Question #204705

If A = ( 1,2,3) (5,6,7) (0,1,4) and B = (1,0,3) ( 5,6,1) (2,1,4) then find i) AB ii) Inverse of A & B


1
Expert's answer
2021-06-09T12:50:34-0400

i)


AB=(123567014)(103561214)AB=\begin{pmatrix} 1 & 2 & 3 \\ 5 & 6 & 7 \\ 0 & 1 & 4 \\ \end{pmatrix}\begin{pmatrix} 1 & 0 & 3 \\ 5 & 6 & 1 \\ 2 & 1 & 4 \\ \end{pmatrix}

=(1(1)+2(5)+3(2)1(0)+2(6)+3(1)1(3)+2(1)+3(4)5(1)+6(5)+7(2)5(0)+6(6)+7(1)5(3)+6(1)+7(4)0(1)+1(5)+4(2)0(0)+1(6)+4(1)0(3)+1(1)+4(4))=\begin{pmatrix} 1(1)+2(5)+3(2) & 1(0)+2(6)+3(1) & 1(3)+2(1)+3(4) \\ 5(1)+6(5)+7(2) & 5(0)+6(6)+7(1) & 5(3)+6(1)+7(4) \\ 0(1)+1(5)+4(2) & 0(0)+1(6)+4(1) & 0(3)+1(1)+4(4) \\ \end{pmatrix}

=(171517494349131017)=\begin{pmatrix} 17 & 15 & 17 \\ 49 & 43 & 49 \\ 13 & 10 & 17 \\ \end{pmatrix}




AB=(171517494349131017)AB=\begin{pmatrix} 17 & 15 & 17 \\ 49 & 43 & 49 \\ 13 & 10 & 17 \\ \end{pmatrix}

i)

Augment the matrix ABAB with the identity matrix:


(171517100494349010131017001)\begin{pmatrix} 17 & 15 & 17 & & 1 & 0 & 0 \\ 49 & 43 & 49 & & 0 & 1 & 0 \\ 13 & 10 & 17 & & 0 & 0 & 1 \\ \end{pmatrix}

R1=R1/17R_1=R_1/17


(115/1711/1700494349010131017001)\begin{pmatrix} 1 & 15/17 & 1 & & 1/17 & 0 & 0 \\ 49 & 43 & 49 & & 0 & 1 & 0 \\ 13 & 10 & 17 & & 0 & 0 & 1 \\ \end{pmatrix}

R2=R249R1R_2=R_2-49R_1


(115/1711/170004/17049/1710131017001)\begin{pmatrix} 1 & 15/17 & 1 & & 1/17 & 0 & 0 \\ 0 & -4/17 & 0 & & -49/17 & 1 & 0 \\ 13 & 10 & 17 & & 0 & 0 & 1 \\ \end{pmatrix}

R3=R313R1R_3=R_3-13R_1


(115/1711/170004/17049/1710025/17413/1701)\begin{pmatrix} 1 & 15/17 & 1 & & 1/17 & 0 & 0 \\ 0 & -4/17 & 0 & & -49/17 & 1 & 0 \\ 0 & -25/17 & 4 & & -13/17 & 0 & 1 \\ \end{pmatrix}

R2=(17/4)R2R_2=-(17/4)R_2


(115/1711/170001049/417/40025/17413/1701)\begin{pmatrix} 1 & 15/17 & 1 & & 1/17 & 0 & 0 \\ 0 & 1 & 0 & & 49/4 & -17/4 & 0 \\ 0 & -25/17 & 4 & & -13/17 & 0 & 1 \\ \end{pmatrix}

R1=R1(15/17)R2R_1=R_1-(15/17)R_2


(10143/415/4001049/417/40025/17413/1701)\begin{pmatrix} 1 & 0 & 1 & & -43/4 & 15/4 & 0 \\ 0 & 1 & 0 & & 49/4 & -17/4 & 0 \\ 0 & -25/17 & 4 & & -13/17 & 0 & 1 \\ \end{pmatrix}

