Video Transcript
Consider the matrices π΄ equals negative four, negative two, four, negative eight, π΅ equals negative eight, four, eight, negative four. If π΄ times πΆ equals π΅, determine πΆ.
Weβve been given a matrix equation that is π΄ time πΆ is equal to π΅. And weβve been told the value of two matrices, π΄ and π΅. Weβre not going to use the traditional route for solving equations, and that will be to divide both sides by π΄. Instead, we recall that the product of the inverse of π΄ and π΄ is the identity matrix. So instead, weβre going to multiply both sides of our equation by the inverse of π΄, remembering, of course, that order is important.
We know that the product of the inverse of π΄ and π΄ is the identity matrix. And the identity matrix multiplied by πΆ is simply πΆ. So, we can see that πΆ is equal to the inverse of π΄ multiplied by π΅. Our job then is to work out what the inverse of π΄ is. We say that the inverse of a two-by-two matrix given as π, π, π, π is one over the determinant of that matrix multiplied by π, negative π, negative π, π. Essentially, we switch these two elements, and we change the sign on these two. Now, the determinant of π΄ is ππ minus ππ. And it follows that this cannot be equal to zero since the inverse of π΄ would be one over zero times our matrix, which is obviously undefined.
So, letβs begin by finding the determinant of our matrix π΄. Itβs π times π, thatβs negative four times negative eight, minus π times π, thatβs negative two times four. Negative four multiplied by negative eight is 32. And negative two multiplied by four is negative eight. So, we have 32 minus negative eight, which is 40 as the determinant of π΄. So, the inverse of π΄ is one over the determinant, thatβs one over 40, multiplied by this matrix. We switch the elements in the top left and bottom right. So, we switch negative eight with negative four. We then change the sign of the elements in the top right and bottom left. So, we have two and negative four.
And finally, to make the next up easier, we can multiply each element by one fortieth. And we find that the inverse of π΄ is negative eight fortieths, two fortieths, negative four fortieths, and negative four fortieths. And that simplifies to negative one-fifth, one twentieth, negative one-tenth, and negative one-tenth. We saw that πΆ is equal to the inverse of π΄ multiplied by π΅. So, thatβs negative one-fifth, one twentieth, negative one-tenth, negative one-tenth multiplied by negative eight, four, eight, negative four.
To find the first element in the product of these two matrices, we take the dot product of the first row in the first matrix and the first column in the second. So, thatβs negative one-fifth multiplied by negative eight plus one twentieth multiplied by eight, which is simply two. To find the second element in the first row, we find the dot product of the first row in the first matrix and the second column in the second. So, thatβs negative one-fifth multiplied by four plus one twentieth multiplied by negative four, which is negative one.
Weβre now going to repeat this process, but this time using the second row in the first matrix. So, the first element in the second row is negative one-tenth multiplied by negative eight plus negative one-tenth multiplied by eight, which is zero. Finally, we find the dot product of the second row in the first matrix and the second column in the second. So, thatβs negative one-tenth times four plus negative one-tenth times negative four, which is also zero. And so, we found πΆ. Itβs the matrix two, negative one, zero, zero.