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Matrices, GEN1 2024 NHT 28 MC

Matrix \(D\) is a \(2 \times 2\) matrix where each element is given by \(d_{ij}\)

Which rule will result in a binary matrix?

  1. \(d_{i j}=i+j\)
  2. \(d_{i j}=i-j\)
  3. \(d_{i j}=i \times j\)
  4. \(d_{i j}=i \ ÷ \  j\)
  5. \(d_{i j}=(i-j)^2\)
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\(E\)

Show Worked Solution

\(\text{Binary matrix: each element is either 0 or 1}\)

\(\text{Consider the matrix in option}\ E:\)

\(M_E=\begin{bmatrix}(1-1)^2,(1-2)^2 \\ (2-1)^2,(2-2)^2\end{bmatrix}=\begin{bmatrix}0,1 \\ 1,0\end{bmatrix}\)

\(\Rightarrow E\)

Matrices, GEN2 2024 VCAA 10

To access the southern end of the construction site, Vince must enter a security code consisting of five numbers.

The security code is represented by the row matrix \(W\).

The element in row \(i\) and column \(j\) of \(W\) is \(w_{i j}\).

The elements of \(W\) are determined by the rule  \((i-j)^2+2 j\).

  1. Complete the following matrix showing the five numbers in the security code.   (1 mark)

To access the northern end of the construction site, Vince enters a different security code, consisting of eight numbers.

This security code is represented by the row matrix \(X\).

The element in row \(i\) and column \(j\) of \(X\) is \(x_{i j}\).

The elements of \(X\) are also determined by the rule  \((i-j)^2+2 j\).

  1. What is the last number in this security code to access the northern end of the construction site?  (1 mark)

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a.    \(W=\ [2\quad 5\quad 10\quad 17\quad 26\ ]\)

b.    \(x_{18}=65\)

Show Worked Solution

a.     \(W\) \(=\ [(0)^2+2\ \ \  (-1)^2+4\ \ \  (-2)^2+36\ \ \  (-3)^2+8\ \ \  (-4)^2+10\ ]\)
    \(=\ [\ 2\quad 5\quad 10\quad 17\quad 26\ ]\)

 

b.    \(x_{18}=(1-8)^2+2\times 8=65\)

Mean mark (a) 54%.
♦ Mean mark (b) 50%

Matrices, GEN1 2023 VCAA 29 MC

Matrix \(K\) is a \(3 \times 2\) matrix.

The elements of \(K\) are determined by the rule  \(k_{i j}=(i-j)^2\).

Matrix \(K\) is


A.

 \begin{bmatrix} 0 & 1 & -2 \\
1 & 0 & -1
\end{bmatrix}


B.

 \begin{bmatrix}
0 & 1 & 4 \\
1 & 0 & 1
\end{bmatrix}


C.

\begin{bmatrix}
0 & -1\\
1 & 0\\
4 & 1
\end{bmatrix}


D.

 \begin{bmatrix}
0 & 1\\
1 & 0\\
2 & 1
\end{bmatrix}


E.

 \begin{bmatrix}
0 & 1\\
1 & 0\\
4 & 1
\end{bmatrix}		    
Show Answers Only

\(E\)

Show Worked Solution

\(K\ \text{is a 3 × 2 matrix (eliminate A)}\)

\(k_{ij}\ \text{will all be}\ \geq 0\ \text{(eliminate C)}\)

\(k_{31} = (3-1)^2 = 4\ \text{(eliminate B and D)}\)

\(\Rightarrow E\)

Matrices, GEN1 2023 VCAA 25 MC

The daily maximum temperature at a regional town for two weeks is displayed in the table below.

\begin{array} {|c|c|c|}
\hline \rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \text{Monday} & \text{Tuesday} & \text{Wednesday} & \text{Thursday} & \text{Friday} & \text{Saturday} & \text{Sunday} \\
\hline \rule{0pt}{2.5ex} \text{Week 1} \rule[-1ex]{0pt}{0pt} & \text{20 °C} & \text{17 °C} & \text{23 °C} & \text{20 °C} & \text{18 °C} & \text{19 °C} & \text{30 °C} \\
\hline \rule{0pt}{2.5ex} \text{Week 2} \rule[-1ex]{0pt}{0pt} & \text{29 °C} & \text{27 °C} & \text{28 °C} & \text{21 °C} & \text{20 °C} & \text{20 °C} & \text{22 °C} \\
\hline
\end{array}

