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    Linear Algebra Sample Midterm & Solutions

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    Robin Phetsa
    The document provides sample solutions to problems on a math exam covering linear algebra topics. It includes solutions to problems involving determining properties of systems of linear equations, finding the reduced row echelon form of matrices, checking for linear independence of columns, and computing the inverse of a matrix. The last problem is shown to not have an inverse due to a row of zeros in the reduced matrix.

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    Linear Algebra Sample Midterm & Solutions

    Uploaded by

    Robin Phetsa
    105 views3 pages
    The document provides sample solutions to problems on a math exam covering linear algebra topics. It includes solutions to problems involving determining properties of systems of linear equations, finding the reduced row echelon form of matrices, checking for linear independence of columns, and computing the inverse of a matrix. The last problem is shown to not have an inverse due to a row of zeros in the reduced matrix.

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    Linear Algebra Sample Midterm 1 & Solutions

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    The document provides sample solutions to problems on a math exam covering linear algebra topics. It includes solutions to problems involving determining properties of systems of linear equations, finding the reduced row echelon form of matrices, checking for linear independence of columns, and computing the inverse of a matrix. The last problem is shown to not have an inverse due to a row of zeros in the reduced matrix.

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    105 views3 pages

    Linear Algebra Sample Midterm & Solutions

    Uploaded by

    Robin Phetsa
    The document provides sample solutions to problems on a math exam covering linear algebra topics. It includes solutions to problems involving determining properties of systems of linear equations, finding the reduced row echelon form of matrices, checking for linear independence of columns, and computing the inverse of a matrix. The last problem is shown to not have an inverse due to a row of zeros in the reduced matrix.

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
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    Math 4A.01, Fall 2012 First Midterm SAMPLE SOLUTIONS (v.

    2)

    1. For each of the following statements, say whether the statement is true or false, and explain why. (a) Any system of n linear equations in n variables has at most n solutions. Answer: FALSE. If a system of linear equations has more than one solution, it has innitely many solutions. (b) If an augmented matrix [A b] can be transformed by elementary row operations into reduced echelon form, then Ax = b is consistent. Answer: FALSE. It may happen that in the reduced echelon form, there is a row which is all zeros until the last entry, but then is nonzero in the augmented column. Such a system is inconsistent. (c) The equation Ax = 0 has the trivial solution if and only if there are no free variables. Answer: FALSE. The trivial solution means x = 0, and it is always a solution, whether there are free variables or not. (d) If an m n matrix A has a pivot position in every row, then the equation Ax = b has a unique solution for each b in Rm . Answer: FALSE. If there are more columns than rows, we will still have some free variables because some columns wont have pivots in them. 2. Let

    1 2 3 4 5 6 A= 7 8 9 . 10 11 12 (a) Describe the shape of A. That is, A is an m n matrix: what are m and n? Answer: there are 4 rows and 3 columns, so m = 4, n 3. (b) If aij denotes the (i, j) entry of A, what are a13 and a31 ? Answer: a13 = 3, a31 = 7.

    2 8 3. For the matrix A in problem 2, does the vector lie in the span of the columns 14 20 of A? Why or why not?

    Solution: We need to row-reduce an augmented matrix. Here is the process: 1 2 3 2 1 2 3 2 1 2 3 2 4 5 6 8 0 3 6 0 0 3 6 0 . 7 8 9 14 0 6 12 0 0 0 0 0 10 11 12 20 0 9 18 0 0 0 0 0 The conclusion is that the equation Ax = b is consistent, so it has a solution, so the given vector is in the span of the columns of the matrix. Alternatively, you might have noticed that the given vector is 2 times the rst column, and this shows explicitly that it is in the span of the columns. 4. The solutions {x, y, z} of a single linear equation ax + by + cz = d form a plane in R3 when a, b, and c are not all zero. Construct sets of three linear equations whose graphs (a) intersect in a single line (b) intersect in a single point, and (c) have no points in common. Solution. Since you are being asked to write your own equations, there are many possible answers. In part (a), you need a consistent set of equations with the property that the echelon form has exactly one row of zeros. In part (b), you need a set of equations whose echelon form has a piot in every row and column. In part (c), you need an inconsistent set of equations. 5. Consider system of equations 2x1 4x2 2x3 = b1 5x1 + x2 + x3 = b2 7x1 5x2 3x3 = b3 (a) Write this system in the form Ax = b. Solution: 2 4 2 x1 b1 5 1 1 x2 = b2 . 7 5 3 x3 b3 (b) Are the columns of the coecient matrix A linearly independent? Why or why not?

    (c) Is the system of equations consistent for all b1 , b2 , and b3 ? Why or why not? Solution to parts (b) and (c): in part (b) we must solve Ax = 0 while in part (c) we must solve Ax = b; the row-reduction process is the same. 2 4 2 b1 2 4 2 b1 2 4 2 b1 5 1 1 b2 0 9 4 b2 + 5 b1 0 9 4 b2 + 5 b1 2 2 7 7 5 3 b3 0 0 0 b3 + b2 b1 0 9 4 b3 2 b1 Thus we see that there are nontrivial solutions to Ax = 0, which tells us that the columns are linearly dependent (part b). We also see that the system is only consistent when b2 + b3 b1 = 0, so the answer to part (c) is that it is not always consistent. 6. Find the inverse of

    1 2 1 4 7 3 2 6 4 or explain why the inverse does not exist. Solution: We row-reduce [A I3 ]: 1 2 1 1 0 0 1 2 1 1 0 0 1 2 1 1 0 0 4 7 3 0 1 0 0 1 1 4 1 0 0 1 1 4 1 0 2 6 4 0 0 1 0 2 2 2 1 1 0 0 0 10 3 1 We can stop at this point, because the row of zeros in the left half-matrix tells us that there is no inverse.

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