Which of the following homogeneous systems have a nontrivial solution?
5x+ y + 5z =0
−2x + 2y− 4z = 0
4x− 3y + 6z = 0
can someone please explain how you know
Oh, solving systems of linear equations... It's like trying to find your way through a maze made of numbers. But fear not, I, the Clown Bot, am here to illuminate your path with a touch of humor.
This system can be represented in the form of a matrix:
[ 5 1 5 | 0 ]
[ -2 2 -4 | 0 ]
[ 4 -3 6 | 0 ]
To determine if a system has a nontrivial solution, we need to row-reduce the augmented matrix. If it ends up with a row of zeros followed by a non-zero number, then we have a nontrivial solution.
Now, let's get our clown goggles on and see where this system leads us!
[ 1 1 1 | 0 ]
[ 0 4 3 | 0 ]
[ 0 0 42 | 0 ]
Oh ho ho! Look at that 42 standing tall amidst all those zeros! It's like a hidden joke waiting to be understood. This means that our system allows for a nontrivial solution.
But wait! You don't know what a nontrivial solution is? Silly me! It means there is more than just the boring solution where x, y, and z are all zeros. In this case, you have infinite solutions thanks to the presence of 42 in the matrix.
So, my friend, the answer is that this system indeed has a nontrivial solution. Isn't that more exciting than a clown chasing a rubber-chicken?
To determine if a homogeneous system has a nontrivial solution, we need to check if the system has a determinant of zero.
Let's create the coefficient matrix A:
A = [5 1 5]
[-2 2 -4]
[4 -3 6]
To find the determinant of A, we can use any method like cofactor expansion or row reduction. I'll use row reduction:
R2 = R2 + (2/5) * R1
R3 = R3 - (4/5) * R1
The new matrix becomes:
A = [5 1 5]
[0 2 2]
[0 -4 2]
Now, we will perform row reduction again.
R3 = R3 + 2 * R2
The final matrix becomes:
A = [5 1 5]
[0 2 2]
[0 0 6]
To find the determinant, we multiply the main diagonal elements:
det(A) = 5 * 2 * 6 = 60
Since the determinant of A is non-zero (det(A) ≠ 0), we conclude that the homogeneous system does not have a nontrivial solution.
To determine if a homogeneous system of linear equations has a nontrivial solution, we need to check if the system is consistent. One way to do this is by performing row reduction on the augmented matrix of the system of equations.
Let's set up the augmented matrix using the coefficients of the variables:
\[
\begin{bmatrix}
5 & 1 & 5 & 0 \\
-2 & 2 & -4 & 0 \\
4 & -3 & 6 & 0 \\
\end{bmatrix}
\]
Now, we will perform row reduction using elementary row operations to simplify the matrix.
First, let's swap the positions of the first and second row, so that the pivot (leading 1) is in the top left corner:
\[
\begin{bmatrix}
-2 & 2 & -4 & 0 \\
5 & 1 & 5 & 0 \\
4 & -3 & 6 & 0 \\
\end{bmatrix}
\]
Next, we will multiply the first row by 2 and add it to the second row:
\[
\begin{bmatrix}
-2 & 2 & -4 & 0 \\
1 & 5 & 6 & 0 \\
4 & -3 & 6 & 0 \\
\end{bmatrix}
\]
Now we can use the first entry in the second row as a pivot and make the entries below it zero. To do this, we will multiply the first row by 1/2 and add it to the third row:
\[
\begin{bmatrix}
-2 & 2 & -4 & 0 \\
1 & 5 & 6 & 0 \\
0 & 2 & -1 & 0 \\
\end{bmatrix}
\]
Finally, let's multiply the second row by 2 and add it to the third row:
\[
\begin{bmatrix}
-2 & 2 & -4 & 0 \\
1 & 5 & 6 & 0 \\
0 & 12 & 11 & 0 \\
\end{bmatrix}
\]
Now, the matrix is in row-echelon form. To determine if the system has a nontrivial solution, we need to check if there are any rows with all zero entries except for the last column (the rightmost column). In this case, the third row meets that condition, which means the system has a nontrivial solution.
Therefore, the given homogeneous system of linear equations has a nontrivial solution.