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Lab problem 8. Solution of systems of linear algebraic equations Calculate the determinant of the coefficient matrix 2) Solve the system of linear equations by the ...

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Lab problem 8. Solution of systems of linear algebraic equations Calculate the determinant of the coefficient matrix 2) Solve the system of linear equations by the Gaussian method 3) Solve the system of linear equations by the method of matrix inversion (4n +42+.3+214 =2 4 + 2r2" 2x3 +24=0 41 +32 2X3 - ~4 =2 2n1 312 - 313 214 =0 211-X2 +513+344 =-1 311 + 512 Ar, 2x4 = [ 411 +Sr2 +43 -414 = 8 41 Tx2 6x3 " 214 4- + 2X3 " 7=l 2n1 +42 +3 +4=2 211 +313 +44=4 211 2x2 " +213 +314 F

Lab problem 8. Solution of systems of linear algebraic equations Calculate the determinant of the coefficient matrix 2) Solve the system of linear equations by the Gaussian method 3) Solve the system of linear equations by the method of matrix inversion (4n +42+.3+214 =2 4 + 2r2" 2x3 +24=0 41 +32 2X3 - ~4 =2 2n1 312 - 313 214 =0 211-X2 +513+344 =-1 311 + 512 Ar, 2x4 = [ 411 +Sr2 +43 -414 = 8 41 Tx2 6x3 " 214 4- + 2X3 " 7=l 2n1 +42 +3 +4=2 211 +313 +44=4 211 2x2 " +213 +314 F | n+2 +313 ~X4=2 211 + 2x2 " +3r3 +414 =0 211 +2 + 5x3 -244 =3 211 212 +3X3 4514 T1 +2r2 ~X3-4=0 10 . (2n 2X2 +3 | .4=1 211 + 312 ~X3+=3 211 +312 213 +14 =] 211 Sx2 +2x3 +44=3 4n S12 +4x3 +314 =1 (3.1 +S12 +.3+214 =5 611 S12 " +.3+.4=5 4++3+.4=2 113 (2n1 3r2 " 3r3 314 4+212 + 2x3 214 F 4 211 2x2 " 2*3 +314 =5 211 + 3r2 +4x3 +414 =7 21 + 2x2 +3 + 214 =4 311 Ax 5X3 614 = 9 211 212 +3+.4=3 ++X3 =] 12. 41 2x2 13 2x4 F1 T1 +2r2 2r; +4=0 41 +412 2r3 +44=2 211 3x2 Ax; 214 = 0 211 6.2 43 + 3x4 F5 3. + 4r2 513 +314 =0 21 +Sr2 + 213 +214 =3 211+2+,3+44=2 13 4 + 2r2 " ~14 =0 211 + 2x2 +3+4=3 21 + X2 +3x3 14=3 211 + 2r2 7 2x, - +.4 =3 4 + 2r2 ` +3r3 ~4 =0 211 212 2x3 +314 =1 21, + 2x2 7 +S.3 ~4=1



