Matrices Help

[Mathlete]

I was working through an old Precalculus state test, when I came across this problem:



I'm not sure how to solve this and haven't been able to find a good search result.

I can't figure out how to solve a 3x3 Matrix. I tried checking a few sites, but I cannot find a guide that I understand.

I would be very grateful for help.

 

[Bob Brown MSEE]

determinant = 0

Lookup determinant.

 

[soroban]

Hello, Mathlete!

\(\displaystyle \text{6. Solve for }x\!:\;\;\begin{vmatrix}x & 14 & 15 \\ 0 & x\!-\!2 & 20 \\ 0&0&x\!-\!3\end{vmatrix} \:=\:0\)

\(\displaystyle (a)\;\{0,\,2,\,\text{-}3\} \quad (b)\;\{14,\,20,\,0\} \quad (c)\;\{3,\,\text{-}2,\,0\} \quad (d)\;\{0,\,2,\,3\} \quad (e)\;\{\text{-}3,\,\text{-}2,\,0\}\)

I assume that you found a site that explains determinants.
I will use the Co-Factor Method, down the first column.

. . . . . . . . . . . \(\displaystyle \begin{vmatrix}x & 14 & 15 \\ 0 & x\!-\!2 & 20 \\ 0&0&x\!-\!3\end{vmatrix} \;=\;0\)

\(\displaystyle x\,\begin{vmatrix}x\!-\!2 & 20 \\ 0 & x\!-\!3\end{vmatrix} - 0\,\begin{vmatrix}14&15 \\ 0 & x\!-\!3\end{vmatrix} + 0\,\begin{vmatrix}14&15 \\ x\!-\!2 & 20\end{vmatrix} \;=\;0 \)

. . . . . \(\displaystyle x\big[(x-2)(x-3) - 0\big] - 0 + 0 \;=\;0\)

. . . . . . . . . \(\displaystyle x(x-2)(x-3) \;=\;0\)

. . . . . . . . .\(\displaystyle x \;=\;0,\,2,\,3\;\;\text{ answer (d)}\)

 

[stapel]

I'm not sure how to solve this and haven't been able to find a good search result.

I can't figure out how to solve a 3x3 Matrix.

Like the first reply suggested, the keyword you need is "determinant". The absolute-value-type bars on the matrix indicate that this is actually the determinant of the matrix, and not the matrix itself.

The second reply gave out the answer (sorry!), but if you'd like to understand how to do determinants yourself, there are loads of online resources on this topic (link).

 

[Mathlete]

Thank you! Very helpful!

I was able to find out how to do it by searching for "determinant".

 

[HallsofIvy]

By the way, this was an "upper triangular" matrix, having only "0" below the main diagonal. "Expanding on the first column" will immediately show you that the determinant of such a matrix is just the product of the numbers on the main diagonal.