# Question Video: Evaluating a Third-Order Determinant in Upper Triangle Form Mathematics • 10th Grade

Find the value of |−5, 2, −4 and 0, 5, 0 and 3, 0, 0|.

03:53

### Video Transcript

Find the value of the determinant of the matrix negative five, two, negative four, zero, five, zero, three, zero, zero.

So, to solve this problem, what we need to know is that to find the value of the determinant of a three-by-three matrix. We need to find the expansion of the determinant by the first row. So, what does this mean in practice? Well, it means that we take each of the terms in the first row and we multiply it by the corresponding two-by-two matrix determinant. It’s worth noting as well that we multiply each of the terms from the first row by either positive or negative one. And this is determined by a pattern that corresponds to the column number.

So, we start with a positive column and then a negative and then a positive. So therefore, if we multiply positive one by negative five, we’re gonna get negative five. And this is multiplied by the determinant of the two-by-two submatrix five, zero, zero, zero. And we’ve found this submatrix by eliminating both the column and the row that negative five were in.

So, next we get negative two multiplied by the determinant of the submatrix zero, zero, three, zero. And we get negative two. It’s because we multiply two by negative one, so that gives negative two. So, now we move on to the last value in our first row. So, then we have negative four multiplied by the determinant of the corresponding submatrix which, this time, is zero, five, three, zero. And again, this time we got negative four because we had negative four as the last term in the first row. And then we multiply this by positive one, so it gives us our negative four.

So, now we take a look at how we work out the determinant of a two-by-two submatrix. Well, to work it out, what we do is we multiply the top-left term by the bottom-right term. So, in my example here, that’ll be 𝑎 multiplied by 𝑑. And then we subtract the top-right term multiplied by the bottom-left term. So, in our example again, that would be 𝑏 multiplied by 𝑐.

So, now if we take a look at the first submatrix in our question, we’re gonna have negative five multiplied by. And then we’ve got five multiplied by zero minus zero multiplied by zero. And then, we move on. And we have negative two multiplied by zero multiplied by zero minus zero multiplied by three. And then, finally, we have negative four multiplied by zero multiplied by zero minus five multiplied by three.

And then we’re gonna have a zero. And that’s because we had five multiplied by zero minus zero multiplied by zero, which is just zero. And negative five multiplied by zero is zero, and then minus zero. Again, it’s cause we had zero multiplied by zero minus zero multiplied by three. And then multiply that by negative two. So, again, we get our second zero. And then we have minus, we’ve got four multiplied by negative 15. And that’s because we had five multiplied by three, which is 15. And it was negative cause we had zero minus that. So, we get negative four multiplied by negative 15 at the end.

So, this is gonna be equal to positive 60. And we got positive 60 because we had a negative multiplied by a negative, which gives us a positive. Cause we had negative four multiplied by negative 15. So therefore, we can say that the value of the determinant of the matrix negative five, two, negative four, zero, five, zero, three, zero, zero is 60.
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}