\$\$ eginpmatrix 1&-1 \ -1&2 \ -1&0 endpmatrix eginpmatrix x\y endpmatrix = eginpmatrix 3 \ 4 \ 5endpmatrix \$\$

So Assuming this goes by \$ Avecx=vecb\$

Then using.... \$A^T A vecx = A^T vecb \$

\$\$ eginpmatrix 1&-1&-1 \ -1&2&0 endpmatrix eginpmatrix 1&-1 \ -1&2 \ -1&0 endpmatrix vecx= eginpmatrix 1&-1&-1 \ -1&2&0 endpmatrix eginpmatrix 3 \ 4 \ 5endpmatrix \$\$

Which solves to be \$\$ eginpmatrix 3&-3 \ -3&5 endpmatrix vecx = eginpmatrix -6 \ 5 endpmatrix \$\$

which i solve to get \$\$ vecx = eginpmatrix -7/3 \ 3/2 endpmatrix \$\$

linear-algebra least-squares
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edited Mar 10 "17 at 1:38

dantopa
asked Nov 3 "15 at 4:00

PatrickPatrick
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| Sexactly how 5 more comments

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Because you are only asking for the solution...

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\$\$ left( eginvmatrix 1 & -1 \ -1 & 2 \ -1 & 0 endvmatrix^intercal eginvmatrix 1 & -1 \ -1 & 2 \ -1 & 0 endvmatrix ight) eginpmatrix x \ y endpmatrix = left( eginvmatrix 1 & -1 \ -1 & 2 \ -1 & 0 endvmatrix^intercal eginpmatrix 3 \ 4 \ 5 endpmatrix ight) \$\$

\$\$ eginvmatrix 3 & -3 \ -3 & 5 endvmatrix eginpmatrix x \ y endpmatrix = eginpmatrix -6 \ 5 endpmatrix \$\$

\$\$ eginpmatrix x = -frac52 \ y = -frac12 endpmatrix \$\$

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answered Nov 3 "15 at 4:37

John AlexiouJohn Alexiou
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The normal equations are correct:\$\$lutz-heilmann.infobfA^*lutz-heilmann.infobfA x = lutz-heilmann.infobfA^*b.\$\$The error is in the solution which need to be\$\$ x = left( lutz-heilmann.infobfA^*lutz-heilmann.infobfA ight)^-1 lutz-heilmann.infobfA^*b.\$\$The inverse is\$\$left( lutz-heilmann.infobfA^*lutz-heilmann.infobfA ight)^-1 = frac1 6left(eginarraycc 5 & 3 \ 3 & 3 \endarray ight)\$\$which provides\$\$ x = -frac12left(eginarrayr 5 \ 1endarray ight).\$\$

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answered Mar 9 "17 at 15:35

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