# Matrices- conditions for unique and no solution

• MHB
• lyd123
In summary, the conversation discusses the process for solving a system of equations and determining if it has a unique solution or no solution. The conversation also covers the different scenarios that may arise and how to handle them. The key steps include subtracting rows and solving for variables. Ultimately, the conclusion is that when b is not equal to 2 and a is not equal to 0, there is a unique solution, whereas if b is not equal to 2 and a is equal to 0, there is no solution. If b is equal to 2 and a is not equal to 0, there are infinitely many solutions, and if both b and a are equal to 0, there is no solution.
lyd123
Hi,how do I go about answering the attached question? I know that for a matrix to have no solution, there needs to be a contradiction in some row. Unique solutions is when m* ${x}_{3}$ =c , where m* ${x}_{3}$ $\ne$ 0.

One way I tried was if a=0,
then from row (1) : b* ${x}_{3}$ =2
${x}_{3}$= 2/b (from row 2 : 2/b = 1 , so b=2)

then from row (2) : 4* ${x}_{3}$ =4
${x}_{3}$= 1

then from row (3) : 2* ${x}_{3}$ =b
${x}_{3}$= b/2 (from row 2 : b/2 = 1 , so b=2)

This seems to work, so when a=0, and b =1 , you have unique solutions?

How do I answer this question? Thanks.View attachment 8733

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I see no reason not to just go ahead and try to solve the system. Since it is already in "augmented matrix form", row reduce the matrix. Start by subtracting the first row from the second row to get $$\begin{bmatrix}a & 0 & b & | & 2 \\ 0 & a & 4- b & | & 2 \\ 0 & a & 2 & | & b \end{bmatrix}$$.

Now subtract the second row from the third to get $$\begin{bmatrix}a & 0 & b & | & 2 \\ 0 & a & 4- b & | & 2 \\ 0 & 0 & b- 2 & | & b- 2\end{bmatrix}$$.

We now have the equations (b- 2)z= b- 2, ay+ (4- b)z= 2, and ax+ bz= 2. From the first equation we can say if $$b\ne 2$$ then z= 1. In that case the second equation becomes ay+ 4- b= 2 so ay= b- 2 and, if $$a\ne 0$$, $$y= \frac{b- 2}{a}$$. In that case the third equation becomes ax+ b= 2 so again, if $$a\ne 0$$, $$z= \frac{2- b}{a}$$. If $$b\ne 2$$ and $$a\ne 0$$ there is a unique solution.

Now, what if, in the case that $$b\ne 2$$, a is 0? In that case, the second equation becomes b- 2= 0 which, since this is the case $$b\ne 0$$ is impossible. In the case that $$b\ne 2$$, a= 0, there is no solution.

What if b= 2? In that case the first equation is satisfied for z any number. If a is not 0, we have $$y= \frac{2- (4- b)z}{a}$$ and $$x= \frac{2- bz}{a}$$ for all values of z so there are infinitely many solutions.

Last, what if b= 2 and a= 0? The equations reduce to 0= , 0= 2, and 2z= 2. 0= 2 is false no matter what x, y, and z are so there is no solution.

Makes perfect sense to me now!

## 1. What are the conditions for a matrix to have a unique solution?

The conditions for a matrix to have a unique solution are that the number of variables (unknowns) must be equal to the number of equations, and the determinant of the matrix must be non-zero. In other words, the matrix must be square and its determinant must not equal zero.

## 2. What does it mean if a matrix has no solution?

If a matrix has no solution, it means that there is no set of values for the variables that can satisfy all of the equations in the system. In other words, the system of equations is inconsistent and there is no point where all of the equations intersect.

## 3. How can I determine if a matrix has a unique solution or no solution?

You can determine if a matrix has a unique solution or no solution by using Gaussian elimination or row reduction to put the matrix in reduced row echelon form. If the resulting matrix has a row of all zeros and the last column is non-zero, then the system has no solution. If the resulting matrix has a row of all zeros but the last column is also zero, then the system has infinitely many solutions. If the resulting matrix has no rows of all zeros, then the system has a unique solution.

## 4. Can a matrix have more than one solution?

Yes, a matrix can have infinitely many solutions. This occurs when the system of equations is consistent (meaning there is at least one solution) and there are more variables than equations. In this case, there are infinitely many combinations of values for the variables that can satisfy the equations.

## 5. What is the significance of the determinant in determining unique and no solutions?

The determinant of a matrix plays a crucial role in determining the number of solutions a system of equations has. If the determinant is non-zero, then the matrix has a unique solution. If the determinant is zero, then the matrix has either no solution or infinitely many solutions, depending on the consistency of the system. The determinant can also give us information about the linear independence of the equations in the system.

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