Matrices, vectors, and the algebra of linear systems

(a) Row by row. For $i=1{:}$ $1-3=-2$ and $1-6=-5.$ For $i=2{:}$ $4-3=1$ and $4-6=-2.$ For $i=3{:}$ $9-3=6$ and $9-6=3.$ So $$ \mathbf{B}=\begin{pmatrix} -2 & -5\\ 1 & -2\\ 6 & 3\end{pmatrix}\,. $$

(b) Row first. $b_{32}$ is row $3,$ column $2{:}$ it exists and equals $9-6=3.$ $b_{23}$ would be row $2,$ column $3,$ and $\mathbf{B}$ has only two columns: it does not exist (the formula would give $4-9=-5,$ but there is no such entry).

(c) $\mathbf{B}$ has three rows and two columns, so it is not square, and only a square matrix has a main diagonal.

(a) The system. Flour and work give $$ \begin{aligned} 4x_{1}+2x_{2}+3x_{3}&=34\,,\\ 2x_{1}\phantom{{}+2x_{2}}+x_{3}&=10\,, \end{aligned} $$ two equations in three unknowns, so $\mathbf{A}$ is $2\times3{:}$ $$ \begin{aligned} \mathbf{A}&=\begin{pmatrix} 4 & 2 & 3\\ 2 & 0 & 1\end{pmatrix}\,,\\ \mathbf{b}&=\begin{pmatrix} 34\\ 10\end{pmatrix}\,. \end{aligned} $$ The $0$ is the second kind's work.

(b) The first list. $4\cdot3+2\cdot5+3\cdot4=12+10+12=34$ and $2\cdot3+4=10{:}$ both equations hold, so $(3,5,4)$ is a solution.

(b) The second list. $4\cdot4+2\cdot5+3\cdot3=16+10+9=35\ne34,$ so $(4,5,3)$ is not a solution, although it has the same three numbers (Watch out 12.6).

(c) Another solution. With $x_{1}=1$ the second equation gives $x_{3}=10-2=8,$ and the first then $4+2x_{2}+24=34,$ so $x_{2}=3.$ The list $(1,3,8)$ is a second solution: the system is consistent and has more than one solution (fewer equations than unknowns often leaves room).

(a) Sum and change. Entry by entry, $$ \begin{aligned} \mathbf{A}+\mathbf{B}&=\begin{pmatrix} 250 & 150\\ 95 & 126\\ 380 & 315\end{pmatrix}\,,\\ \mathbf{B}-\mathbf{A}&=\begin{pmatrix} 10 & -10\\ 5 & 6\\ -20 & 15\end{pmatrix}\,. \end{aligned} $$ The sum is the two months together, per commodity and outlet; the difference is the change from September to October.

(b) The falls. The negative entries of $\mathbf{B}-\mathbf{A}{:}$ commodity $1$ at outlet $2$ (down $10$) and commodity $3$ at outlet $1$ (down $20$), in hundreds of kronor.

(c) A scalar multiple. Ten per cent more everywhere is $1.1\,\mathbf{B},$ every entry multiplied by $1.1.$ Its entry in row $3,$ column $2$ is $1.1\cdot165=181.5.$

Add $-\mathbf{A}$ to both sides. Associativity, then $\mathbf{A}+(-\mathbf{A})=\mathbf{0}$ and $\mathbf{X}+\mathbf{0}=\mathbf{X}$ (with commutativity to put the terms in place) give $$2\mathbf{X}=3\mathbf{B}-\mathbf{A}\,.$$

Compute the right-hand side. $$ \begin{aligned} 3\mathbf{B}&=\begin{pmatrix} 3 & 0\\ 6 & 6\end{pmatrix}\,,\\ 3\mathbf{B}-\mathbf{A}&=\begin{pmatrix} 2 & 2\\ 6 & 2\end{pmatrix}\,. \end{aligned} $$

Divide by $2.$ Multiplying both sides by the scalar $\tfrac12$ (by Definition 12.8, entry by entry), $$ \mathbf{X}=\begin{pmatrix} 1 & 1\\ 3 & 1\end{pmatrix}\,. $$

Check. $$ \begin{aligned} 2\mathbf{X}+\mathbf{A}&=\begin{pmatrix} 2 & 2\\ 6 & 2\end{pmatrix}\\ &\quad+\begin{pmatrix} 1 & -2\\ 0 & 4\end{pmatrix}\\ &=\begin{pmatrix} 3 & 0\\ 6 & 6\end{pmatrix}=3\mathbf{B}\,. \end{aligned} $$ By Definition 12.2 the matrix equation is four ordinary equations, and all four hold.

