Question

Solve the system of given equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\(\begin{cases}3a-b-4c=3\\2a-b+2c=-8\\a+2b-3c=9\end{cases}\)

Answer 1

Step 1
We need to convert it into Reduced echelon form :
First write the given equation in the form of augmented matrix:
\(\begin{bmatrix}3&-1&-4&|&3 \\2&-1&2&|&-8\\1&2&-3&|&9 \end{bmatrix}\)
Let's reduce it to row echelon form:
Step 2
Divide row1 by 3:
\(\begin{bmatrix}1&-\frac{1}{3}&-\frac{4}{3}&|&1 \\2&-1&2&|&-8\\1&2&-3&|&9 \end{bmatrix}\)
Now,
\(R_2=R_2-2R_1\)
\(\begin{bmatrix}1&-\frac{1}{3}&-\frac{4}{3}&|&1 \\0&-\frac{1}{3}&\frac{14}{3}&|&-10\\1&2&-3&|&9 \end{bmatrix}\)
Step 3
Now, \(R_3=R_3-R_1\)
\(\begin{bmatrix}1&-\frac{1}{3}&-\frac{4}{3}&|&1 \\0&-\frac{1}{3}&\frac{14}{3}&|&-10\\0&\frac{7}{3}&-\frac{5}{3}&|&8 \end{bmatrix}\)
Next,
\(R_1=R_1-R_2\)
\(\begin{bmatrix}1&0&-6&|&11 \\0&-\frac{1}{3}&\frac{14}{3}&|&-10\\0&\frac{7}{3}&-\frac{5}{3}&|&8 \end{bmatrix}\)
Step 4
Next,
\(R_3=R_3+7(R_2)\)
\(\begin{bmatrix}1&0&-6&|&11 \\0&-\frac{1}{3}&\frac{14}{3}&|&-10\\0&0&31&|&-62 \end{bmatrix}\)
Next, \(R_2=-3(R_2)\)
\(\begin{bmatrix}1&0&-6&|&11 \\0&1&-14&|&-30\\0&0&31&|&-62 \end{bmatrix}\)
Step 5
Further,
\(R_3=\frac{R_3}{31}\)
\(\begin{bmatrix}1&0&-6&|&11 \\0&1&-14&|&-30\\0&0&1&|&-\frac{62}{31} \end{bmatrix}\)
Next,
\(R_1=R_1+6(R_3)\)
\(\begin{bmatrix}1&0&0&|&-1 \\0&1&0&|&-58\\0&0&1&|&-\frac{62}{31} \end{bmatrix}\)
Hence,
We have reduced this matrix into echelon form and gauss elimination,
Now,
We get the values as:
a=-1 , b=-58 , c=-2