ekiim's blog
  1. Find the inverses (no special system required) ofA1=[0230],A2=[2042],A3=[cosθsinθsinθcosθ] A_{1}=\left[ \begin{array}{ll}{0} & {2} \\ {3} & {0}\end{array}\right], \quad A_{2}=\left[ \begin{array}{ll}{2} & {0} \\ {4} & {2}\end{array}\right], \quad A_{3}=\left[ \begin{array}{rr}{\cos \theta} & {-\sin \theta} \\ {\sin \theta} & {\cos \theta}\end{array}\right]

    1. Find the inverse of the permuation matrices P=[001010100] and P=[001100010] P=\left[ \begin{array}{lll}{0} & {0} & {1} \\ {0} & {1} & {0} \\ {1} & {0} & {0}\end{array}\right] \quad \text { and } \quad P=\left[ \begin{array}{lll}{0} & {0} & {1} \\ {1} & {0} & {0} \\ {0} & {1} & {0}\end{array}\right]
    2. Explain why for permutations P1P^{-1} is always the same as PTP^T, by using showing that the 11’s are in the right places to give PPT=IPP^T = I.
  2. From AB=CAB=C find a formula A1A^{-1}. Do the same for PA=LUPA =LU.

    1. If A is invertible and AB=ACAB =AC prove quickly that B=CB=C.
    2. If A=[1000]A = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}, find an example with AB=ACAB= AC but BCB \neq C.
  3. If the inverse of A2A^2 is BB show that the inverse of AA is ABAB. (Thus whenever A2A^2 is invertible)

  4. Use the Gauss-Jordan method to invert A1=[100111001],A2=[210121012],A3=[001011111] A_{1}=\left[ \begin{array}{lll}{1} & {0} & {0} \\ {1} & {1} & {1} \\ {0} & {0} & {1}\end{array}\right], \quad A_{2}=\left[ \begin{array}{rrr}{2} & {-1} & {0} \\ {-1} & {2} & {-1} \\ {0} & {-1} & {2}\end{array}\right], \quad A_{3}=\left[ \begin{array}{ccc}{0} & {0} & {1} \\ {0} & {1} & {1} \\ {1} & {1} & {1}\end{array}\right]

  5. Find three 2 by 2 matrices other than A=IA=I and A=IA = -I that are their own inverse A2=IA^2 = I.

  6. Show that [1133]\begin{bmatrix} 1 & 1\\3 & 3\end{bmatrix}, has no inverse by trying to solve [1133][abcd]=[1001] \left[ \begin{array}{ll}{1} & {1} \\ {3} & {3}\end{array}\right] \left[ \begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right]=\left[ \begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]

  7. When elimination fails for singular matrices like A=[2146038500070009] A=\left[ \begin{array}{llll}{2} & {1} & {4} & {6} \\ {0} & {3} & {8} & {5} \\ {0} & {0} & {0} & {7} \\ {0} & {0} & {0} & {9}\end{array}\right] show that AA cannot be invertible. The third row of A1A^{-1}, multiplying A, should give the third row of A1A=IA^{-1}A = I. Why is this impossible?

  8. Find the inverses (in any legal way) of:A1=[0001002003004000],A2=[1000121000231000341],A3=[ab00cd0000ab00cd] A_{1}=\left[ \begin{array}{llll}{0} & {0} & {0} & {1} \\ {0} & {0} & {2} & {0} \\ {0} & {3} & {0} & {0} \\ {4} & {0} & {0} & {0}\end{array}\right], \quad A_{2}=\left[ \begin{array}{rrrr}{1} & {0} & {0} & {0} \\ {-\frac{1}{2}} & {1} & {0} & {0} \\ {0} & {-\frac{2}{3}} & {1} & {0} \\ {0} & {0} & {-\frac{3}{4}} & {1}\end{array}\right], \quad A_{3}=\left[ \begin{array}{cccc}{a} & {b} & {0} & {0} \\ {c} & {d} & {0} & {0} \\ {0} & {0} & {a} & {b} \\ {0} & {0} & {c} & {d}\end{array}\right]

  9. give examples of AA and BB such that:

