Math 315: Linear Algebra Solutions to Midterm Exam I


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1 Math 35: Linear Algebra s to Midterm Exam I # Consider the following two systems of linear equations (I) ax + by = k cx + dy = l (II) ax + by = 0 cx + dy = 0 (a) Prove: If x = x, y = y and x = x 2, y = y 2 are both solutions of (I), then x = x x 2, y = y y 2 is a solution of (II) (b) Prove: If x = x, y = y is a solution of (I) and x = x 0, y = y 0 is a solution of (II), then x = x + x 0, y = y + y 0 is a solution of (I) (a) If x = x, y = y and x = x 2, y = y 2 are solutions of (I), then (I) is satisfied if one plugs in these values to the system, ie ax + by = k ax2 + by 2 = k cx + dy = l cx 2 + dy 2 = l Subtracting the top equation of the second system from the top of the first system and doing the same with the second pair of equations results a(x x 2 ) + b(y y 2 ) = 0 c(x x 2 ) + d(y y 2 ) = 0 This implies that x = x x 2, y = y y 2 is a solution of (II) (b) Similarly, if x = x, y = y is a solution of (I) and x = x 0, y = y 0 is a solution of (II), then the corresponding systems are satisfied if one plugs in the given values there, that is ax + by = k cx + dy = l ax0 + by 0 = 0 cx 0 + dy 0 = 0
2 Summing up the first equations and the second ones, we get a(x + x 0 ) + b(y + y 0 ) = k c(x + x 0 ) + d(y + y 0 ) = l This implies that x = x + x 0, y = y + y 0 is a solution of (I) #2 Let Ax = 0 be a homogeneous system of n linear equations in n unknowns, and let Q be an invertible n n matrix Prove: Ax = 0 has just the trivial solution is and only if (QA)x = 0 has just the trivial solution A homogeneous system of linear equations Ax = 0 has only the trivial solution if and only if the matrix A is invertible (Theorem 64) Similarly, the homogeneous system (QA)x = 0 has just the trivial solution if and only if the matrix QA is invertible Therefore, we have to show that A is invertible if and only if QA is invertible If A is invertible and Q is invertible, then QA is invertible as the product of two invertible matrices (Theorem 46) On the other hand, if QA is invertible, then both Q and Q are invertible (Theorem 65), which completes the proof #3 ( a b Prove: If A = c d ), then adj(adj A) = A ( ) d c For the matrix A one has C = d, C 2 = c, C 2 = b, and C 22 = a, so adj A = b a Similarly, for the cofactors C ij of the matrix B = adj A it holds: C = a, C 2 = b, C 2 = c, and ( ) a b C 22 = d Hence, adj B = adj(adj A) = = A c d #4 (a) Check whether it is true or not: an n n matrix A is invertible if and only if adj(a) is invertible It it is true, prove it, otherwise provide a counterexample (b) Prove: If A is invertible then [adj(a)] = det(a) A = adj(a ) 2
3 (a) We know that if A is invertible, then A = adja (Theorem 242) This implies adja = det(a) det(a) A To prove that adja is invertible, we show that its determinant is nonzero (cf Theorem 233): det(adja) = det(a) det(a ) = det(a) det(a) = To prove the converse, assume adja is invertible and assume for the purpose of getting a contradiction that A is not invertible First note, that A does not have a row entirely consisting of zeros Indeed, if this is the case for the ith row of A, then for any j i all the cofactors of the elements of row j contain a zero row and, hence, are zeros This implies that the jth column of adja consists entirely of zeros, so the matrix adja is not invertible This contradicts our assumption concerning adja Therefore, there is a nonzero element in the first row of A, let it be the element a j Now, by the Theorem of Laplace (Theorem 24), multiplying the elements of the first row with its cofactors and summing up the products results in det(a) Furthermore, as we know, doing the same with the elements of the first row and the cofactors of any other row results in 0 (Example 5 on page 07) In other terms, a C + + a j C j + + a n C n = det(a) = 0 a C a j C 2j + + a n C 2n = = 0 a C n + + a j C nj + + a n C nn = 0 After dividing these equalities by a j 0 and denoting b i = a i /a j for i =,, n we obtain b C + + C j + + b n C n = det(a) = 0 b C C 2j + + b n C 2n = = 0 b C n + + C nj + + b n C nn = 0 Finally, denoting by r i = (C i, C 2i,, C ni ) the ith row of adja for i =,, n we can rewrite the previous scalar equalities in the vector form b r + + r j + + b n r n = 0 This vector form equality can be interpreted as follows: we sequentially add to the ith row of adja the multiple b of the first row, then the multiple b 2 of the second row,, and the multiple b n of the last row and get a matrix with a row of zeros Adding to a row a multiple of another row is an elementary row operation that does not change the determinant (Theorem 223(c)) So, a series of elementary row operations applied to the matrix adja lead to a matrix with a row of zeros, whose determinant is then zero This means that the determinant of the original matrix (that is adja) is also zero This, in turn, implies that adja is not invertible, which is a contradiction and completes the proof 3
4 (b) To prove the first equality, we have to show that [adja][adja] = [adja] det(a) A = I n Since A is invertible, we have A = adja (Theorem 242) Equivalently, adja = det(a) det(a)a Substituting this into the above formula results so the first equality is established [det(a)a ] det(a) A = I n, To prove the second equality, apply the formula A = adja for det(a) A (which does exist) One gets (A ) = det(a ) adj(a ), which is equivalent to A = det(a)adj(a ), from where the required equality immediately follows #5 Prove: If A is an n n matrix and r is a number, then det(ra) = r n det(a) Multiplying a matrix A by a constant r results in a matrix whose all entries are the multiplies r of the original ones Consider an elementary matrix product of A: a i a 2,i2 a nin After multiplying each element by r this elementary matrix product transforms to r n a i a 2,i2 a nin Therefore, every elementary matrix product of the matrix ra is a multiple r n of the corresponding product of A The same holds for the signed elementary matrix products So, the sum of all signed elementary products of ra (in other words, det(ra) is a multiple r n of the corresponding sum for A (which is det(a)) This proves the required equality #6 Let A be an invertible matrix and A be its inverse How A will change after applying to A the following elementary row operations? (a) Swapping rows i and j (b) Multiplying row i through by a constant α 0 (c) Adding to row i a multiple α 0 of row j Justify your answers Applying an elementary row operation to a matrix A is equivalent to multiplying A on the left by 4
5 the corresponding elementary matrix E This results in the matrix B = EA The question is what is the relationship between A and B = A E In case of swapping the rows, E is obtained from I n by swapping rows i and j In this case E = E and B = A E It is easily seen that B is obtained from A by swapping columns i and j If the elementary operation from (b) is applied to I n to get E, then E is obtained from E by replacing its entry α with /α in column i In this case B is obtained from A by multiplying the ith column of A through by /α If the elementary operation from (c) is applied to I n to get E, then E is obtained from E by negating its (the only) nondiagonal entry α (see Table on page 5) In this case B is obtained from A by adding α times column i to column j To see this, let us take a closer look on the matrix product A E We denote the elements of A by a ij a a i a j a n a i a ii a ij a in A E = = a ji a ji a jj a jn 0 α 0 a ni a ni a nj a nn a a i a j αa i a n a i a ii a ij αa ii a in a ji a ji a jj αa ji a jn a ni a ni a nj αa ni a nn Similar matrix analysis can be done in cases (a) and (b) It is omitted here for brevity and because these operations are less complex than the third one 5
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