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,
![A = \left[\begin{array}{ccc}1&3&2\\-2&1&-1\\2&0&3\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%262%5C%5C-2%261%26-1%5C%5C2%260%263%5Cend%7Barray%7D%5Cright%5D)
![B =\left[\begin{array}{ccc}5\\-2\\11\end{array}\right]](https://tex.z-dn.net/?f=B%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C-2%5C%5C11%5Cend%7Barray%7D%5Cright%5D)
![X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]](https://tex.z-dn.net/?f=X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D)
i.e AX=B
We can write as augmented matrix
![\left[\begin{array}{ccc|c}1&3&2&5\\-2&1&-1&-2\\2&0&3&11\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%262%265%5C%5C-2%261%26-1%26-2%5C%5C2%260%263%2611%5Cend%7Barray%7D%5Cright%5D)
![\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]](https://tex.z-dn.net/?f=%5Cfrac%7BR_2%5Crightarrow%20R_2%2B2R_1%7D%7BR_3%5Crightarrow%20R_3-2R_1%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%262%265%5C%5C0%267%263%268%5C%5C0%26-6%26-1%261%5Cend%7Barray%7D%5Cright%5D)
![\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]](https://tex.z-dn.net/?f=%5Cfrac%7BR_3%5Crightarrow%20R_3%2B%5Cfrac%7B6R_2%7D%7B7%7D%20%7D%7BR_1%5Crightarrow%20R_1-%5Cfrac%7B3R_2%7D%7B7%7D%20%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%265%2F7%2611%2F7%5C%5C0%267%263%268%5C%5C0%260%2611%2F7%2655%2F7%5Cend%7Barray%7D%5Cright%5D)
![\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]](https://tex.z-dn.net/?f=%5Cfrac%7BR_2%5Crightarrow%5Cfrac%7BR_2%7D%7B7%7D%7D%7BR_3%5Crightarrow%5Cfrac%7B7%7D%7B11%7DR_3%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%265%2F7%2611%2F7%5C%5C0%261%263%2F7%268%2F7%5C%5C0%260%2611%2F7%2655%2F7%5Cend%7Barray%7D%5Cright%5D)
![\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]](https://tex.z-dn.net/?f=%5Cfrac%7BR_1%5Crightarrow%20R_1%20-%5Cfrac%7B5%7D%7B7%7DR_3%7D%7BR_2%5Crightarrow%20R_2%20-%5Cfrac%7B3%7D%7B7%7DR_3%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%26-2%5C%5C0%261%260%26-1%5C%5C0%260%261%265%5Cend%7Barray%7D%5Cright%5D)
Since Rank (A|B) = Rank (A) = 3 = number of variables
<h3>⇒ systems has unique solution and </h3><h3>x = -2 , y = -1 , z = 5</h3>