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Answers for Exercises

Section 1: Determinants

 

Section 2: Cofactors

 

Section 3: Cramer's Rule

 

Section 4: Partitioned Matrices

 

Section 5: Elementary Matrices

 

Section 6: Inverse Matrices

    1. The determinant of matrix A is -1. As det(A)≠0, matrix A is invertible.
    2. The determinant of matrix B is -1. As det(B)≠0, matrix B is invertible.
    3. The determinant of matrix C is 1. As det(C)$\neq$0, matrix C is invertible.\\
    4. The determinant of matrix D is 0. As det(D)=0, matrix D is not invertible.\\
    1. Adjoin the matrix with an identity matrix. [12−3|10005−1|010−122|001] Add the row 1 to row 3. Multiply row 2 by 1/5. [12−3|10001−15|015004−1|101] Add -4 times row 2 to row 3. [12−3|10001−15|015000−15|1−451] Multiply row 3 by -5. [12−3|10001−15|0150001|−54−5] Add 1/5 times row 3 to row 2. Add 3 times row 3 to row 1. [120|−1412−15010|−11−1001|−54−5] Add -2 times row 2 to row 1. [100|−1210−13010|−11−1001|−54−5]$ inv(A)=[−1210−13−11−1−54−5]
    2. Adjoin the matrix with an identity matrix. [−332|100−1−2−1|010021|001] Multiply row 1 by -1/3. [1−1−23|−1300−1−2−1|010021|001] Add row 1 to row 2. [1−1−23|−13000−3−53|−1310021|001] Multiply row 2 by -1/3. [1−1−23|−13000159|19−130021|001] Add -2 times row 2 to row 3. [1−1−23|−13000159|19−130001|2−6−9] Multiply row 3 by -9. Add -5/9 times row 3 to row 2. Add 2/3 times row 3 to row 1. [1−10|1−4−6010|−135001|2−6−9] Add row 2 to row 3. [100|0−1−1010|−135001|2−6−9] inv(B)=[0−1−1−1352−6−9]
    1. det(A)=-1

      cof(A11)=$(-1)^{(i+j)}$det(A11)=$(-1)^{(1+1)}((0*-2)-(1*3))=-3$

      cof(A12)=$(-1)^{(i+j)}$det(A12)=$(-1)^{(1+2)}((-2*-2)-(2*3))=2$

      cof(A13)=$(-1)^{(i+j)}$det(A13)=$(-1)^{(1+3)}((-2*1)-(0*2))=-2$

      cof(A21)=$(-1)^{(i+j)}$det(A21)=$(-1)^{(2+1)}((3*-2)-(1*-1))=5$

      cof(A22)=$(-1)^{(i+j)}$det(A22)=$(-1)^{(2+2)}((3*-2)-(2*-1))=-4$

      cof(A23)=$(-1)^{(i+j)}$det(A23)=$(-1)^{(2+3)}((3*1)-(2*3))=3$

      cof(A31)=$(-1)^{(i+j)}$det(A31)=$(-1)^{(3+1)}((3*1)-(2*3))=9$

      cof(A32)=$(-1)^{(i+j)}$det(A32)=$(-1)^{(3+2)}((3*1)-(2*3))=-7$

      cof(A33)=$(-1)^{(i+j)}$det(A33)=$(-1)^{(3+3)}((3*1)-(2*3))=6$

      inv(A)=1det(A)$adj(A)=$1det(A)$M(A)′=−1[−32−25−439−76]′=−1[−3592−4−7−236]=[3−5−9−2472−3−6]$

    2. det(B)=1

      cof(B11)=${(-1)}^(i+j)$det(B11)=$(-1)^{(1+1)}((0*1)-(1*1))=-1$

      cof(B12)=${(-1)}^(i+j)$det(B12)=$(-1)^{(1+2)}((2*0)-(2*1))=2$

      cof(B13)=${(-1)}^(i+j)$det(B13)=$(-1)^{(1+3)}((2*1)-(2*1))=0$

      cof(B21)=${(-1)}^(i+j)$det(B21)=$(-1)^{(2+1)}((-1*0)-(1*-3))=-3$

      cof(B22)=${(-1)}^(i+j)$det(B22)=$(-1)^{(2+2)}((-3*0)-(2*-3))=6$

      cof(B23)=${(-1)}^(i+j)$det(B23)=$(-1)^{(2+3)}((-3*1)-(2*-1))=1$

      cof(B31)=${(-1)}^(i+j)$det(B31)=$(-1)^{(3+1)}((-1*1)-(1*-3))=2$

      cof(B32)=${(-1)}^(i+j)$det(B32)=$(-1)^{(3+2)}((-3*1)-(2*-3))=-3$

      cof(B33)=${(-1)}^(i+j)$det(B33)=$(-1)^{(3+3)}((-3*1)-(2*-1))=-1$

      inv(B)=1det(A)adj(B)=1det(B),M(B)′=1[−120−3612−3−1]′=1[−1−3226−301−1]=[−1−3226−301−1]

 

Section 7: Elimination: A = LU

    1. (a)[10003100−1010−34−21][1−2−2−30−36000240001],(b)

     

    Section 8: PLU Factorization

      1. (a)[10003/41001/2−2/7101/4−3/71/31][879507/49/417/400−6/7−2/70002/3],[0010000101001000], \[ \mbox{(b)} \quad \begin{bmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} ,

       

      Section 9: Reflection

       

      Section 10: Givens Rotation

       

      Section 11: Special Matrices