smc-616-10-Basic Calculations

Matrices, GEN1 2024 VCAA 26 MC

A market stall sells three types of candles.

The cost of each type of candle is shown in matrix \(C\) below.

\begin{align}
C=\left[\begin{array}{lll}
25 & 32 & 43
\end{array}\right] \end{align}

Towards the end of the day, the cost of each item is discounted by 15%.

Which one of the following expressions can be used to determine each discounted price?

\(0.15C\)
\(0.85C\)
\(8.5C\)
\(15C\)

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\(B\)

Show Worked Solution

\(\text{Discounted price is 85}\%\ \text{of the original price.}\)

\(\therefore, 0.85 is the multiplier}\)

\(\Rightarrow B\)

MATRICES, FUR2 2021 VCAA 1

Elena imports three brands of olive oil: Carmani (`C`) Linelli (`L`) and Ohana (`O`).

The number of 1 litre bottles of these oils sold in January 2021 is shown in matrix `J` below.

`J = {:[(2800),(1700),(2400)]:} {:(C),(L),(O):}`

What is the order of matrix `J`? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Elena expected that in February 2021 the sales of all three brands of olive oil would increase by 5%.
She multiplied matrix `J` by a scalar value, `k` , to determine the expected volume of sales for February.
What is the value of the scalar `k`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

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`3 xx 1`
`1.05`

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a. `3 xx 1`

b. `text{All brands} \ ↑ 5%`

`:. k = 1.05`

MATRICES, FUR1 2008 VCAA 1 MC

If `[(1, 0), (0, 1)] + [(3, 7), (8, d)] = [(4, 7), (8, 11)]` then `d` is equal to

A. `– 11`

B. `– 10`

C. `7`

D. `10`

E. `11`

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`D`

Show Worked Solution

`1 + d`	`= 11`
`:. d`	`= 10`

`=> D`

MATRICES, FUR1 2009 VCAA 2 MC

The matrix `[[12,15,3],[−6,0,24]]` can also be written as

A. `[12,15,3] + [−6,0,24]`

B. `[[12],[−6]] + [[15],[0] ]+ [[3],[24]]`

C. `[[3],[6]] [[4,5,1],[−1,0,4]]`

D. `1/3 × [[4,5,1],[−2,0,8]]`

E. `3 × [[4,5,1],[−2,0,8]]`

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`E`

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`=> E`

MATRICES, FUR1 2009 VCAA 1 MC

`3[[2,1],[0,3]] + 2[[−1,0],[2,−7]]` equals

A. `[[4,3],[4,−5]]`

B. `6[[1,1],[2,−4]]`

C. `[[4,3],[4,2]]`

D. `5[[1,1],[2,−4]]`

E. `[[3,6],[7,4]]`

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`A`

Show Worked Solution

`3[(2,1),(0,3)] + 2[(−1,0),(2,−7)]`

`= [(6,3),(0,9)] + [(−2,0),(4,−14)]`

`= [(4,3),(4,−5)]`

`=> A`

MATRICES, FUR1 2012 VCAA 1 MC

`2 xx [(2,8), (4, -1), (3,5)] -[(3,7), (4,2), (2,3)] quad text (equals)`

`[(1,1), (0,-3), (4,2)]`
`[(-2,2), (0,-6), (2,4)]`
`[(1,9), (12,0), (8,13)]`
`[(1,9), (4,-4), (4,7)]`
`[(-1,1), (0,-3), (1,2)]`

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`D`

Show Worked Solution

`2 xx [(2,8),(4,−1),(3,5)] – [(3,7),(4,2),(2,3)]`

`= [(4,16),(8,−2),(6,10)] – [(3,7),(4,2),(2,3)]`

`= [(1,9),(4,−4),(4,7)]`

`rArr D`

MATRICES, FUR1 2013 VCAA 1 MC

`[(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)][(2), (0), (0), (2)] - 2× [(0), (-1), (-1), (0)] \ text (equals)`

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`B`

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`I xx [(2),(0),(0),(2)] – 2[(0),(−1),(−1),(0)]`

`= [(2),(0),(0),(2)] – [(0),(−2),(−2),(0)] = [(2),(2),(2),(2)]`

`rArr B`

MATRICES, FUR1 2006 VCAA 1 MC

The matrix `[(12, 36), (0, 24)]` is equal to

A.	`12[(0,3),(0,2)]`	B.	`12[(1,3),(0,2)]`

C.	`12[(0,24),(−12,12)]`	D.	`12[(0,24),(0,12)]`

E.	`12[(1,3),(−12,2)]`

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`B`

Show Worked Solution

`12[(1,3),(0,2)]`	`= [(12 xx 1, 12 xx 3),(12 xx 0,12 xx 2)]`
	`= [(12,36),(0,24)]`

`rArr B`

MATRICES, FUR1 2007 VCAA 1 MC

The matrix sum `[(0, -4), (2, 5)] + [(5, 4), (-2, 2)]` is equal to

A. `[(5, 0), (0, 7)]`	B. `[(0, 0), (0, 7)]`

C. `[(5, -4), (0, 7)]`	D. `[(0, 5, -4,4), (2, -2, 5, 2)]`

E. `[(0, -4, 5, 4), (2, 5, -2, 2)]`

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`A`

Show Worked Solution

`[(0,−4),(2,5)] + [(5,4),(−2,2)]`

`= [(0 + 5,−4 + 4),(2 + −2,5 + 2)]`

`= [(5,0),(0,7)]`

`=> A`