10 use cramers rule to solve the following system x1 2x2 x3...

Name: 10 use cramers rule to solve the following system x1 2x2 x3 5 x1 4x2 3x3 2 x1 x3 3 32691
Uploaded: 2022-02-21T07:10:58-08:00
Duration: 6 min 3 s
Channel: Haricharan Gupta
Description: 10 use cramers rule to solve the following system x1 2x2 x3 5 x1 4x2 3x3 2 x1 x3 3 32691

Transcript

00:01 Hello students we are given some equations we need to solve them by kramer's rule if it is possible okay first of all we are supposed to know that we need to calculate delta which is nothing but the determinant of coefficient matrix so basically we will write four minus one and one okay second row is nothing but two to three and here five minus two and six okay now what we can do here? we can just write delta is equals to here four times of so basically two times of six is nothing but twelve minus minus minus minus minus plus it is six after the minus minus plus one times of 12 okay minus 5 3 15 okay after that plus one times of it is nothing but minus 4 and minus 10 now delta is equal to it is nothing but 18 18 times of 4 is nothing but 72 it is nothing but minus 3 and here it is nothing but 40 so basically delta comes out as what we can do 72 minus 3 comes out as 69 69 minus 14 comes out as now we need to calculate delta x which can be written as how we will write delta x okay, so basically delta x can be written as as delta x1 because here variables are x1 x2 x3 so that did a determinant of okay we need to put the first row as d1 d2 3 which is nothing but the consulate of minus 5 10 and 1 okay minus 5 10 and 1 second column can you written as b2 b3 okay b1 b2 b3 okay b1 b2 b3 comes out as minus 1 2 minus 2 minus 2 and minus 2 okay third column can be written as c1 c2 c3 c1 c3 which comes out as 1 3 6 1 3 6 okay 1 3 and 6 now delta x1 comes out as minus 5 times of it is 12 minus minus minus plus 16 sorry 6 minus minus plus 1 times of 60 minus 3 plus 1 times of minus 20 and minus 2 okay so basically delta x1 comes out as 12 plus 6 comes out as 18 18 times of minus 5 comes out as minus 90 okay here plus it is nothing but 57 and it is minus 22 okay so basically delta x1 comes out as minus 90 minus 90 minus 22 comes out as minus 112 and plus 57 which is nothing but minus 55 it's all about delta x1 now what we can do we just need to calculate delta x2 we how we will write delta x2 the first column can be written as a 1 a 2 a 3 which is nothing but 4 2 and 5 3 which is nothing but d1 d2 d3 which is nothing but minus 5 10 1 10 and 1 okay students after the c1 c2 c3 so c1 c3 is nothing but 1 3 and 6 okay see 1 3 and 6 okay see 1 1 3 and 6 now 1 3 and 6 now how we will calculate we will have to solve it by calculation delta x2 comes out as okay 4 times of c 60 minus 3 minus minus plus 5 times of it is nothing but 12 minus 15 okay after that plus 1 times of 2 minus 50 okay now after the delta x2 comes out as okay 60 minus 3 comes out of 57 and times 4 which is nothing but 2 to 8 okay 12 minus 15 comes out as minus 3 minus so basically it is minus 15 and here minus 48 so delta x2 comes out as 228 minus 15 comes out as 216 5 okay now we need to calculate similarly delta x3 so basically how we will write first column can be written as a 1 a2 a3 okay a 1 a2 a 3 which is nothing but 4 2 and 5 okay see 4 2 and 5 and b sorry the second column can be written as b1 b2 b3 b1 b3 comes out as minus 1 minus 1 minus 2 minus 2 minus 2 and minus 2 okay see but we can see again minus 1 2 and minus 2 okay the third column can be written as d1 d2 d3 which is nothing but minus 5 10 1 okay minus 5 10 and 1 now what we can do delta x3 comes out as 4 times of it is 2 minus minus plus 20 minus minus plus 1 it is nothing but 2 minus 50 okay after that minus 5 times of minus 4 minus 10 okay now we can see the delta x 3 comes out as it is nothing but 22 and 22 times of 4 comes out as 88 after that minus 48 minus 4 minus 10 comes out as minus 14 times of minus 5 comes out as plus 70 okay so delta x 3 comes out as 88 minus 48 comes out as 40 40 plus 70 40 plus 70 comes out as 1 110 okay now we need to calculate the exact solution so now we can write x1 is nothing but delta x1 divided by delta okay so delta x1 how we will find it is nothing but minus 55 okay so basically we will put minus 55 divided by what was our delta delta is nothing but 55 so basically it is nothing but now minus 155 x2 can be written as delta x2 divided by delta so x2 is nothing but sorry delta x2 is nothing but 165 okay and divided by 55 which comes out as 3 x3 is nothing but delta x3 divided by delta okay which is nothing but one one 10 divided by 55 comes out as two now hence we can state that hence solution is okay solution is nothing but minus 1 1 3 and 2...

Question

(10) Use Cramer's Rule to solve the following system: $x_1 + 2x_2 - x_3 = 5$ $x_1 - 4x_2 + 3x_3 = -2$ $x_1 + x_3 = 3$

Question

(10) Use Cramer's Rule to solve the following system: $x_1 + 2x_2 - x_3 = 5$ $x_1 - 4x_2 + 3x_3 = -2$ $x_1 + x_3 = 3$

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