    1. A+BA+B is not invertible although AA and BB are invertible
    2. A+BA + B is invertible although AA and BB are not invertible
    3. all of A,BA, B and A+BA+B are invertible In the last case use A1(A+B)B1=B1+A1A^{-1}(A+B)B^{-1} = B^{-1} + A^{-1} to show that B1+A1B^{-1} + A^{-1} is also invertible — and find a formula for its inverse.
  10. Which properties of a matrix AA are preserved by its inverse (assuming A1A^{-1} exists)? (1) AA is trinagular, (2) A is Symetric, (3) A is tridiagonal, (4) all entries are whole numbers, (5) all entries are fraction (including whole numbers like 31\frac31)

  11. If A=[31]A = \begin{bmatrix}3 \\ 1\end{bmatrix} and B=[22]B = \begin{bmatrix}2 \\ 2\end{bmatrix} compute ATB,BTA,ABT,BATA^TB,B^TA, AB^T, BA^T.

  12. (Important) Prove that even for rectangular matrices AATAA^T and ATAA^TA are always symmetric. Show by example that they may not be equal even for square matrices.

  13. Show that for any wquare matrix B,A=B+BTB, A = B + B^T is always symmetric and K=BBTK = B - B^T is aways skew-smmetric — which means that KT=KK^T = -K. Find these matrices when B=[1311]B = \begin{bmatrix} 1 & 3 \\ 1 & 1\end{bmatrix}, and write BB as the sum of a symmetric matrix and a skewsymetrix matrix.

    1. How many entries can be chosen indeppendently, in a symmetric matrix of order nn?
    2. Hoe many entries can be chosen independently in a skew-symmetric matrix of order nn?
    1. If A = LDU, which 1’s on the diagonals of LL and UU, what is the corresponding factorization of ATA^T? Note that AA and ATA^T (squres matrices with no row exchanges share the same pivots.)
    2. What triangula system will give solution to ATy=bA^Ty = b
  14. If A=L1D1U1A=L_1D_1U_1 and A=L2D2U2A = L_2D_2U_2, prove that L1=L2,D1=D2L_1 = L_2, D_1=D_2 and U1=U2U_1 = U_2, If AA is invertible the factorization is unique.

    1. Derive the equation for L11L2D2=D1U1U21L_1^{-1}L_2D_2 = D_1U_1U_2^{-1} and explain why one side is lower triangular and the other side is upper triangular.
    2. Compare the main diagonals in the eauation, and then compare the offdiagonals.
  15. Under what conditions on its entries AA is invertible if: A=[abcde0f00] or A=[ab0cd000e]? A=\left[ \begin{array}{lll}{a} & {b} & {c} \\ {d} & {e} & {0} \\ {f} & {0} & {0}\end{array}\right] \quad \text { or } \quad A=\left[ \begin{array}{lll}{a} & {b} & {0} \\ {c} & {d} & {0} \\ {0} & {0} & {e}\end{array}\right] ?

  16. If the 3 by 3 matrix AA has row 1 + row 2 = row 3, show that it is impossible to solve Ax=[124]TAx = [ 1 \quad 2 \quad 4]^T. Can AA be invertible?

  17. Compute the symmetric LDLTLDL^{T} factorization of A=[1353121851830] and A=[abbd] A=\left[ \begin{array}{ccc}{1} & {3} & {5} \\ {3} & {12} & {18} \\ {5} & {18} & {30}\end{array}\right] \quad \text { and } \quad A=\left[ \begin{array}{ll}{a} & {b} \\ {b} & {d}\end{array}\right]

  18. Find the inverse of A=[100014100131310121211] A=\left[ \begin{array}{cccc}{1} & {0} & {0} & {0} \\ \frac{1}4 & {1} & {0} & {0} \\ {\frac{1}{3}} & {\frac{1}{3}} & {1} & {0} \\ {\frac{1}{2}} & {\frac{1}{2}} & {1} & {1}\end{array}\right]

  19. If AA and BB are suqre matrices, show that IABI - AB is invertible if IBAI - BA is invertible. Start from B(IAB)=(IBA)BB(I-AB) = (I-BA)B.