Answers

For each of the systems of equations that follow, use Gaussian elimination to obtain an equivalent system whose coefficient matrix is in row echelon form. Indicate whether the system is consistent. If the system is consistent and involves no free variables, use back substitution to find the unique solution. If the system is consistent and there are free variables, transform it to reduced row echelon form and find all solutions. $$ \begin{array}{ll}{\text { (a) } x_{1}-2 x_{2}=3} & {\text { (b) } 2 x_{1}-3 x_{2}=5} \\ {2 x_{1}-x_{2}=9} & {-4 x_{1}+6 x_{2}=8}\end{array} $$ $$ \begin{array}{rlrl} \text { (c) } {x_{1}+x_{2}} & {=0} & {\text { (d) } 3 x_{1}+2 x_{2}-x_{3}} & {=4} \\ {2 x_{1}+3 x_{2}} & {=0} & {x_{1}-2 x_{2}+2 x_{3}} & {=1} \\ {3 x_{1}-2 x_{2}} & {=0} & {11 x_{1}+2 x_{2}+x_{3}} & {=14}\end{array} $$ $$ \begin{aligned} \text { (e) } 2 x_{1}+3 x_{2}+x_{3} &=1 \\ x_{1}+x_{2}+x_{3} &=3 \\ 3 x_{1}+4 x_{2}+2 x_{3} &=4 \end{aligned} $$ $$ \begin{aligned} \text { (f) } x_{1}-x_{2}+2 x_{3} &=4 \\ 2 x_{1}+3 x_{2}-x_{3} &=1 \\ 7 x_{1}+3 x_{2}+4 x_{3} &=7 \end{aligned} $$ $$
\begin{array}{r} \text { (g) }{x_{1}+x_{2}+x_{3}+x_{4}=0} \\ {2 x_{1}+3 x_{2}-x_{3}-x_{4}=2} \\ {3 x_{1}+2 x_{2}+x_{3}+x_{4}=5} \\ {3 x_{1}+6 x_{2}-x_{3}-x_{4}=4}\end{array} $$ $$
\begin{aligned} \text { (h) } x_{1}-2 x_{2} &=3 \\ 2 x_{1}+x_{2} &=1 \\-5 x_{1}+8 x_{2} &=4 \end{aligned} $$ $$ \begin{aligned} \text { (i) }-x_{1}+2 x_{2}-x_{3}=& 2 \\-2 x_{1}+2 x_{2}+x_{3}=& 4 \\ 3 x_{1}+2 x_{2}+2 x_{3}=& 5 \\-3 x_{1}+8 x_{2}+5 x_{3}=& 17 \end{aligned} $$ $$ \begin{aligned} \text { (j) } & x_{1}+2 x_{2}-3 x_{3}+x_{4}=1 \\ &-x_{1}-x_{2}+4 x_{3}-x_{4}=6 \\ &-2 x_{1}-4 x_{2}+7 x_{3}-x_{4}=1 \end{aligned} $$ $$ \begin{array}{l}{\text { (k) } x_{1}+3 x_{2}+x_{3}+x_{4}=3} \\ {\quad 2 x_{1}-2 x_{2}+x_{3}+2 x_{4}=8} \\ {\quad x_{1}-5 x_{2} \quad+\quad x_{4}=5}\end{array} $$ $$
\begin{aligned} \text { (l) } x_{1}-3 x_{2}+x_{3} &=1 \\ 2 x_{1}+x_{2}-x_{3} &=2 \\ x_{1}+4 x_{2}-2 x_{3} &=1 \\ 5 x_{1}-8 x_{2}+2 x_{3} &=5 \end{aligned} $$

In this problem of determinants solution of linear system, we have given that the determinant of a three by three matrix, this is a three by three matrix they didn't mind over three with three medals can also be found using the method of diagonals. So in diagonals we have to write in the form of like this. So this is the metrics. See a 11 A 2, 1 and 831. That's what we a This will be here 1, 2, a 22 and a 32 and this will be a 13 that's what we A 2, 3, and that's what we ate 3, 3. So what we have to do is Mr1, we have to rewrite column one and 2 automatic. Say to the right of metrics, so that's what we hear A 11 column, a 21 and a 31 And now this will be second column that they want to 8 2, 2 and a 32 enough. What we have to do is identify the diagonals. Divan Through the six. So that's what we like this. This will be the first to diagonal. This will be second diagonal. So to say this is the one this would be needed to. And now this would be literally similarly here in this portion, this will be the fourth diagonal. So that's what we're here for. Two diagonals. This will be 50 diagonal. And now this would be the 6th diagonals. And now we have identified it. So here are six and a little and now we have to find these summer products D. One, D. Two, D. Three. So D. One D. Two D. Three. So this will be Divan plus D. Two plus D. Three. That means even +18228337 to 8 to 2 to +3831 like this. And then the step four subtract the some of their products, The 45 basics from the some. So we have to subtract the submission of the food. D. five and d. six From this. d. one d. 2 d. three. So from this we can see that he had each day is a project. So this will be like this 811 8 to 2 833 plus a 1 to 8 to three and +831 plus a 13 A 21 and 832 So this is the one day 23 minus. This will be the value of determinant. A. This will give minus before. So therefore would be a 3182273. So this would be 8318 or two and even three Bliss A 3-8-3. Even one 83 to 8 to three and a 11 plus. This will be a 338 to 1 and a 12 So that's what we ate quickly. E to one and a 12 So when we see it this is the same formula that we have discussed earlier to find the value of determinant. A. So this is the same. So we can say this is the Same as the previous one. So we can say that we can also find the value of determinant by using the diagnosis method.

