### Video Transcript

Without expanding the determinant, prove that the determinant of the matrix π₯, π, π, π, π₯, π, π, π, π₯ is equal to π₯ plus π plus π multiplied by π₯ minus π multiplied by π₯ minus π.

Letβs begin by recalling what we know about the determinant of a three-by-three matrix. For the three-by-three matrix with elements π, π, π, π, π, π, π, β, π, its determinant is π multiplied by ππ minus πβ minus π multiplied by ππ minus ππ plus π multiplied by πβ minus ππ. This question specifically tells us not to expand the determinant. So weβre going to need to manipulate our matrix somewhat.

The first fact we recall is that if thereβs a common factor in all the elements of any column in the determinant, then this factor can be taken outside the determinant. We also recall that if we add the elements of any column or row to the multiples of the elements of any other column or row, the value of the determinant remains unchanged.

Notice how each row contains an π₯, an π, and a π. Weβre therefore going to add elements in column one to the elements in column two and column three. The elements in column two and column three remain unchanged. But column one now becomes π₯ plus π plus π on all three rows. And we can use the first fact that we stated to take out this factor of π₯ plus π plus π. And each of the elements in our first column become one. Next, our aim is to have two elements in a row or a column that are equal to zero.

Now, earlier, we showed how to find the value of determinant of a three-by-three matrix by evaluating in terms of the elements of its first row. Actually though, the value of the determinant is unchanged if we evaluated in terms of the elements of any of its columns. So weβre going to make two of the elements in the first column of our determinant equal to zero. But first, weβll achieve that by subtracting the elements from row one from the elements from row two. By doing so on row two, we now have zero, π₯ minus π, zero. And then, weβll repeat this process with row three, subtracting each of the elements on row one from each of the elements on row three. And in row three, we now have zero, zero, π₯ minus π.

We can now evaluate the determinant in terms of the elements in this first column. Itβs one multiplied by π₯ minus π multiplied by π₯ minus π minus zero multiplied by zero, which is simply π₯ minus π multiplied by π₯ minus π. And weβve proven that the determinant of our three-by-three matrix is π₯ plus π plus π multiplied by π₯ minus π multiplied by π₯ minus π.