R3=R3+(25/17)R2R_3=R_3+(25/17)R_2


(10143/415/4001049/417/4000469/425/41)\begin{pmatrix} 1 & 0 & 1 & & -43/4 & 15/4 & 0 \\ 0 & 1 & 0 & & 49/4 & -17/4 & 0 \\ 0 & 0 & 4 & & 69/4 & -25/4 & 1 \\ \end{pmatrix}

R3=R3/4R_3=R_3/4


(10143/415/4001049/417/4000169/1625/161/4)\begin{pmatrix} 1 & 0 & 1 & & -43/4 & 15/4 & 0 \\ 0 & 1 & 0 & & 49/4 & -17/4 & 0 \\ 0 & 0 & 1 & & 69/16 & -25/16 & 1/4 \\ \end{pmatrix}

R1=R1R3R_1=R_1-R_3


(100241/1685/161/401049/417/4000169/1625/161/4)\begin{pmatrix} 1 & 0 & 0 & & -241/16 & 85/16 & -1/4 \\ 0 & 1 & 0 & & 49/4 & -17/4 & 0 \\ 0 & 0 & 1 & & 69/16 & -25/16 & 1/4 \\ \end{pmatrix}

On the left is the identity matrix. On the right is the inverse matrix.



(AB)1=(241/1685/161/449/417/4069/1625/161/4)(AB)^{-1}=\begin{pmatrix} -241/16 & 85/16 & -1/4 \\ 49/4 & -17/4 & 0 \\ 69/16 & -25/16 & 1/4 \\ \end{pmatrix}

Augment the matrix AA with the identity matrix:


(123100567010014001)\begin{pmatrix} 1 & 2 & 3 & & 1 & 0 & 0 \\ 5 & 6 & 7 & & 0 & 1 & 0 \\ 0 & 1 & 4 & & 0 & 0 & 1 \\ \end{pmatrix}

R2=R25R1R_2=R_2-5R_1


(123100048510014001)\begin{pmatrix} 1 & 2 & 3 & & 1 & 0 & 0 \\ 0 & -4 & -8 & & -5 & 1 & 0 \\ 0 & 1 & 4 & & 0 & 0 & 1 \\ \end{pmatrix}

R2=R2/4R_2=-R_2/4


(1231000125/41/40014001)\begin{pmatrix} 1 & 2 & 3 & & 1 & 0 & 0 \\ 0 & 1 & 2 & & 5/4 & -1/4 & 0 \\ 0 & 1 & 4 & & 0 & 0 & 1 \\ \end{pmatrix}

R1=R12R2R_1=R_1-2R_2


(1013/21/200125/41/40014001)\begin{pmatrix} 1 & 0 & -1 & & -3/2 & 1/2 & 0 \\ 0 & 1 & 2 & & 5/4 & -1/4 & 0 \\ 0 & 1 & 4 & & 0 & 0 & 1 \\ \end{pmatrix}

R3=R3R2R_3=R_3-R_2


(1013/21/200125/41/400025/41/41)\begin{pmatrix} 1 & 0 & -1 & & -3/2 & 1/2 & 0 \\ 0 & 1 & 2 & & 5/4 & -1/4 & 0 \\ 0 & 0 & 2 & & -5/4 & 1/4 & 1 \\ \end{pmatrix}

R3=R3/2R_3=R_3/2


(1013/21/200125/41/400015/81/81/2)\begin{pmatrix} 1 & 0 & -1 & & -3/2 & 1/2 & 0 \\ 0 & 1 & 2 & & 5/4 & -1/4 & 0 \\ 0 & 0 & 1 & & -5/8 & 1/8 & 1/2 \\ \end{pmatrix}

R1=R1+R3R_1=R_1+R_3


(10017/85/81/20125/41/400015/81/81/2)\begin{pmatrix} 1 & 0 & 0 & & -17/8 & 5/8 & 1/2 \\ 0 & 1 & 2 & & 5/4 & -1/4 & 0 \\ 0 & 0 & 1 & & -5/8 & 1/8 & 1/2 \\ \end{pmatrix}

R2=R22R3R_2=R_2-2R_3


(10017/85/81/20105/21/210015/81/81/2)\begin{pmatrix} 1 & 0 & 0 & & -17/8 & 5/8 & 1/2 \\ 0 & 1 & 0 & & 5/2 & -1/2 & -1 \\ 0 & 0 & 1 & & -5/8 & 1/8 & 1/2 \\ \end{pmatrix}

On the left is the identity matrix. On the right is the inverse matrix.



