𝑇 multiplied by 𝐢 minus 𝐡. Let’s begin then by subtracting 𝐡 from 𝐢. Remember, to do this, we simply subtract the 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. So 𝐢 minus 𝐡 is three, negative nine, 12, 15.

Next, we said we’ll need to multiply this by the inverse of 𝑇. For a two-by-two matrix, 𝐴 is equal to π‘Ž, 𝑏, 𝑐, 𝑑. Its inverse is given by one over the determinant of π‘Ž multiplied by 𝑑, negative 𝑏, negative 𝑐, π‘Ž. And of course, the determinant is found by subtracting 𝑏 multiplied by 𝑐 from π‘Ž multiplied by 𝑑.

Let’s use this formula to help us find the inverse of the matrix 𝑇. The determinant of 𝑇 is three multiplied by three minus zero multiplied by zero which is nine. And therefore the inverse of 𝑇 is one-ninth multiplied by three, zero, zero, three. Now, you’re probably thinking this is a bit strange because that part of the matrix looks identical to the matrix 𝑇. That’s because when we swap π‘Ž with 𝑑, we’re swapping three with three. And they stay the same. And when we change the sign of 𝐡 and 𝐢, negative zero is just zero. So it will actually look really similar.

Finally, we’ll multiply every element inside this matrix by one-ninth. And we can see that the inverse of 𝑇 is a third, zero, zero, a third. 𝑋 is therefore equal to a third, zero, zero, a third multiplied by three, negative nine, 12, 15. To multiply two matrices, we first find the dot product of the elements from the first row in the first matrix and the elements of the first column in the second matrix.

For this first element then, that’s a third multiplied by three plus zero multiplied by 12 which is one. Next, we multiply each element in the first row of the first matrix by the elements from the second column in the second matrix. This time that’s a third multiplied by negative nine plus zero multiplied by 15 which is negative three. Next, we’ll multiply each element in the second row of the first matrix by elements from the first column in the second matrix. That’s zero multiplied by three plus a third multiplied by 12 which is four.

Finally, we’ll multiply each element in the second row of the first matrix by the elements from the second column in the second matrix. That’s zero multiplied by negative nine plus a third multiplied by 15 which is five. And our equation is solved for 𝑋. 𝑋 is equal to one, negative three, four, five.

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