Solve the System of Linear Equations Using the Inverse Matrix of the Coefficient Matrix

Linear algebra problems and solutions

Problem 442

Consider the following system of linear equations
\begin{align*}
2x+3y+z&=-1\\
3x+3y+z&=1\\
2x+4y+z&=-2.
\end{align*}

(a) Find the coefficient matrix $A$ for this system.

(b) Find the inverse matrix of the coefficient matrix found in (a)

(c) Solve the system using the inverse matrix $A^{-1}$.

 
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Solution.

(a) Find the coefficient matrix $A$ for this system.

The system can be written as
\[A\mathbf{x}=\mathbf{b},\] where
\[A=\begin{bmatrix}
2 & 3 & 1 \\
3 &3 &1 \\
2 & 4 & 1
\end{bmatrix}\] is the coefficient matrix of the system and
\[\mathbf{x}=\begin{bmatrix}
x \\
y \\
z
\end{bmatrix} \text{ and } \mathbf{b}=\begin{bmatrix}
-1 \\
1 \\
-2
\end{bmatrix}.\]

 

(b) Find the inverse matrix of the coefficient matrix

We apply the elementary row operations to the augmented matrix $[A\mid I]$, where $I$ is the $3\times 3$ identity matrix.
\begin{align*}
&[A\mid I]= \left[\begin{array}{rrr|rrr}
2 & 3 & 1 & 1 &0 & 0 \\
3 & 3 & 1 & 0 & 1 & 0 \\
2 & 4 & 1 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_2-R_1\\R_3-R_1}}
\left[\begin{array}{rrr|rrr}
2 & 3 & 1 & 1 &0 & 0 \\
1 & 0 & 0 & -1 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 1 \\
\end{array} \right]\\[6pt] &\xrightarrow[\text{then } R_2 \leftrightarrow R_3]{R_1\leftrightarrow R_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & -1 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 1 \\
2 & 3 & 1 & 1 &0 & 0
\end{array} \right] \xrightarrow{R_3-2R_1}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & -1 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 1 \\
0 & 3 & 1 & 3 & -2 & 0
\end{array} \right]\\[6pt] &\xrightarrow{R_3-3R_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & -1 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 1 \\
0 & 0 & 1 & 6 & -2 & -3
\end{array} \right].
\end{align*}

Now the left $3\times 3$ part has become the identity matrix.
So the matrix $A$ is invertible and the inverse is given by the right $3\times 3$ part.
Hence we obtain
\[A^{-1}=\begin{bmatrix}
-1 & 1 & 0 \\
-1 &0 &1 \\
6 & -2 & -3
\end{bmatrix}.\]

 

(c) Solve the system using the inverse matrix $A^{-1}$.

As noted in (a), the system can be written using matrices as
\[A\mathbf{x}=\mathbf{b}.\] Multiplying by the inverse $A^{-1}$ on the left, we have
\begin{align*}
A^{-1}A\mathbf{x}=A^{-1}\mathbf{b}\\
I\mathbf{x}=A^{-1}\mathbf{b}\\
\mathbf{x}=A^{-1}\mathbf{b}.
\end{align*}
Therefore the solution $\mathbf{x}$ of the system is given by
\begin{align*}
\mathbf{x}&=A^{-1}\mathbf{b}\\
&=\begin{bmatrix}
-1 & 1 & 0 \\
-1 &0 &1 \\
6 & -2 & -3
\end{bmatrix}\begin{bmatrix}
-1 \\
1 \\
-2
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
2 \\
-1 \\
-2
\end{bmatrix}.
\end{align*}
Thus, the solution of the system is
\[x=2, y=-1, z=-2.\]


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