Consider the following system of linear equations: Instructions: Solve the system by reducing its augmented matrix to reduced row echelon form (RREF). Yes, you must reduce it all the way to RREF. Write out the matrix at each step of the procedure, and be specific as to what row operations you use in each step. At the end of the procedure, clearly state the solution to the system outside of a matrix. 1. If the solution is unique, express the solution in real numbers. 3. If there are infinitely many solutions, express the solution in parameter(s). 3. If there is no solution, say so, and explain why.All of the following are possible ranks of a 4x3 matrix except:0123 4 How is the number of parameters in the general solution of a consistent linear system related to the rank of its coefficient matrix? Let r= number of rows in the coefficient matrix c= number of columns in the coefficient matrix p= number of parameters in the general solution R=rank of the coefficient matrix 1. R=p+r 2. R=C+p 3. R=r-p 4. R=C-p 5. R=p-

Question

06-06-2020
Mathematics

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Consider the following system of linear equations: Instructions: Solve the system by reducing its augmented matrix to reduced row echelon form (RREF). Yes, you must reduce it all the way to RREF. Write out the matrix at each step of the procedure, and be specific as to what row operations you use in each step. At the end of the procedure, clearly state the solution to the system outside of a matrix. 1. If the solution is unique, express the solution in real numbers. 3. If there are infinitely many solutions, express the solution in parameter(s). 3. If there is no solution, say so, and explain why.All of the following are possible ranks of a 4x3 matrix except:0123 4 How is the number of parameters in the general solution of a consistent linear system related to the rank of its coefficient matrix? Let r= number of rows in the coefficient matrix c= number of columns in the coefficient matrix p= number of parameters in the general solution R=rank of the coefficient matrix 1. R=p+r 2. R=C+p 3. R=r-p 4. R=C-p 5. R=p-

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shallomisaiah19 shallomisaiah19 · Answer 1 · 2020-06-08T06:21:02+02:00

Answer:

Consider the following system of linear equations: 2 + 3y + 2z = 5 - 2x + y - z= -2 2x + 3z = 11 Instructions: • Solve the system by reducing its augmented matrix to reduced row echelon form (RREF). Yes, you must reduce it all the way to RREF. • Write out the matrix at each step of the procedure, and be specific as to what row operations you use in each step. • At the end of the procedure, clearly state the solution to the system outside of a matrix. • If the solution is unique, express the solution in real numbers. • If there are infinitely many solutions, express the solution in parameter(s). . If there is no solution, say so, and explain why.

All of the following are possible ranks of a 4x3 matrix EXCEPT O 1 2 3 4

How is the number of parameters in the general solution of a consistent linear system related to the rank of its coefficient matrix? Let r= number of rows in the coefficient matrix c= number of columns in the coefficient matrix p= number of parameters in the general solution R=rank of the coefficient matrix 1. R=p+r 2. R=C+p 3. R=r-p 4. R=C-p 5. R=p-r

Step-by-step explanation:

x + 3y +2z = 5

-2x + y - z = -2

2x + 3z = 11

Here,

[tex]A = \left[\begin{array}{ccc}1&3&2\\-2&1&-1\\2&0&3\end{array}\right][/tex]

[tex]B =\left[\begin{array}{ccc}5\\-2\\11\end{array}\right][/tex]

[tex]X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex]

i.e AX=B

We can write as augmented matrix

[tex]\left[\begin{array}{ccc|c}1&3&2&5\\-2&1&-1&-2\\2&0&3&11\end{array}\right][/tex]

[tex]\frac{R_2\rightarrow R_2+2R_1}{R_3\rightarrow R_3-2R_1} \left[\begin{array}{ccc|c}1&3&2&5\\0&7&3&8\\0&-6&-1&1\end{array}\right][/tex]

[tex]\frac{R_3\rightarrow R_3+\frac{6R_2}{7} }{R_1\rightarrow R_1-\frac{3R_2}{7} } \left[\begin{array}{ccc|c}1&0&5/7&11/7\\0&7&3&8\\0&0&11/7&55/7\end{array}\right][/tex]

[tex]\frac{R_2\rightarrow\frac{R_2}{7}}{R_3\rightarrow\frac{7}{11}R_3} \left[\begin{array}{ccc|c}1&0&5/7&11/7\\0&1&3/7&8/7\\0&0&11/7&55/7\end{array}\right][/tex]

[tex]\frac{R_1\rightarrow R_1 -\frac{5}{7}R_3}{R_2\rightarrow R_2 -\frac{3}{7}R_3} \left[\begin{array}{ccc|c}1&0&0&-2\\0&1&0&-1\\0&0&1&5\end{array}\right][/tex]

Since Rank (A|B) = Rank (A) = 3 = number of variables

Respuesta :

⇒ systems has unique solution and

x = -2 , y = -1 , z = 5

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