(A)1=(17/85/81/25/21/215/81/81/2)(A)^{-1}=\begin{pmatrix} -17/8 & 5/8 & 1/2 \\ 5/2 & -1/2 & -1 \\ -5/8 & 1/8 & 1/2 \\ \end{pmatrix}

Augment the matrix bb with the identity matrix:


(103100561010214001)\begin{pmatrix} 1 & 0 & 3 & & 1 & 0 & 0 \\ 5 & 6 & 1 & & 0 & 1 & 0 \\ 2 & 1 & 4 & & 0 & 0 & 1 \\ \end{pmatrix}

R2=R25R1R_2=R_2-5R_1


(1031000614510214001)\begin{pmatrix} 1 & 0 & 3 & & 1 & 0 & 0 \\ 0 & 6 & -14 & & -5 & 1 & 0 \\ 2 & 1 & 4 & & 0 & 0 & 1 \\ \end{pmatrix}

R3=R32R1R_3=R_3-2R_1


(1031000614510012201)\begin{pmatrix} 1 & 0 & 3 & & 1 & 0 & 0 \\ 0 & 6 & -14 & & -5 & 1 & 0 \\ 0 & 1 & -2 & & -2 & 0 & 1 \\ \end{pmatrix}

R2=R2/6R_2=R_2/6


(103100017/35/61/60012201)\begin{pmatrix} 1 & 0 & 3 & & 1 & 0 & 0 \\ 0 & 1 & -7/3 & & -5/6 & 1/6 & 0 \\ 0 & 1 & -2 & & -2 & 0 & 1 \\ \end{pmatrix}

R3=R3R2R_3=R_3-R_2


(103100017/35/61/60001/37/61/61)\begin{pmatrix} 1 & 0 & 3 & & 1 & 0 & 0 \\ 0 & 1 & -7/3 & & -5/6 & 1/6 & 0 \\ 0 & 0 & 1/3 & & -7/6 & -1/6 & 1 \\ \end{pmatrix}

R3=3R3R_3=3R_3


(103100017/35/61/600017/21/23)\begin{pmatrix} 1 & 0 & 3 & & 1 & 0 & 0 \\ 0 & 1 & -7/3 & & -5/6 & 1/6 & 0 \\ 0 & 0 & 1 & & -7/2 & -1/2& 3 \\ \end{pmatrix}

R1=R13R3R_1=R_1-3R_3


(10023/23/29017/35/61/600017/21/23)\begin{pmatrix} 1 & 0 & 0 & & 23/2 & 3/2 & -9 \\ 0 & 1 & -7/3 & & -5/6 & 1/6 & 0 \\ 0 & 0 & 1 & & -7/2 & -1/2& 3 \\ \end{pmatrix}

R2=R2+(7/3)R3R_2=R_2+(7/3)R_3


(10023/23/290109170017/21/23)\begin{pmatrix} 1 & 0 & 0 & & 23/2 & 3/2 & -9 \\ 0 & 1 & 0 & & -9 & -1 & 7 \\ 0 & 0 & 1 & & -7/2 & -1/2& 3 \\ \end{pmatrix}

On the left is the identity matrix. On the right is the inverse matrix.



(B)1=(23/23/299177/21/23)(B)^{-1}=\begin{pmatrix} 23/2 & 3/2 & -9 \\ -9 & -1 & 7 \\ -7/2 & -1/2 & 3 \\ \end{pmatrix}



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