Exercise 1.4.2. Multiply the matrices below:
![\left[ \begin{array}{rr} 4&1 \\ 5&1 \\ 6&1 \end{array} \right] \left[ \begin{array}{r} 1 \\ 3 \end{array} \right] \rm and \left[ \begin{array}{rrr} 1&2&3 \\ 4&5&6 \\ 7&8&9 \end{array} \right] \left[ \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \right] \rm and \left[ \begin{array}{rr} 4&3 \\ 6&6 \\ 8&9\end{array} \right] \left[ \begin{array}{r} \frac{1}{2} \\ \frac{1}{3} \end{array} \right]](https://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Barray%7D%7Brr%7D+4%261+%5C%5C+5%261+%5C%5C+6%261+%5Cend%7Barray%7D+%5Cright%5D+%5Cleft%5B+%5Cbegin%7Barray%7D%7Br%7D+1+%5C%5C+3++%5Cend%7Barray%7D+%5Cright%5D+%5Crm+and+%5Cleft%5B+%5Cbegin%7Barray%7D%7Brrr%7D+1%262%263+%5C%5C+4%265%266+%5C%5C+7%268%269+%5Cend%7Barray%7D+%5Cright%5D+%5Cleft%5B++%5Cbegin%7Barray%7D%7Br%7D+0+%5C%5C+1+%5C%5C+0+%5Cend%7Barray%7D+%5Cright%5D+%5Crm+and+%5Cleft%5B++%5Cbegin%7Barray%7D%7Brr%7D+4%263+%5C%5C+6%266+%5C%5C+8%269%5Cend%7Barray%7D+%5Cright%5D+%5Cleft%5B++%5Cbegin%7Barray%7D%7Br%7D+%5Cfrac%7B1%7D%7B2%7D+%5C%5C+%5Cfrac%7B1%7D%7B3%7D++%5Cend%7Barray%7D+%5Cright%5D&bg=ffffff&fg=333333&s=0&c=20201002)
Work by columns instead of by rows.
Answer: The first example multiplies a 3×2 matrix by a 2×1 matrix (column vector), producing a 3×1 matrix (column vector). Working by rows gives us the following:
![\left[ \begin{array}{rr} 4&1 \\ 5&1 \\ 6&1 \end{array} \right] \left[ \begin{array}{r} 1 \\ 3 \end{array} \right] = \left[ \begin{array}{r} 4 \cdot 1 + 1 \cdot 3 \\ 5 \cdot 1 + 1 \cdot 3 \\ 6 \cdot 1 + 1 \cdot 3 \end{array} \right] = \left[ \begin{array}{r} 7 \\ 8 \\ 9 \end{array} \right]](https://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Barray%7D%7Brr%7D+4%261+%5C%5C+5%261+%5C%5C+6%261++%5Cend%7Barray%7D+%5Cright%5D+%5Cleft%5B+%5Cbegin%7Barray%7D%7Br%7D+1+%5C%5C+3++%5Cend%7Barray%7D+%5Cright%5D+%3D+%5Cleft%5B+%5Cbegin%7Barray%7D%7Br%7D+4+%5Ccdot+1+%2B+1+%5Ccdot+3+%5C%5C+5+%5Ccdot+1+%2B+1+%5Ccdot+3+%5C%5C+6+%5Ccdot+1+%2B+1+%5Ccdot+3++%5Cend%7Barray%7D++%5Cright%5D+%3D+%5Cleft%5B+%5Cbegin%7Barray%7D%7Br%7D+7+%5C%5C+8+%5C%5C+9++%5Cend%7Barray%7D++%5Cright%5D&bg=ffffff&fg=333333&s=0&c=20201002)
Working by columns is done as follows:
For the second example we’re multiplying a 3×3 matrix times a 3×1 matrix to produce a 3×1 matrix (column vector). Working by rows gives us the following:
Working by columns is done as follows:
Note that this is a special case of the general case that multiplying an nxn matrix by a column vector of the form (0, …, 0, 1, 0, …, 0) (where the jth entry is 1 and the others zero) produces a column vector corresponding to the jth column of the matrix.
The third example multiplies a 3×2 matrix by a 2×1 matrix to form a 3×1 matrix. Working by rows gives us the following:
Working by columns is done as follows:
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition
My Math Homework 