(a) Two inner products. $$ \begin{aligned} \mathbf{p}\cdot\mathbf{x}&=480+500+350\\ &=1330\,,\\ \mathbf{c}\cdot\mathbf{x}&=280+300+300\\ &=880\,. \end{aligned} $$ Revenue $1330$ kronor, cost $880$ kronor.

(b) The margin per unit. $\mathbf{p}-\mathbf{c}=(5,8,5),$ so $$ \begin{aligned} (\mathbf{p}-\mathbf{c})\cdot\mathbf{x}&=200+200+50\\ &=450\,. \end{aligned} $$ By rules (a), (c) and (b), $$ \begin{aligned} &(\mathbf{p}-\mathbf{c})\cdot\mathbf{x}\\ &=\mathbf{p}\cdot\mathbf{x}+\bigl((-1)\mathbf{c}\bigr)\cdot\mathbf{x}\\ &=\mathbf{p}\cdot\mathbf{x}-\mathbf{c}\cdot\mathbf{x}\,, \end{aligned} $$ revenue minus cost: the profit, $1330-880=450$ kronor.

(c) A scalar. The new prices are $1.05\,\mathbf{p},$ and by rule (b) the new revenue is $$ \begin{aligned} &(1.05\,\mathbf{p})\cdot\mathbf{x}\\ &=1.05\,(\mathbf{p}\cdot\mathbf{x})\\ &=1.05\cdot1330=1396.5 \end{aligned} $$ kronor.

(a) $\mathbf{A}\mathbf{B}.$ Rows of $\mathbf{A}$ against columns of $\mathbf{B}{:}$ $$ \begin{aligned} c_{11}&=1\cdot2+2\cdot1=4\,,\\ c_{12}&=1\cdot(-1)+2\cdot1=1\,,\\ c_{21}&=0\cdot2+3\cdot1=3\,,\\ c_{22}&=0\cdot(-1)+3\cdot1=3\,, \end{aligned} $$ so $\mathbf{A}\mathbf{B}=\begin{pmatrix} 4 & 1\\ 3 & 3\end{pmatrix}.$

(a) $\mathbf{B}\mathbf{A}.$ Rows of $\mathbf{B}$ against columns of $\mathbf{A}{:}$ $$ \begin{aligned} d_{11}&=2\cdot1+(-1)\cdot0=2\,,\\ d_{12}&=2\cdot2+(-1)\cdot3=1\,,\\ d_{21}&=1\cdot1+1\cdot0=1\,,\\ d_{22}&=1\cdot2+1\cdot3=5\,, \end{aligned} $$ so $\mathbf{B}\mathbf{A}=\begin{pmatrix} 2 & 1\\ 1 & 5\end{pmatrix}.$

(b) "Even when both exist they are usually different matrices": both products exist, both are $2\times2,$ and three of their four entries differ. The order of the factors is part of the question.

(a) Substitute. $$ \begin{aligned} z_{1}&=2(x_{1}+2x_{2})-(3x_{1}-x_{2})\\ &\quad+(-x_{1}+x_{2})\\ &=-2x_{1}+6x_{2}\,,\\ z_{2}&=(x_{1}+2x_{2})+4(-x_{1}+x_{2})\\ &=-3x_{1}+6x_{2}\,. \end{aligned} $$

(b) The two matrices. $$ \begin{aligned} \mathbf{B}&=\begin{pmatrix} 1 & 2\\ 3 & -1\\ -1 & 1\end{pmatrix}\,,\\ \mathbf{A}&=\begin{pmatrix} 2 & -1 & 1\\ 1 & 0 & 4\end{pmatrix}\,. \end{aligned} $$ $(2\times3)(3\times2)=2\times2,$ and row by column $$ \begin{aligned} c_{11}&=2-3-1=-2\,,\\ c_{12}&=4+1+1=6\,,\\ c_{21}&=1+0-4=-3\,,\\ c_{22}&=2+0+4=6\,. \end{aligned} $$

(b) The comparison. $\mathbf{A}\mathbf{B}=\begin{pmatrix} -2 & 6\\ -3 & 6\end{pmatrix}$ holds exactly the coefficients found in (a), each in its place: row $i$ is $z_{i},$ column $k$ is $x_{k}.$ This is the reason for Definition 12.17.

(c) Two ways. $y_{1}=1+4=5,$ $y_{2}=3-2=1,$ $y_{3}=-1+2=1,$ so $z_{1}=10-1+1=10$ and $z_{2}=5+4=9.$ Directly, $$ \begin{aligned} &\begin{pmatrix} -2 & 6\\ -3 & 6\end{pmatrix} \begin{pmatrix} 1\\ 2\end{pmatrix}\\ &=\begin{pmatrix} -2+12\\ -3+12\end{pmatrix} =\begin{pmatrix} 10\\ 9\end{pmatrix}\,. \end{aligned} $$ The same numbers.