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Using paper and pencil, find the QR factorization of the matrices in Exercises 15 through 28. Compare with Exercises 1 through 14.

17.[425003-25]

Short Answer

Expert verified

The QR factorization of the matrix is 425003-25=1543003-455035.

Step by step solution

01

Determine column u→1 and entries r11 of R

Consider the matrixM=425003-25wherev1=403and localid="1659954267209" v2=250-25.

By the theorem of QR method, the value ofu1andr11is defined as follows:

r11=v1u1=1r11v1

Simplify the equationr11=v1as follows:

r11=v1r11=403r11=42+02+32r11=5

Substitute the values 5 for r11and 403forv1 in the equationu1=1r11v1 as follows:

u1=1r11v1u1=15403u1=4/503/5

Therefore,ui=4/503/5 and r11=5.

02

Determine column v→2⊥ and r12 entries  of R

As r11=u1v2, substitute the values 250-25for v2and 45035Tfor in the equation r11=u1v2as follows:

r12=u1.v2r12=45035T.250-25r12=20-0-15r12=5

Substitute the values250-25 for v2, 5 forr12 and 4/503/5for u1in the equation v2=v2-r12u1as follows:

v2=v2-r12u1v2=250-25-54/503/5v2=250-25-403v2=210-28

Therefore,role="math" localid="1659954948498" v2=210-28 and r12=5.

03

Determine column u→2 and entries r22 of R

The value of and is defined as follows:

r22=v2u2=1r22v2

Simplify the equationrole="math" localid="1659956549783" r22=v2 as follows:

role="math" localid="1659956569123" r22=v2r22=210-28r22=(21)2+02+(-28)2r22=35

Substitute the values 35 forr22 and210-28 forv2 in the equationu2=1r22v2 as follows:

u2=1r22v2u2=135210-28u2=21/350-28/35u2=3/50-4/5

The valuesu2=3/50-4/5 and r22=35.

Therefore, the matricesQ=4/53/5003/5-4/5 and R=55035.

Hence, the QR factorization of the given matrix is, 425003-25=1543003-455035.

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