In Exercises 53–54, evaluate each determinant.

Question

In Exercises 53–54, evaluate each determinant. $\begin{vmatrix}\begin{vmatrix} 3 & 1 \ -2 & 3 \end{vmatrix} & \begin{vmatrix} 7 & 0 \ 1 & 5 \end{vmatrix} \\begin{vmatrix} 3 & 0 \ 0 & 7 \end{vmatrix} & \begin{vmatrix} 9 & -6 \ 3 & 5 \end{vmatrix}\end{vmatrix}$

Accepted Answer

Welcome back. I am so glad you're here. Were asked to evaluate the following determinant. We have four matrices inside of one gigantic matrix. So the way to figure out our determinant is we're going to figure out the determinant of each of these four matrices. And then once we have those four determinants we'll find we'll have a two by two matrix again And we can find the determinant of those determinants. So let's start off by writing each of these inner matrices. I'm going to start reading down the columns. So 23 is the first column negative 54 is the second column. And then going down the column we have the first column, 01 and then 40. And then starting on the right hand, the second column we've got 0462. And then our last inner matrix is 9182. Okay, we've got our four matrices. Now we'll figure out the determinant of each one of these. So starting we recall from previous lessons on finding the determinant of a two by two matrix. We're going to start in the upper left hand corner and multiply going down to the bottom right two times four is eight. And next we're going to start in the bottom left corner and go up to the upper right and multiply those two terms and we get negative 15. And then we are going to subtract those two terms from each other. 8 -20. Well that's eight plus 15. So that's 23. Now we'll do the same thing for the one below that. We've got zero times zero which is zero and we've got one times four which is four. And then we subtract those from each other. Zero minus four is negative four. Now moving to the second column of matrices, we've got zero times two. Well that's zero And four times 6 is 24. So now we are going to subtract those from each other. 0 -24 is -24. And our final matrix within a matrix we've got nine times two is 18 And we have eight times one is eight. Now we subtract those from each other and 18 -8 is 10. Okay, so the matrix of determinants that we have starting in the upper left hand corner is 23. Going down the column negative four and then we have negative 24 10 for our second column. And finding the determinant of these determinants were going to go 23 times 10 which is 230 and then we've got negative four times negative 24 which is positive 96. And now we are going to subtract those two from each other and 230 minus 96 is 134. So our determinant is answer choice B 134. Well done. We'll catch you on the next one

In Exercises 53–54, evaluate each determinant. $\begin{vmatrix}\begin{vmatrix} 3 & 1 \\ -2 & 3 \end{vmatrix} & \begin{vmatrix} 7 & 0 \\ 1 & 5 \end{vmatrix} \\\begin{vmatrix} 3 & 0 \\ 0 & 7 \end{vmatrix} & \begin{vmatrix} 9 & -6 \\ 3 & 5 \end{vmatrix}\end{vmatrix}$

Key Concepts

Determinant of a 2x2 Matrix

Properties of Determinants

Matrix Notation and Evaluation

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