Matrix Equations and Linear Equations: Exercises

Problem 1

Express the following system of linear equations as a matrix equation.

\[ \begin{array}{rcrcrcrcr} 4u & + & x & - & 2y & + & 10z & = & 4 \\ -2u & + & 3x & -& 4y & + & 5z & = & 1 \\ u & - & x & + & 3y & - & 5z & = & -6 \end{array} \]

Answer

Problem 2

Express the systems of linear equations below as matrix equations. Then find the solutions by using the inverse of the coefficient matrix.

\[ \begin{array}{rcrcrcr} x & - & 2y & + & 2z & = & 1 \\ 2x & - & 3y & + & 6z & = & 4 \\ x & + & y & + & 7z & = & -2 \end{array} \quad \text{and} \quad \begin{array}{rcrcrcr} x & - & 2y & + & 2z & = & 3 \\ 2x & - & 3y & + & 6z & = & 1 \\ x & + & y & + & 7z & = & 1 \end{array} \]

Solution

Problem 3

Solve the matrix equation \(\Mtrx{A}\Mtrx{X} = \Mtrx{B}\) if

\[ A = \left[ \begin{array}{rrr} 1 & 0 & 3 \\ 2 & -1 & 4 \\ -1 & 2 & 2 \end{array}\right], \quad B = \left[ \begin{array}{r} 1 \\ -2 \\ 1 \end{array}\right], \quad X = \left[ \begin{array}{r} x \\ y \\ z \end{array}\right] \]

Hint Answer

Solution for Problem 2

The matrix equations correspondig to the given system of linear equations are

\[ \left[ \begin{array}{rrr} 1 & -2 & 2 \\ 2 & -3 & 6 \\ 1 & 1 & 7 \end{array}\right]\ \begin{bmatrix} x \\ y \\ z \end{bmatrix}\ =\ \left[ \begin{array}{r} 1 \\ 4 \\ -2 \end{array}\right] \qquad \left[ \begin{array}{rrr} 1 & -2 & 2 \\ 2 & -3 & 6 \\ 1 & 1 & 7 \end{array}\right]\ \begin{bmatrix} x \\ y \\ z \end{bmatrix}\ =\ \left[ \begin{array}{r} 3 \\ 1 \\ 1 \end{array}\right] \]

The inverse of the common coefficient matrix \(\Mtrx{A}\) is

\[ A^{-1}\ =\ \left[ \begin{array}{rrr} 27 & -16 & 6 \\ 8 & -5 & 2 \\ -5 & 3 & -1 \end{array}\right] \]

Consequently, the unique solution of first system is linear equations is given by

\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix}\ =\ \left[ \begin{array}{rrr} 27 & -16 & 6 \\ 8 & -5 & 2 \\ -5 & 3 & -1 \end{array}\right]\, \left[\begin{array}{r} 1 \\ 4 \\ -2 \end{array}\right]\ =\ \left[\begin{array}{r} -49 \\ -16 \\ 9 \end{array}\right] \]

While the unique solution of the second system of linear equations is given by

\[ \begin{bmatrix} x \\ y \\ z \end{bmatrix}\ =\ \left[ \begin{array}{rrr} 27 & -16 & 6 \\ 8 & -5 & 2 \\ -5 & 3 & -1 \end{array}\right]\, \left[ \begin{array}{r} 3 \\ 1 \\ 1 \end{array}\right]\ =\ \left[ \begin{array}{r} 71 \\ 21 \\ -13 \end{array}\right] \]