# Matrix solutions with variable coefficient

• MHB
• TheFallen018
In summary, the conversation discusses finding the RREF of a matrix and determining the values of a that result in a unique solution, no solution, and infinitely many solutions. The RREF is found by subtracting 3 times the first row from the second row, resulting in a new matrix with a+3 in the second row. When a is not equal to -3, the RREF is $\begin{bmatrix}1 &0 &|1 \\ 0 &1 &|0 \end{bmatrix}$, indicating a unique solution. However, when a equals -3, the RREF is $\begin{bmatrix}1 &-1 &|1 \\ 0 &0 &|0 \end{b TheFallen018 Hey guys, I've got this question, that I think I have figured out, but I'm not completely sure. Basically, I've got the following matrix to put into RREF. Not really a problem, but I'm not sure if I'm handling the variable incorrectly.$\begin{bmatrix}
1 &-1 &|1 \\
3 &a &|3
\end{bmatrix}$So, I subtracted 3*row1 from row2, which resulted in this$\begin{bmatrix}
1 &-1 &|1 \\
0 &a+3 &|0
\end{bmatrix}$The next thing I did is where I'm not sure about. I divided row2 by (a+3), and then added the resulting row into row 1 to get the matrix in RREF.$\begin{bmatrix}
1 &0 &|1 \\
0 &1 &|0
\end{bmatrix}$The remaining parts of the question ask me to find values of a where there is a unique solution, no solutions, and infinitely many solutions. My thoughts were that there's no value a can take that will make the bottom row inconsistent, so there's no value where there's no solution. There can be infinitely many solutions if we go back to the part where we still have (a+3) in the second row. If we make a equal to -3, then$x_{2}$becomes a free variable and gives us infinitely many solutions. If a is any other value, then there is a unique solution. Does that seem right? Thanks TheFallen018 said: Hey guys, I've got this question, that I think I have figured out, but I'm not completely sure. Basically, I've got the following matrix to put into RREF. Not really a problem, but I'm not sure if I'm handling the variable incorrectly.$\begin{bmatrix}
1 &-1 &|1 \\
3 &a &|3
\end{bmatrix}$So, I subtracted 3*row1 from row2, which resulted in this$\begin{bmatrix}
1 &-1 &|1 \\
0 &a+3 &|0
\end{bmatrix}$The next thing I did is where I'm not sure about. I divided row2 by (a+3), and then added the resulting row into row 1 to get the matrix in RREF.$\begin{bmatrix}
1 &0 &|1 \\
0 &1 &|0
\end{bmatrix}$The remaining parts of the question ask me to find values of a where there is a unique solution, no solutions, and infinitely many solutions. My thoughts were that there's no value a can take that will make the bottom row inconsistent, so there's no value where there's no solution. There can be infinitely many solutions if we go back to the part where we still have (a+3) in the second row. If we make a equal to -3, then$x_{2}$becomes a free variable and gives us infinitely many solutions. If a is any other value, then there is a unique solution. Does that seem right? Thanks Your conclusions are correct (solution is unique except when$a=-3$, when there are infinitely many solutions). But the RREF is only equal to$\begin{bmatrix}1 &0 &|1 \\ 0 &1 &|0 \end{bmatrix}$when$a\ne-3$. In fact, if$a=-3$then you cannot divide by$a+3$, so the final step of your calculation of the RREF is invalid in that case. If$a=-3$then the previous step in the calculation becomes$\begin{bmatrix}1 &-1 &|1 \\ 0 &0 &|0 \end{bmatrix}$, and that is the RREF when$a=-3\$.

## 1. What is a matrix solution with variable coefficient?

A matrix solution with variable coefficient is a set of values that satisfy a system of equations with variables represented as a matrix. These solutions can be expressed as a single matrix or as a set of matrices.

## 2. How is a matrix solution with variable coefficient represented?

A matrix solution with variable coefficient can be represented by using the coefficient matrix, variable matrix, and constant matrix. The coefficient matrix contains the numerical coefficients of the variables, the variable matrix contains the variables themselves, and the constant matrix contains the constants in the system of equations.

## 3. What types of problems can be solved using matrix solutions with variable coefficient?

Matrix solutions with variable coefficient can be used to solve systems of linear equations, simultaneous equations, and systems of equations with complex numbers. They can also be applied to solve problems in economics, physics, and engineering.

## 4. How do you find the matrix solution with variable coefficient?

To find the matrix solution with variable coefficient, you can use various methods such as Gaussian elimination, Cramer's rule, or matrix inversion. These methods involve manipulating the coefficient, variable, and constant matrices to arrive at a solution.

## 5. What are the benefits of using matrix solutions with variable coefficient?

Matrix solutions with variable coefficient offer a more efficient and organized way to solve systems of equations with multiple variables. They also allow for quick and accurate solutions to complex problems, making them useful in various fields of study and research.

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