 This information can also be represented by matrix \(M\), shown below.

\[\textit{M}=\begin{bmatrix}
20 & 17 & 23 & 20 & 18 & 19 & 30 \\
29 & 27 & 28 & 21 & 20 & 20 & 22
\end{bmatrix}\]

Element \(m_{21}\) indicates that

  1. the temperature was 29 °C on Monday in week 2.
  2. the temperature was 17 °C on Tuesday in week 1.
  3. the lowest temperature for these two weeks was 17 °C.
  4. the highest temperature for these two weeks was 29 °C.
  5. week 2 had a higher average maximum temperature than week 1.
Show Answers Only

\(A\)

Show Worked Solution

\(m_{21}\ \text{refers to the data in row 2, column 1 = 29 °C on Monday in week 2}\)

\(\Rightarrow A\)

MATRICES, FUR1 2020 VCAA 6 MC

The element in row  `i`  and column  `j`  of matrix  `M`  is  `m_(ij)`.

`M`  is a  3 × 3  matrix. It is constructed using the rule  `m_(ij) = 3i + 2j`.

`M`  is

A.  

`[(5,7,9),(7,9,11),(11,13,15)]`

 

 B.  

`[(5,7,9),(8,10,12),(11,13,15)]`

 
C.  

`[(5,7,10),(8,10,13),(11,13,16)]`

 

 D.  

`[(5,8,11),(7,10,13),(9,12,15)]`

 
E.  

`[(5,8,11),(8,11,14),(11,14,17)]`

 

    
Show Answers Only

`B`

Show Worked Solution

`text(By Elimination):`

`m_13 = 3 xx 1 + 2 xx 3 = 9`

`:.\ text(Eliminate)\ C, D and E`
 

`m_21 = 3 xx 2 + 2 xx 1 = 8`

`:.\ text(Eliminate)\ A`

`=>  B`

MATRICES, FUR1-NHT 2019 VCAA 1 MC

The number of individual points scored by Rhianna (`R`), Suzy (`S`), Tina (`T`), Ursula (`U`) and Vicki (`V`) in five basketball matches `(F, G, H, I, J)` is shown in matrix `P` below.
 

`{:(),(),(P=):}{:(qquadqquadqquad\ text(match)),((quadF,G,H,I,J)),([(2,\ 0,\ 3,\ 1,\ 8),(4,7,2,5,3),(6,4,0,0,5),(1,6,1,4,5),(0,5,3,2,0)]):}{:(),(),({:(R),(S),(T),(U),(V):}):}{:(),(),(text(player)):}`
 

Who scored the highest number of points and in which match?

  1. Suzy in match  `I`
  2. Tina in match  `H`
  3. Vicki in match  `F`
  4. Ursula in match  `G`
  5. Rhianna in match  `J`
Show Answers Only

`E`

Show Worked Solution

`text(Highest points = 8 =)\ e_15`

`e_15 \ => \ text(Rhianna in match)\ J`

`=>\ E`

MATRICES, FUR1 2019 VCAA 3 MC

Consider the matrix `P`, where  `P = [(3, 2, 1), (5, 4, 3)]`.

The element in row `i` and column `j` of matrix `P` is  `p_(ij)`.

The elements in matrix `P` are determined by the rule

  1. `p_(ij) = 4 - j`
  2. `p_(ij) = 2i + 1`
  3. `p_(ij) = i + j + 1`
  4. `p_(ij) = i + 2j`
  5. `p_(ij) = 2i - j + 2`
Show Answers Only

`E`

Show Worked Solution

`text(Consider option E):`

`P_11 = 2 xx 1 – 1 + 2 = 3`

`P_12 = 2 xx 1 – 2 + 2 = 2`

`P_13 = 2 xx 1 – 3 + 2 = 1`

`P_21 = 2 xx 2 – 1 + 2 = 5`

`text(etc…)`

`=>  E`

MATRICES, FUR1 2017 VCAA 6 MC

The table below shows information about two matrices, `A` and `B`.
 

 
The element in row `i` and column `j` of matrix `A` is `a_(ij)`.

The element in row `i` and column `j` of matrix `B` is `b_(ij)`.

The sum  `A + B`  is

A.

