Question

Given \(A\) and \(b\) in Exercises 11 and 12, write the augmented matrix for the linear system that corresponds to the matrix equation \(Ax = b\). Then solve the system and write the solution as a vector.

12. \({\mathop{\rm A}\nolimits} = \left[ {\begin{array}{*{20}{c}}1&2&1\\{ - 3}&{ - 1}&2\\0&5&3\end{array}} \right],b = \left[ {\begin{array}{*{20}{c}}0\\1\\{ - 1}\end{array}} \right]\)

Short answer

The solution as a vector is \(x = \left[ {\begin{array}{*{20}{c}}{\frac{3}{5}}\\{ - \frac{4}{5}}\\1\end{array}} \right]\).

Step-by-step solution

Writing the matrix in the augmented form

The augmented form of the matrix is

\(\left[ {\begin{array}{*{20}{c}}1&2&1&0\\{ - 3}&{ - 1}&2&1\\0&5&3&{ - 1}\end{array}} \right]\).

Applying the row operation

Perform an elementary row operationto produce the first augmented matrix.

The sum of 3 times row one and row twois written in row two.

\(\left[ {\begin{array}{*{20}{c}}1&2&1&0\\0&5&5&1\\0&5&3&{ - 1}\end{array}} \right]\)

Applying the row operation

Perform an elementary row operationto produce the second augmented matrix.

Multiply row two by \(\frac{1}{5}\).

\(\left[ {\begin{array}{*{20}{c}}1&2&1&0\\0&1&1&{\frac{1}{5}}\\0&5&3&{ - 1}\end{array}} \right]\)

Applying the row operation

Perform an elementary row operationto produce the third augmented matrix.

The sum of \( - 2\) times row two and row one is written in row one.

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&{ - \frac{2}{5}}\\0&1&1&{\frac{1}{5}}\\0&5&3&{ - 1}\end{array}} \right]\)

Applying the row operation

Perform an elementary row operationto produce the fourth augmented matrix.

The sum of \( - 5\)times row two and row three is written in row three.

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&{ - \frac{2}{5}}\\0&1&1&{\frac{1}{5}}\\0&0&{ - 2}&{ - 2}\end{array}} \right]\)

Applying the row operation

Perform an elementary row operationto produce the fifth augmented matrix.

Multiply row three by \( - \frac{1}{2}\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&{ - \frac{2}{5}}\\0&1&1&{\frac{1}{5}}\\0&0&1&1\end{array}} \right]\)

Applying the row operation

Perform an elementary row operationto produce the sixth augmented matrix.

The sum of row one and row three are written in row one.

\(\left[ {\begin{array}{*{20}{c}}1&0&0&{\frac{3}{5}}\\0&1&1&{\frac{1}{5}}\\0&0&1&1\end{array}} \right]\)

The sum of \( - 1\) times row three and row twois written in row two.

\(\left[ {\begin{array}{*{20}{c}}1&0&0&{\frac{3}{5}}\\0&1&0&{ - \frac{4}{5}}\\0&0&1&1\end{array}} \right]\)

Converting matrix into the equation form

To obtain the solution of the vector equation, to convert the augmented matrix into vector equations.

Write the obtained matrix \(\left[ {\begin{array}{*{20}{c}}1&0&0&{\frac{3}{5}}\\0&1&0&{ - \frac{4}{5}}\\0&0&1&1\end{array}} \right]\)in the equation notation.

\(\begin{array}{l}{x_1} = \frac{3}{5}\\{x_2} = - \frac{4}{5}\\{x_3} = 1\end{array}\)

Writing the solution of the system as a vector

The solution of the system as a vector is shown below.

\(\begin{array}{c}x = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{\frac{3}{5}}\\{ - \frac{4}{5}}\\1\end{array}} \right]\end{array}\)

Thus, the solution as a vector is \(x = \left[ {\begin{array}{*{20}{c}}{\frac{3}{5}}\\{ - \frac{4}{5}}\\1\end{array}} \right]\).