Chapter 8: Problem 88
Solve for \(x\). $$\left|\begin{array}{rr} x+4 & -2 \\ 7 & x-5 \end{array}\right|=0$$
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 8: Problem 88
Solve for \(x\). $$\left|\begin{array}{rr} x+4 & -2 \\ 7 & x-5 \end{array}\right|=0$$
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
The matrix $$P=\left[\begin{array}{lll} 0.6 & 0.1 & 0.1 \\ 0.2 & 0.7 & 0.1 \\ 0.2 & 0.2 & 0.8 \end{array}\right]$$ is called a stochastic matrix. Each entry \(p_{i j}(i \neq j)\) represents the proportion of the voting population that changes from party \(i\) to party \(j,\) and \(p_{i i}\) represents the proportion that remains loyal to the party from one election to the next. Compute and interpret \(P^{2}\).
Determine whether the statement is true or false. Justify your answer. In Cramer's Rule, the numerator is the determinant of the coefficient matrix.
Write a brief paragraph explaining the difference between a square matrix and its determinant.
Determine whether the statement is true or false. Justify your answer. Multiplication of an invertible matrix and its inverse is commutative.
A property of determinants is given (\(A\) and \(B\) are square matrices). State how the property has been applied to the given determinants and use a graphing utility to verify the results. If \(B\) is obtained from \(A\) by multiplying a row by a nonzero constant \(c\) or by multiplying a column by a nonzero constant \(c,\) then \(|B|=c|A|\). (a) \(\left|\begin{array}{cc}5 & 10 \\ 2 & -3\end{array}\right|=5\left|\begin{array}{cc}1 & 2 \\ 2 & -3\end{array}\right|\) (b) \(\left|\begin{array}{rrr}1 & 8 & -3 \\ 3 & -12 & 6 \\ 7 & 4 & 9\end{array}\right|=12\left|\begin{array}{rrr}1 & 2 & -1 \\ 3 & -3 & 2 \\ 7 & 1 & 3\end{array}\right|\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.