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Bess [88]
3 years ago
12

The system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the ge

neral solution of the system, if a solution exists. (If there is no solution, enter NO SOLUTION. If there are infinitely many solutions, express your answers in terms of z as in Example 3.) 5x + 4y + 5z = −1 x + y + 2z = 1 2x + y − z = −3
Mathematics
1 answer:
Leno4ka [110]3 years ago
4 0

Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

5x+4y+5z=-1

x+y+2z=1

2x+y-z=-3

Firs we find determinant of system of equations

Let a matrix A=\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right] and B=\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]

\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}

\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0

Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply R_1\rightarrow R_1-4R_2 and R_3\rightarrow R_3-2R_2

\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]:\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]

ApplyR_2\rightarrow R_2-R_1

\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]:\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]

Apply R_3\rightarrow R_3+R_2

\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]:\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]

Apply R_3\rightarrow- \frac{1}{2} and R_2\rightarrow R_2-R_3

\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]:\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]

Apply R_1\rightarrow R_1-R_3

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]:\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

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