Video: Evaluating Determinants

Find the value of |βˆ’2, 5, βˆ’8 and 5, 1, 8 and 0, 0, βˆ’5|.

01:40

Video Transcript

Find the value of the determinant of this three-by-three matrix.

Remember, the determinant of a three-by-three matrix π‘Ž, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓, 𝑔, β„Ž, 𝑖 shown is equal to π‘Ž multiplied by 𝑒𝑖 minus π‘“β„Ž minus 𝑏 multiplied by 𝑑𝑖 minus 𝑓𝑔 plus 𝑐 multiplied by π‘‘β„Ž minus 𝑒𝑔. Another way to think about this is to look at each of the elements on the top right. We multiply the first element on the top row by the determinant of the two-by-two matrix that’s not in π‘Žs row or column.

And to find the determinant of a two-by-two matrix, we multiply the top left and bottom right element. And then we subtract the product of the top right and bottom left element. We then subtract 𝑏 multiplied by the two-by-two matrix that’s not in 𝑏s row or column.

And finally, we add 𝑐 multiplied by the determinant of the two-by-two matrix that’s not in 𝑐s row or column. That’s negative two multiplied by one multiplied by negative five minus eight multiplied by zero. Then, we subtract five multiplied by five multiplied by negative five minus eight multiplied by zero. And finally, we add negative eight multiplied by five multiplied by zero minus one multiplied by zero.

We have quite a lot of products which give us an answer of zero here. So actually, our expression for the determinant of our matrix simplifies to negative two multiplied by negative five minus five multiplied by negative 25. Negative two multiplied by negative five is 10. And negative five multiplied by negative 25 is positive 125.

And we found the determinant of our matrix. It’s 135.

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