### Video Transcript

Given that π΄ is a matrix of order three by two, where π subscript one one equals zero, π subscript one two equals π subscript three one minus three, π subscript two one equals four, π subscript two two equals a half π subscript one one, π subscript three one equals eight, and π subscript three two equals a quarter π subscript two one, determine the matrix π΄.

The first piece of information weβre given in the question is that this matrix π΄ is of order three by two. Remember that the order of a matrix is its dimensions, and the convention is that we list rows first and columns second. So, a matrix of order three by two has three rows and two columns. We can go ahead then and write down the structure of our matrix π΄. It has three rows and two columns. And so, we see that there are six elements that we need to complete.

Now, weβve been given some further information about each of the elements in our matrix π΄. And in order to use this information, we need to recall that a lower case π followed by the subscript ππ denotes the element of matrix upper case π΄, which is in row π and column π. As with giving the order of a matrix, the convention is that we list rows first and columns second. So, letβs look at the information weβve been given.

Firstly, weβre told that π subscript one one is equal to zero. So, this is the element in row one and column one of the matrix π΄. Itβs this element here. So, we can go ahead and fill in zero, and we found one element of our matrix. Now, the next statement tells us about the element π subscript one two in terms of the element π subscript three one which we donβt know yet. So, weβll ignore this statement for now. The next statement tells us that π subscript two one is equal to four. So, thatβs the element in the second row and first column of our matrix π΄. Itβs this element here. We can therefore fill this element in.

Now, the next statement tells us about the element π subscript two two in terms of the element π subscript one one which we do know. But weβll come back to this statement in the moment because the next statement, π three one equals eight, tells us about another element of our matrix explicitly. It tells us that the element in the third row and first column is equal to eight, thus this element here. So, we can fill it in. And we now have the complete first column of our matrix π΄. Now, letβs consider each of the statements we havenβt used.

Firstly, π subscript one two is equal to π subscript three one minus three. π subscript one two is the element in the first row and second column of our matrix; itβs this element here. And π subscript three one is the element in the third row and first column. Itβs this one here. So, we can say that element π one two is equal to eight minus three, which is equal to five. Next, we can use the statement π subscript two two is equal to a half π subscript one one. So, this means that the element in the second row and second column, thatβs this element here, is a half the element in the first row and first column. We have a half multiplied by zero, which is equal to zero.

Finally, we can use the last statement; π subscript three two is equal to a quarter π subscript two one. So, this tells us that the element in the third row and second column, thatβs this element here, is equal to a quarter the element in the second row and first column. Thatβs this element here. We have then a quarter of four or a quarter times four, which is equal to one. So, weβve completed the problem. The matrix π΄ is equal to the matrix zero, five, four, zero, eight, one.