`[(5,7,9),(8,10,12),(11,13,15)]`

 

 B.

`[(5,8,11),(7,10,13),(9,12,15)]`

 
C.

`[(3,6,9),(3,6,9),(3,6,9)]`

 

 D.

`[(3,3,3),(6,6,6),(9,9,9)]`

 
E.

`[(3,6,3),(6,3,9),(3,9,3)]`

 

    
Show Answers Only

`D`

Show Worked Solution
`a_(ij) + b_(ij)` `= 2i + j + i – j`
  `= 3i`
`A + B` `= [(3xx1,3xx1,3xx1),(3xx2,3xx2,3xx2),(3xx3,3xx3,3xx3)]`
  `= [(3,3,3),(6,6,6),(9,9,9)]`

`=> D`

MATRICES, FUR1 2017 VCAA 1 MC

Kai has a part-time job.

Each week, he earns money and saves some of this money.

The matrix below shows the amounts earned (`E`) and saved (`S`), in dollars, in each of three weeks.
 

`{:(qquadqquadqquadqquadquadEquadqquadS),({:(text(week 1)),(text(week 2)),(text(week 3)):}[(300,100),(270,90),(240,80)]):}`

 

How much did Kai save in week 2?

  1.   `$80`
  2.   `$90`
  3. `$100`
  4. `$170`
  5. `$270`
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`B`

Show Worked Solution

`text(Kai saved $90 in week 2.)`

`=> B`

MATRICES, FUR1 2016 VCAA 5 MC

Let `M = [(1,2,3,4),(3,4,5,6)]`.

The element in row `i` and column `j` of `M` is `m_(ij)`.

The elements of `M` are determined by the rule

  1. `m_(ij) = i + j - 1`
  2. `m_(ij) = 2i - j + 1`
  3. `m_(ij) = 2i + j - 2`
  4. `m_(ij) = i + 2j - 2`
  5. `m_(ij) = i + j + 1` 
Show Answers Only

`C`

Show Worked Solution

`text(Check)\ m_14 = 4,`

`text(Option)\ A:\  1 + 4 – 1 = 4\ \ text{(correct)}`

`text(Option)\ B:\  2 – 4 + 1 = −1\ \ text{(incorrect)}`

`text(Option)\ C:\  2 + 4 – 2 = 4\ \ text{(correct)}`

`text(Option)\ D:\  1 + 8 – 2 = 7\ \ text{(incorrect)}`

`text(Option)\ E:\  1 + 4 + 1 = 6\ \ text{(incorrect)}`

 

`text(Check)\ m_22 = 4,`

`text(Option)\ A:\  2 + 2 – 1 = 3\ \ text{(incorrect)}`

`text(Option)\ C:\  4 + 2 – 2 = 4\ \ text{(correct)}`

`=> C`

MATRICES, FUR1 2014 VCAA 6 MC

The order of matrix `X` is `3 xx 2.`

The element in row `i ` and column `j` of matrix `X` is `x_(ij)` and it is determined by the rule

`x_(ij) = i + j`

The matrix `X` is

VCAA MATRICES FUR1 2014 6ii

Show Answers Only

`E`

Show Worked Solution

`[(x_11, x_12), (x_21, x_22), (x_31, x_32)]`

♦ Mean mark 42%.

`:. X = [(2, 3), (3, 4), (4, 5)]`

`=>E`

MATRICES, FUR1 2015 VCAA 8 MC

The order of matrix `X` is `2 xx 3`.

The element in row `i` and column `j` of matrix `X` is `x_(ij)` and it is determined by the rule

`x_(ij) = i - j`

Which one of the following calculations would result in matrix `X`?

A.   `[(1,1,1),(2,2,2)] - [(1,2,3),(1,2,3)]`

B.   `[(1,2,3),(1,2,3)] - [(1,1,1),(2,2,2)]`

C.   `[(2,2,2),(2,2,2)] - [(3,3,3),(3,3,3)]`

D.   `[(1,2),(1,2),(1,2)] - [(1,1),(2,2),(3,3)]`

E.   `[(1,1),(2,2),(3,3)] - [(1,2),(1,2),(1,2)]`

Show Answers Only

`A`

Show Worked Solution

`x_(ij) = i – j`

♦ Mean mark 43%.

`:. X = [(0,−1,−2),(1,0,−1)]`

`=> A`