### Video Transcript

Consider matrix π΄ is equal to π sub ππ, where π is equal to one, two, and three and π is equal to one and two. True or False: This matrix is a square matrix.

In this question, weβre given a matrix π΄ where the entry in row π, column π is given by π sub ππ. And weβre told that π can vary from one to three and π can vary from one to two. We need to use this to determine if matrix π΄ is a square matrix. To do this, letβs start by recalling what we mean by a square matrix. This is when a matrix has the same number of rows as its columns. And in fact, this means we can answer our question directly.

Since π sub ππ is the entry in row π, column π of our matrix and our values of π range from one to three and the values of π range from one to two, our matrix must have three rows and two columns. Since these values are not equal, we can conclude π΄ is not a square matrix, which means we can just conclude that the answer is false. π΄ is not a square matrix.

However, it can be useful to see exactly why this is true. So, letβs write out the matrix π΄ by using its definition. First, we know π΄ is a matrix and π sub ππ tells us the entry in row π, column π of our matrix. So, π sub one one is the entry in row one, column one of our matrix. Similarly, π sub one two is the entry in row one, column two. And we canβt go any further because in the question weβre told that values of π only range between one and two. So, instead, we need to increase the value of π, which means we move on to a new row. The entry in row two, column one is π sub two one, and the entry in row two, column two is π sub two two. We can then do exactly the same for the third row. And then we know our values of π can only range from one to three. So we canβt add any more rows to our matrix. This then gives us our matrix π΄. And this confirms that π΄ is not a square matrix. It has three rows and two columns. Therefore, the answer is false.