Math 74: Practice Midterm Show your work and explain your reasoning as appropriate. No calculators. One page of handwritten notes is allowed for the exam, as well as one blank page of scratch paper.. Consider the 3 3 matrix A = (a) Check that A is nonsingular, and find A. An n n matrix is nonsingular if and only if it is invertible, so it will be enough to see that A exists. To do this, form the n n (n = 3) matrix and use Gaussian elimination to get to reduced row echelon form. The result is 5, 3 so we conclude that A = (b) Solve the system of linear equations 5 3 x +x x 3 =, x x +x 3 =, x x 3 =. The coefficient matrix for this system is the matrix A from (a), and the constant vector is b = The solution to Ax = b is x = A b, which is x x = x = 5 x 3 = 5 5 3
(c) Solve the system of linear equations x +x x 3 =, x x +x 3 =, x x 3 =. Just as in (b), the coefficient matrix is A, but now the constant vector is b = The solution is A b, which is x x = x = x 3 = 5 3 6 4
. Consider the vectors v =, v =, v 3 = where a R is some real number. Find all values of a such that {v, v, v 3 } is a linearly independent set. We want values of a so that the only solution to c v +c v +c 3 v 3 = is c = c = c 3 =. Form the matrix for this system, and use Gaussian elimination: a a a+ a, if a+ (!!) So, as long as a, we get c = c = c 3 = as the only solution to the system, and therefore the vectors are independent for these values of a. On the other hand, if a =, then v 3 = v v. So one of them is a linear combination of the others, and therefore the set is linearly dependent. Conclusion: the vectors are linearly independent exactly when a. 3. True or false? If a k n matrix A has rank equal to k, then for some b in R k, the system of linear equations Ax = b is inconsistent. This is false. The rank of A is, by definition, the dimension of its range. The only k-dimensional subspace of R k is all of R k, so every b in R k lies in the range of A. But, by definition, the range of A is the set of all b so that Ax = b is consistent. Therefore Ax is consistent for every b in R k.
4. (a) Consider the matrix A = 4 3 6 7 Solve the homogeneous system Ax =. (Give a parametric solution if appropriate.) Using row operations, put A in reduced row echelon form. The result is 3. So, taking x 3 to be a free variable (since the third column doesn t have a leading ), the solutions are given by x = 3t x = t x 3 = t.. (b) Find a basis for the range of A. The range of A is the span of the columns of A. To find a basis, you can write these columns as row vectors and use Gaussian elimination to get to (reduced) echelon form. That is the same as finding reduced echelon form of the transposed matrix A T. The result is 6 5 3 3 Taking the nonzero rows, and transposing them back to column vectors, we get a basis consisting of the two vectors 6 5, 3 3. (c) What is rank(a)? By definition, rank(a) is the dimension of the range of A. This, in turn, is the number of vectors in a basis. From the previous part, then, rank(a) =.
(d) What is the dimension of Null(A)? One approach would be to find a basis for Null(A). But it s easier to use the rank-nullity theorem, which says: dim(range(a)) + dim(null(a)) = n whenever A is a k n matrix. Since dim(range(a)) = by the previous part, this equation is +dim(null(a)) = 3, so dim(null(a)) =. (e) Consider the vectors v =, v = 3, v 3 = Is this set linearly independent? Explain why or why not. 4 6 7. This is the set of columns of A. If it were linearly independent, it would form a basis for Range(A) (since, by definition, it spans this subspace). But we already computed the dimension of the range: it is, so a basis cannot have 3 vectors. Therefore the set is not linearly independent. (You could also directly exhibit one of the vectors as a linear combination of the others, showing that the set is linearly dependent. For instance, v 3 = v +v.)
5. (a) Is V = a subspace of R 3? Give reasons. x x x 3 No. The zero vector is not in this set. (b) Is W = : x +x +x 3 = { [ ] [ x x = : there exist numbers c x,c so that x = c a subspace of R? If so, give its dimension; if not, say why. ] +c [ [ This] is a subspace, [ ] of dimension. It is, by definition, the span of the two vectors and. You can check these vectors are linearly independent, so they form a basis. (As a consequence, this subspace is just all of R.) ]} 6. Consider the matrix [ ] λ A = 4 3 λ where λ is a real number. Find all values of λ such that A is not invertible. Recall that A is not invertible if and only if it is singular. This is the same as saying the columns of A are linearly dependent. So let s see when Ax = has nontrivial solutions. First, assume λ, so we can divide the first row by λ. Using Gaussian elimination, A reduces to [ +λ (3 λ)(+λ) 4(3 λ) (3 λ)(+λ) ] = [ +λ (λ ) (3 λ)(+λ) If λ =, then the lower-right entry is zero, so this is reduced echelon form. The variable x is free, so there are nontrivial solutions. This means that the matrix is singular when λ =. On the other hand, if λ and λ 3, then the lower-right entry is not zero. So you can divide by it and get a leading. Then there are no free parameters, so there is only the trivial solution. This means that when λ is not, 3, or, the matrix is nonsingular. We re left with checking the cases λ = and λ = 3. By plugging in those numbers and doing Gaussian elimination, you find the matrix reduces to the identity matrix in both cases. So both of those cases are nonsingular. To sum up, A is not invertible for λ =. For all other values of λ, A is invertible. ].
There is another way, if you know about determinants: A is singular if and only if det(a) = ( λ)(3 λ) ()( 4) = (λ λ+) =. (We will discuss this after the midterm.) Solving λ λ+ = gives λ =. So A is not invertible if and only if λ =.