# Doubt from Matrices and Determinants

1. The number of 3x3 matrices that can be formed by using 1 and -1 such that the product of elements in each row and column is -1.

Ans : 16

1. If A = \lim_{n \to \infty}(\begin{matrix} 1 & \frac{x}{n} \\ \frac{-x}{n} & 1 \end{matrix})^n then |A| is

Ans : 1

Bro for 2 nd use 1to the power infinity u will get finally e ka power 0 hence ans would be 1

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For first use sarrus rule

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for the second one
expan we get (1+(x/n)^2)^n which is 1ki power infinity …
using that we get e^(x^2/n) and n tends to infinity which gives e^0 which is 1
hence |A|=1

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Thanks guys…
I got the second one but I don’t understand how can you use sarrus rule in the first one?

1st ques.
Case 1: all rows and columns are (-1)
Case 2: one row is [-1 -1 -1] (* 3 for 3 rows) and the other two rows have one (-1) and both can only come together in the same column so *3 for 3 columns ------> 3 * 3=9 for case 2
Case 3: each row has one (-1) and none can come in the same column(only one in one column and row) so 3 * 2 * 1 = 6 cases.
TOTAL = 1+9+6= 16

(I’m unable to upload the photo)

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I too don’t understand how to solve 1st question type questions…

What is this? Can someone tell me? Got on YT.

Its just a method to expand , nothing fancy. I think its not necessary to do it.

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I didn’t know about it. IG it is lengthier than normal one

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Its only used for symmetrical matrices otherwise useless , you can use normal method though.

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Sarrus can be applied to any 3x3 square matrix, right? abhi youtube pe vid dekha

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Yeah but not used frequently for normal 3X3 since its lengthy, it simplifies the symmetrical matrices well and mostly used in their expansion.

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yes i get it now thanks, it will faster for symmetrical matrices

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Did you learn its formula? The cross method one

remembered the method. column wise sarrus rule

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For P1:

Fill in the * positions in this grid first:

(Don’t fill the positions with numbers yet)

**1
**2
345

Clearly 2^4 =16 ways of doing this.

Now observe that position 1, 2, 3 and 4 are fixed. That is, you have only 1 way to fill them.
This also implies that the number to go in position 5 is fixed.
So, a grid is uniquely determined by the top left numbers that we initially filled.