### Video Transcript

Solve for matrix π in the matrix equation ππ plus π΅ equals πΆ, where π is equal to three, zero, zero, three; π΅ is equal to five, seven, negative 10, negative eight; πΆ is equal to eight, negative two, two, seven.

Letβs begin by imagining this was a standard linear equation and solving for π. If this was the case, the first thing we would do is subtract π΅ from both sides. And that tells us that ππ is equal to πΆ minus π΅. Next, we recall that the inverse of a matrix multiplied by that matrix is simply the identity matrix. So to solve this equation, weβll multiply both sides by the inverse of π.

Remember, order does matter. So weβll work out the inverse of π multiplied by the matrix found from subtracting π΅ from πΆ. And of course, since the inverse of a matrix multiplied by that matrix is just the identity matrix, that tells us that π is equal to the inverse of π multiplied by πΆ minus π΅. Letβs begin then by subtracting π΅ from πΆ. 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 πΆ minus π΅ is three, negative nine, 12, 15.

Next, we said weβll need to multiply this by the inverse of π. For a two-by-two matrix, π΄ is equal to π, π, π, π. Its inverse is given by one over the determinant of π multiplied by π, negative π, negative π, π. And of course, the determinant is found by subtracting π multiplied by π from π multiplied by π.

Letβs use this formula to help us find the inverse of the matrix π. The determinant of π is three multiplied by three minus zero multiplied by zero which is nine. And therefore the inverse of π is one-ninth multiplied by three, zero, zero, three. Now, youβre probably thinking this is a bit strange because that part of the matrix looks identical to the matrix π. Thatβs because when we swap π with π, weβre swapping three with three. And they stay the same. And when we change the sign of π΅ and πΆ, negative zero is just zero. So it will actually look really similar.

Finally, weβll multiply every element inside this matrix by one-ninth. And we can see that the inverse of π is a third, zero, zero, a third. π 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, thatβs 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 thatβs a third multiplied by negative nine plus zero multiplied by 15 which is negative three. Next, weβll multiply each element in the second row of the first matrix by elements from the first column in the second matrix. Thatβs zero multiplied by three plus a third multiplied by 12 which is four.

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