Chapter 6: Problem 6
Evaluate each determinant. $$ \left|\begin{array}{rr}1 & -3 \\\\-8 & 2\end{array}\right| $$
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 6: Problem 6
Evaluate each determinant. $$ \left|\begin{array}{rr}1 & -3 \\\\-8 & 2\end{array}\right| $$
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
Use Cramer's rule to solve each system. $$ \begin{aligned}&2 x+2 y+3 z=10\\\&4 x-y+z=-5\\\&5 x-2 y+6 z=1\end{aligned} $$
In Exercises \(5-8,\) find values for the variables so that the matrices in each exercise are equal. $$ \left[\begin{array}{rr} x & y+3 \\ 2 z & 8 \end{array}\right]=\left[\begin{array}{rr} 12 & 5 \\ 6 & 8 \end{array}\right] $$
If \(A B=-B A,\) then \(A\) and \(B\) are said to be anticommutative. Are \(A=\left[\begin{array}{rr}0 & -1 \\ 1 & 0\end{array}\right]\) and \(B=\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right]\) anti- commutative?
In Exercises \(1-4\) a. Give the order of each matrix. b. If \(A=\left[a_{i j}\right],\) identify \(a_{32}\) and \(a_{23}\) or explain why identification is not possible. $$ \left[\begin{array}{rrrr} 1 & -5 & \pi & e \\ 0 & 7 & -6 & -\pi \\ -2 & 1 & 11 & -1 \end{array}\right] $$
In Exercises \(9-16,\) find: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$ A=\left[\begin{array}{ll} 4 & 1 \\ 3 & 2 \end{array}\right], \quad B=\left[\begin{array}{ll} 5 & 9 \\ 0 & 7 \end{array}\right] $$
What do you think about this solution?
We value your feedback to improve our textbook solutions.