Math, asked by jatinparmar412, 1 month ago

If a is a 2x2 matrix such that |3a| 108 then |a| is?​​

Answers

Answered by ycuteboyyy3
2

Step-by-step explanation:

Correct option is

D

[

36

0

−32

4

]

A⋅[

1

0

2

3

] is a scalar matrix and ∣3A∣=108

Let the scalar matrix be [

k

0

0

k

]

A⋅[

1

0

2

3

]=[

k

0

0

k

]

A=[

k

0

0

k

][

1

0

2

3

]

−1

⟹A=[

k

0

0

k

][

1

0

−2/3

1/3

]

⟹A=[

k

0

−2k/3

k/3

]

∣A∣=

3

k

2

∣3A∣=108

As we know kA∣=k

n

∣A∣

∣3A∣=3

2

3

k

2

⟹k

2

=36⟹k=±6

So, putting the value of k in the matrix A

A=[

6

0

(−2/3)(6)

6/3

]

A=[

6

0

−4

2

]

So, A

2

=[

6

0

−4

2

][

6

0

−4

2

]

⟹A

2

=[

36

0

−32

4

]

Hence, this is the answer.

Answered by manjupjha2409
0

Answer:

Given:

Given:

Selling price (S.P.) of a saree = ₹2,600.and

Profit gained by selling = 30%.

\small\underline{\frak{\pmb{ \red{ To \: find : }}}}

Tofind:

Tofind:

The cost price (C.P.) of the saree.

Understanding the concept:

‎ ‎ ‎ ‎ ‎ ‎We're given with the selling price and the profit gained by selling a saree. And we're asked to find the cost price of the saree. For finding this, first let's recall the chapter- "Profit and Loss", which we've studied in previous classes!

Cost Price (C.P.) - The price at which an article is purchased is called it's cost price.

Selling Price (S.P.) - The price at which an article is sold is called it's selling price.good

Profit - If the S.P. of an article is greater than its C.P., we say that there is a profit.

Loss - If the S.P. of an article is less than its C.P., we say that there is a loss.

Overheads - All the expenditure incurred on transportation, repairs, etc are categorised as overheads. Overheads are always included in the C.P. of the article.

In this question, we're only going to deal with the first three sub-topics which are mentioned above. Let's start calculating the required answer!

\small\underline{\frak{\pmb{ \red{ Formula \: to \: be \: used:- }}}}

Formulatobeused:−

Formulatobeused:−

\underline{\boxed{\bf{ C.P. = \bigg(\sf\dfrac{100}{100+Gain \: \% \: (or) \: Loss\:\%} \times S.P \bigg)}}}

C.P.=(

100+Gain%(or)Loss%

100

×S.P)

\small\underline{\frak{\pmb{ \red{ Solution: }}}}

Solution:

Solution:

As per the given data, we've all the required values to substitute them in the formula to find

\begin{gathered}\begin{gathered}\begin{gathered} \\ \longrightarrow\tt{ \pink{C.P. = \bigg(\dfrac{100}{100+Gain \: \% \: (or) \: Loss\:\%} \times S.P \bigg)}}\end{gathered} \end{gathered} \end{gathered}

⟶C.P.=(

100+Gain%(or)Loss%

100

×S.P)

S.P. = 2600

Profit % = 30

Now, on substituting these measures,

\begin{gathered}\begin{gathered}\begin{gathered} \\ \longmapsto { \sf{C.P. = \dfrac{100}{100 + 30} \times 2600}} \\ \\ \longmapsto{ \sf{C.P. = \dfrac{100 \times 260 \cancel{0}}{13 \cancel{0}}}} \\ \\ \longmapsto{ \sf{C.P. \dfrac{ \cancel{26000}}{ \cancel{13}} }} \\ \\ \longmapsto \boxed{ \tt{ \pmb { \red{C.P. = 2,000}}}}\end{gathered}\end{gathered} \end{gathered}

⟼C.P.=

100+30

100

×2600

⟼C.P.=

13

0

100×260

0

⟼C.P.

13

26000

C.P.=2,000

C.P.=2,000

We've obtained the C.P. as ₹2,000. Let's verify it!

\small\underline{\frak{\pmb{ \red{ Verification: }}}}

Verification:

Verification:

To verify, let's ignore the value of S.P. in the formula and insert the obtained C.P. and profit % in it. Then we shall check does the given value of S.P. equals the same that we get here.

\begin{gathered}\begin{gathered}\begin{gathered} \\ \longmapsto { \sf{C.P. = \dfrac{100}{100 + profit \: \%} \times S.P.}} \\ \\ \longmapsto{ \sf{2000 = \dfrac{100}{100 + 30} \times S.P.}} \\ \\ \longmapsto{ \sf{2000 = \dfrac{100}{130} \times S.P. }} \\ \\ \longmapsto{ \sf{2000 \times 130 = 100 \times S.P.}} \\ \\ \longmapsto{ \sf{260000 = 100 \times S.P.}} \\ \\ \longmapsto{ \sf{ \frac{2600 \cancel{00}}{ 1\cancel{00}} = S.P. }} \\ \\ \longmapsto { \underline{ \underline{ \bf{2,600 = S.P.}}}}\end{gathered} \end{gathered} \end{gathered}

⟼C.P.=

100+profit%

100

×S.P.

⟼2000=

100+30

100

×S.P.

⟼2000=

130

100

×S.P.

⟼2000×130=100×S.P.

⟼260000=100×S.P.

1

00

2600

00

=S.P.

2,600=S.P.

Since, the S.P. amount is same, our answer is correct!

\begin{gathered}\begin{gathered}\begin{gathered}\\ \therefore\underline{\sf{\pmb{The\:Cost\: Price\:of\:the\:saree\:is\:\pink{2,000/-}.}}}\end{gathered}\end{gathered} \end{gathered}

TheCostPriceofthesareeis2,000/−.

TheCostPriceofthesareeis2,000/−.

_________________________________________

\begin{gathered}\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{Gain = \sf S.P. \: – \: C.P.} \\ \\ \bigstar \:\bf{Loss = \sf C.P. \: – \: S.P.} \\ \\ \bigstar \: \bf{Gain \: \% = \sf \Bigg( \dfrac{Gain}{C.P.} \times 100 \Bigg)\%} \\ \\ \bigstar \: \bf{Loss \: \% = \sf \Bigg( \dfrac{Loss}{C.P.} \times 100 \Bigg )\%} \\ \\ \\ \bigstar \: \bf{S.P. = \sf\dfrac{100+Gain \: \% \: (or) \: Loss\:\%}{100} \times C.P.} \\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered} \end{gathered} < /p > < p > \end{gathered}

Step-by-step explanation:

Hope it helps

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