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Unformatted text preview: Midterm Problems MAT26500  11,12 Problem 1. Consider the matrix 1 0 0 5 7 21 0 1 0 3 0 1 5 0 0 1 3 1 12 0 0 0 1 2 17 0 0 0 1. Is this matrix in row echelon form? In reduced row echelon form? (Both? Neither?) 2. How many solutions to the corresponding linear system are there (interpret the matrix as the augmented matrix for a lineary system)? 3. Assume the variables are name x 1 ,...,x k . What is k (i.e. how many variables are there)? Which variables correspond to parameters (are ‘free variables’)? Which variables are determined by these parameters (‘free variables’)? Answer: 1. Yes. Yes. (i.e. both) It is in row echelon form because below the leading entry in each row (i.e. the first nonzero entry, we sometimes called it in class the ‘pivot’) every entry is zero. It is in reduced row echelon form because, moreover, for each such leading entry (i.e. for every ‘pivot’) every entry above it is zero as well. 2. We ignore the rows consisting of all zeros. There is no row for which all the coefficients are zero but the entry in the last column is nonzero, therefore solutions exist. Because there are columns without a pivot (leading entry), there are infinitely many solutions. 3. The coefficient matrix for this system has 6 columns, therefore k = 6. Whenever a matrix is in row echelon form, we can determine this in terms of the columns containing a pivot. The variables containing a pivot can be expressed in terms of the remaining variables. In this example, this tells us that for a solution ( x 1 ,...,x 6 ), the values of the variables x 1 ,x 2 ,x 3 ,x 5 are determined by the values of the variables (the ‘parameters’) x 4 ,x 6 . Problem 2. Consider the matrix 0 1 3 0 2 0 0 0 1 2 0 0 0 0 1 1 Is this matrix in row echelon form? In reduced row echelon form? How many solutions to the corresponding linear system are there (interpret the matrix as the augmented matrix for a linear system)? Answer: 1. Yes. No. It is in row echelon form because below the leading entry in each row (i.e. the first nonzero entry, we sometimes called it in class the ‘pivot’) every entry is zero. It is not in reduced row echelon form because for the last such leading entry (i.e. the last ‘pivot’), there are nonzero entries above it. 2. The last row has leading entry in the last column, which represents the equation 0 · x 1 + ... · x 4 = 1; therefore, there are no solutions. Problem 3. Consider the linear system x 1 +2 x 2 x 3 + x 4 = 2 7 x 1 +14 x 2 5 x 3 + x 4 = 1 2 x 1 3 x 2 +6 x 3 x 4 = 3 1. Determine how many solutions this system has. 2. Find the general solution....
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This note was uploaded on 03/31/2012 for the course MA 265 taught by Professor Bens during the Spring '08 term at Purdue UniversityWest Lafayette.
 Spring '08
 Bens
 Linear Algebra, Algebra

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