Consider the following system of simultaneous linear equations.
\(y+z=4\)
\(x-y+z=1\)
\(-x+y=2\)
The solution to these simultaneous equations can be found by calculating
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Consider the following system of simultaneous linear equations.
\(y+z=4\)
\(x-y+z=1\)
\(-x+y=2\)
The solution to these simultaneous equations can be found by calculating
\(E\)
\(
\begin{bmatrix}
0 & 1 & 1 \\ 1 & -1 & 1 \\
-1 & 1 & 0\end{bmatrix}
\times \begin{bmatrix}
x \\ y \\ z\end{bmatrix}
= \begin{bmatrix}
4 \\ 1 \\ 2\end{bmatrix}
\)
\(
\begin{bmatrix}
x \\ y \\ z\end{bmatrix}
= \begin{bmatrix}
0 & 1 & 1 \\
1 & -1 & 1 \\
-1 & 1 & 0
\end{bmatrix}^{-1}
\times \begin{bmatrix}
4 \\ 1 \\ 2
\end{bmatrix}
\)
\(
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
= \begin{bmatrix}
1 & -1 & -2 \\
1 & -1 & -1 \\
0 & 1 & 1
\end{bmatrix}
\times\begin{bmatrix}
4 \\ 1 \\ 2\end{bmatrix}
\)
\(\Rightarrow E\)
`ax + 4y = 10`
`18x + by = 6`
The set of simultaneous linear equations above does not have a unique solution when
`A`
`text{In matrix form:}`
`[(a,4),(18,b)] [(x),(y)] = [(10),(6)]`
`text{No unique solution} =>\ text{det} = 0`
`ab – 4 xx 18` | `= 0` |
`ab` | `= 72` |
`=> A`
Market researchers claim that the ideal number of bookshops (`x`), sports shoe shops (`y`) and music stores (`z`) for a shopping centre can be determined by solving the equations
`2x + y + z = 12`
`x-y+z=1`
`2y-z=6`
`qquad[(qquadqquadqquad),(),()][(quad),(quad),(quad)] = [(quad),(quad),(quad)]`
a. | `[(2,1,1),(1,-1,1),(0,2,-1)][(x),(y),(z)] = [(12),(1),(6)]` |
b. | `text(det)\ [(2,1,1),(1,-1,1),(0,2,-1)] = 1 != 0` |
`:.\ text(A unique solution exists.)`
c. `text(By CAS,)`
`[(2,1,1),(1,-1,1),(0,2,-1)]^(-1) = [(-1,3,2),(1,-2,-1),(2,-4,-3)]`
d. | `[(x),(y),(z)]` | `= [(-1,3,2),(1,-2,-1),(2,-4,-3)][(12),(1),(6)]` |
`= [(3),(4),(2)]` |
`:.\ text(Estimated ideal numbers are:)`
`text(3 bookshops)`
`text(4 shoe shops)`
`text(2 music stores)`
The basketball coach has written three linear equations which can be used to predict the number of points, `p`, rebounds, `r`, and assists, `a`, that Oscar will have in his next game.
The equations are `p + r + a` | `= 33` |
`2p - r + 3a` | `= 40` |
`p + 2r + a` | `= 43` |
Complete the missing information below. (1 mark)
`[(qquadqquad),(qquadqquad),(qquadqquad)][(p),(r),(a)] = [(33),(40),(43)]`
This matrix equation can be solved in the following way.
`[(p),(r),(a)] = [(7,-1,-4),(-1,0,1),(x,1,3)][(33),(40),(43)]`
a. | `[(1,1,1),(2,-1,3),(1,2,1)][(p),(r),(a)] = [(33),(40),(43)]` |
b. | `[(7,-1,-4),(-1,0,1),(x,1,3)]` | `=[(1,1,1),(2,-1,3),(1,2,1)]^(-1)` |
`= [(7,-1,-4),(-1,0,1),(-5,1,3)]` |
`:.x = -5`
c. | `[(p),(r),(a)]` | `= [(7,-1,-4),(-1,0,1),(-5,1,3)][(33),(40),(43)]` |
`= [(19),(10),(4)]` |
`:.\ text(Oscar is predicted to have 10 rebounds in the next game.)`
`x + z` | `= 6` |
`2y + z` | `= 8` |
`2x + y + 2z` | `= 15` |
The solution of the simultaneous equations above is given by
`A`
`[(1,0,1),(0,2,1),(2,1,2)][(x),(y),(z)] = [(6),(8),(15)]`
`:. [(x),(y),(z)]` | `= [(1,0,1),(0,2,1),(2,1,2)]^(−1)[(6),(8),(15)]` |
`= [(−3,−1,2),(−2,0,1),(4,1,−2)][(6),(8),(15)]` |
`rArr A`
Consider the following system of three simultaneous linear equations.
`2x+z=5`
`x-2y=0`
`y-z=-1`
This system of equations can be written in matrix form as
A. `[(2, 1), (1, -2), (1, -1)][(x), (y), (z)] = [(5), (0), (-1)]` | B. `[(2,0,1), (1,-2,0), (0,1, -1)][(x), (y), (z)] = [(5), (0), (-1)]` |
C. `[(2, 1, 5), (1, -2, 0), (1, -1, -1)][(x), (y), (z)] = [(5), (0), (-1)]` | D. `[(2, 1, 0), (1, -2, 0), (1, -1, 0)][(x), (y), (z)] = [(5), (0), (-1)]` |
E. `[(2, 1), (1, -2), (1, -1)][(5), (0), (-1)] = [(x), (y), (z)]` |
`B`
`=> B`
`y - z` | `= 8` |
`5x - y` | `= 0` |
`x + z` | `= 4` |
The system of three simultaneous linear equations above can be written in matrix form as
A. | `[[0,1,-1],[0,5,-1],[1,0,1]][[x],[y],[z]]=[[8],[0],[4]]` | B. | `[[0,1,-1],[5,-1,0],[1,0,1]][[x],[y],[z]]=[[8],[0],[4]]` |
C. | `[[1,-1],[5,-1],[1,1]][[x],[y],[z]]=[[8],[0],[4]]` | D. | `[[0,5,1],[1,-1,0],[-1,0,1]][[x],[y],[z]]=[[8],[0],[4]]` |
E. | `[[0,5,0],[-1,-1,0],[1,1,0]][[x],[y],[z]]=[[8],[0],[4]]` |
`B`
`=>B`