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Linear Algebra: Orthogonal Columns\newlineLet \newlineMM be\newline \begin{matrix} 1 & 3 & 1\ 2 & -2 & a\ 1 & 1 & b \end{matrix} \newlineand let \newlineaa and \newlinebb be such that the columns of \newlineMM are orthogonat. Compute \newlineaa\newlinePick ONE option\newline12-\frac{1}{2}\newline1-1\newline12\frac{1}{2}\newline22\newlineClear Selection

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Q. Linear Algebra: Orthogonal Columns\newlineLet \newlineMM be\newline \begin{matrix} 1 & 3 & 1\ 2 & -2 & a\ 1 & 1 & b \end{matrix} \newlineand let \newlineaa and \newlinebb be such that the columns of \newlineMM are orthogonat. Compute \newlineaa\newlinePick ONE option\newline12-\frac{1}{2}\newline1-1\newline12\frac{1}{2}\newline22\newlineClear Selection
  1. Write Matrix M: First, let's write down the matrix M with the given elements and the unknowns aa and bb:\newlineM = \left| \begin{array}{ccc}\(\newline1 & 2 & 1 (\newline\)3 & -2 & 1 (\newline\)1 & a & b (\newline\)\end{array} \right|\)\newlineFor the columns of M to be orthogonal, the dot product of any two distinct columns must be zero. Let's start by taking the dot product of the first and second columns.\newlineDot product of first and second columns:\newline(1)(2)+(3)(2)+(1)(a)=26+a=a4(1)(2) + (3)(-2) + (1)(a) = 2 - 6 + a = a - 4\newlineWe want this to be equal to zero for orthogonality:\newlinea4=0a - 4 = 0
  2. Find Dot Product: Now, let's solve for aa:a4=0a - 4 = 0a=4a = 4We have found the value of aa that makes the first and second columns orthogonal.

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