Veloute go, Jordan Elimination method here. Now it's Devin. First, we will rate augmented metrics. We get 1111 four door one minus two minus one. Jiro one minus toe, minus one minus toe minus two 32134 Now, Steptoe, we will compute minus off to our one plus r two, then minus off our one plus r three and then minus off. Three R one plus R four to clear the first column. V get one 111 Full. Jiro minus one minus four minus three. Minus it. Geo minus three, minus two minus three. Minus six. Deal minus one minus toe. Jiro minus eight Now step three, you will compute minus off our too, then minus off. Our two plus are one three r two plus r three and then are two plus are for to clear the second column. Then we get one. Jiro minus three, minus toe minus for Jill. 1438 Gentle, gentle. 10 6 It Dean gentle, gentle three. Jiro Now it's their food David City tard and fourth Rose. So it will be one geo minus three, minus two minus four. Joe 1438 Gero Gero Toe three. Jiro and Gero Gero 10 6 and 18 No, it's their fight. We will compute half off our three. Then three are three plus are one then minus for our three plus arto. And then my understand, are three plus our for two. Clear that third column, then we get one. Jiro Dio five Divided by three minus four Joe one Jiro minus 38 Gentle, gentle 13 divided by toe Jiro and Jiro Jiro Jiro minus 9 18 Now Step six Villaluz minus one. Divided my name off our for then minus off five divided by two off our four plus are one, then three Our four plus arto and then minus off three divided by two are four plus R three to clear the fourth column so it will be one Gero Gero Gero Gero Gero Gero Gero Gero Gero gero Gero gero one 123 and minus two w equal to one X equal Toto vehicle to three and Jellicle do. Minus toe is our final answer

Be real huge. Go, Jorden! Elimination method here. Now it's Devlin. We will frustrate augmented medics. So it will be three minus for 119 11 minus one minus one gentle. So one four minus toe. Three minus one. So one minus tree tree. Now Steptoe and a steptoe virile reorder. That was hence. It will be 11 minus one minus one. Joe. So one for minus toe. Three three minus 4119 minus 1 to 1 minus 33 No. Step three. In a step three, we will. Huge minus off to urban plus arto minus off. Three R one plus R three and r one plus R four now figured 11 minus one minus one. Joe Geo minus 16 Jiro three Jiro minus 7449 Joe three Jiro minus for three Now step for here. We will compute minus off our toe, then minus off. Our two plus are want seven are toe plus R three and minus off. Three are do plus are for to clear the second column We get one jo five minus one. Joe gentle one minus six. Geo minus three. Jiro Jiro minus 38 for minus 12. Jiro Jiro 18 minus for Well, no, it's their fight initiative five valued operation are three plus our food in tarred rope. Big one Jo five minus one. Joe Gee, it'll one minus six geo minus three. Gentle, gentle minus 20. Geo geo, Jiro Jiro 18 minus four. And well, no. Step six in this village, minus off our three divided by 20 then minus or five r three plus are one, six are three plus are too and minus of 18 are three plus are full. Then we get one Gero Gero minus one. Joe gentle One gentle, gentle minus three. Gero Gero one Gero Gero Gero Gero Gero minus four and well, now Step seven. We will compute minus often by four off our four, then are four. Blessed are bond to clear the fourth column we get one Gero Gero Gero Gero Gero Gero Gero minus three Gero gero Gero Gero Gero Gero Gero one minus three Soul The blue It pulled Kojiro X equal to minus tree by equal to zero and Jellicle two minus three is ever final answer

We have been given that the determinant of three by three matrix A is defined as follows. And then we have been clearly, we can see that in the given question. The determinant has been expanded and open and now what we have announced this, we need to verify the if the method of the diagonal rule identify the diagnosed even and multiplying them. Okay, so we need to verify the method using the method of diagnosis character. Not with general if it matches the general general solution of a determinant. So first we can just clearly write it down and I just write it. So what we have been given is A is even one even to a 13 and then you do one, a two, 8 to 3 and then a 31 832 and a 33 Now on the right hand side of this matrix all the determinant. We can just write the first renders, you can come bigger part and second and the third column, so you want to 8 to 2 and then it really and then even three. Thank you too. And a 32 oh Ernest. We need to actually reread the 1st and 2nd columns on the right hand side of the determinant. So now uh from the figure we can clearly see that uh to get the first diagnosed even it is, it would remember deploying these three elements and then I will be changing the color so that it would be easy then for the second diagnosed. D. Two. So the second diagnosed can be obtained by multiplying these three elements and then the third diagnosed can be octane. Bye! Multiplying these three elements which I am crossing by the blue arrow and then to obtain the fourth diagnosed and just change them ahead. Then to obtain the fourth diagnosed, we need to my to play these three elements. And then for the first diagnosed we need to multiply these three elements. And then finally for obtaining the sixth diagonal G. Six. We need to multiply these three elements. Okay I think I didn't change the color but I just mentioned it. This is 45 this is for the six. We just need to multiply these three elements and then uh we can just find even day to day three and subsequently to the values of the six. We just simply need to add the even PSD 2 23 and then subtract Divan D. Four D. Five and D. Six at the addition of the four different A. Six. And then we can clearly obtain the values as similar has given in the given question which is even even 1822833 plus even 2823831 plus even 3821832 subtracted by a 31822 even three plus a 3 to 8 to three, even one plus 8338 to 1 even to We will get the exact same answer we originally. So this is how we solve the question. Thank you, friends. I hope you like the video.


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