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Solving $Ax=y$ by Inspection

Sometimes it is possible to solve a system of linear equations by inspection. This requires some degree of familiarity with the column space of a matrix.

The expression $Ax=y$ denotes saying that $y$ is a linear combination of the columns of $A$ or that equivalently it is in the column space of $A$. The amounts by which columns of $A$ contribute to the linear combination are determined by the elements of $x$. Knowing this enables us to solve for $x$ by inspection.

If $y$ is in the column space of $A$ then there is at least one solution; otherwise, there is no solution.

The column space of $A$ is the vector space containing all possible linear combinations of the columns of $A$.

Here are some examples.

Example 1

       $\left(\begin{array}{cccc}3& 3& 1& 2\end{array}\right)$$\left(\begin{array}{c}{x}_0\\ x_1\\ x_2\\ x_3\end{array}\right)$ $=$ $\left(\begin{array}{c}14\end{array}\right)$

Here is one possible way of writing $y$ as a linear combination of the columns $c_i$ of $A$:

       $\left(\begin{array}{c}14\end{array}\right)$ $=$ $2$ $c_0$ +$2$ $c_1$ +$0$ $c_2$ +$1$ $c_3$

       $\left(\begin{array}{c}14\end{array}\right)$ $=$ $2$ $$ \left(\begin{array}{c}3\end{array}\right)$$ +$2$ $$ \left(\begin{array}{c}3\end{array}\right)$$ +$0$ $$ \left(\begin{array}{c}1\end{array}\right)$$ +$1$ $$ \left(\begin{array}{c}2\end{array}\right)$$

       $\left(\begin{array}{c}14\end{array}\right)$ $=$ $x_0$ $$ \left(\begin{array}{c}3\end{array}\right)$$ +$x_1$ $$ \left(\begin{array}{c}3\end{array}\right)$$ +$x_2$ $$ \left(\begin{array}{c}1\end{array}\right)$$ +$x_3$ $$ \left(\begin{array}{c}2\end{array}\right)$$

       ⇒     $\left(\begin{array}{c}{x}_0\\ x_1\\ x_2\\ x_3\end{array}\right)$ =$$ \left(\begin{array}{c}2\\ 2\\ 0\\ 1\end{array}\right)$$

The solution $x$ is obtained by writing the coefficients of the linear combination as a column vector. This is a possible solution; there may be more.

Example 2

       $\left(\begin{array}{cccc}3& 0& 0& 1\\ 2& 3& 1& 1\end{array}\right)$$\left(\begin{array}{c}{x}_0\\ x_1\\ x_2\\ x_3\end{array}\right)$ $=$ $\left(\begin{array}{c}5\\ 8\end{array}\right)$

Here is one possible way of writing $y$ as a linear combination of the columns $c_i$ of $A$:

       $\left(\begin{array}{c}5\\ 8\end{array}\right)$ $=$ $1$ $c_0$ +$1$ $c_1$ +$1$ $c_2$ +$2$ $c_3$

       $\left(\begin{array}{c}5\\ 8\end{array}\right)$ $=$ $1$ $$ \left(\begin{array}{c}3\\ 2\end{array}\right)$$ +$1$ $$ \left(\begin{array}{c}0\\ 3\end{array}\right)$$ +$1$ $$ \left(\begin{array}{c}0\\ 1\end{array}\right)$$ +$2$ $$ \left(\begin{array}{c}1\\ 1\end{array}\right)$$

       $\left(\begin{array}{c}5\\ 8\end{array}\right)$ $=$ $x_0$ $$ \left(\begin{array}{c}3\\ 2\end{array}\right)$$ +$x_1$ $$ \left(\begin{array}{c}0\\ 3\end{array}\right)$$ +$x_2$ $$ \left(\begin{array}{c}0\\ 1\end{array}\right)$$ +$x_3$ $$ \left(\begin{array}{c}1\\ 1\end{array}\right)$$

       ⇒     $\left(\begin{array}{c}{x}_0\\ x_1\\ x_2\\ x_3\end{array}\right)$ =$$ \left(\begin{array}{c}1\\ 1\\ 1\\ 2\end{array}\right)$$

The solution $x$ is obtained by writing the coefficients of the linear combination as a column vector. This is a possible solution; there may be more.

Example 3

       $\left(\begin{array}{cccc}1& 3& 2& 2\\ 0& 3& 1& 0\\ 1& 2& 3& 2\end{array}\right)$$\left(\begin{array}{c}{x}_0\\ x_1\\ x_2\\ x_3\end{array}\right)$ $=$ $\left(\begin{array}{c}8\\ 2\\ 10\end{array}\right)$

Here is one possible way of writing $y$ as a linear combination of the columns $c_i$ of $A$:

       $\left(\begin{array}{c}8\\ 2\\ 10\end{array}\right)$ $=$ $2$ $c_0$ +$0$ $c_1$ +$2$ $c_2$ +$1$ $c_3$

       $\left(\begin{array}{c}8\\ 2\\ 10\end{array}\right)$ $=$ $2$ $$ \left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)$$ +$0$ $$ \left(\begin{array}{c}3\\ 3\\ 2\end{array}\right)$$ +$2$ $$ \left(\begin{array}{c}2\\ 1\\ 3\end{array}\right)$$ +$1$ $$ \left(\begin{array}{c}2\\ 0\\ 2\end{array}\right)$$

       $\left(\begin{array}{c}8\\ 2\\ 10\end{array}\right)$ $=$ $x_0$ $$ \left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)$$ +$x_1$ $$ \left(\begin{array}{c}3\\ 3\\ 2\end{array}\right)$$ +$x_2$ $$ \left(\begin{array}{c}2\\ 1\\ 3\end{array}\right)$$ +$x_3$ $$ \left(\begin{array}{c}2\\ 0\\ 2\end{array}\right)$$

       ⇒     $\left(\begin{array}{c}{x}_0\\ x_1\\ x_2\\ x_3\end{array}\right)$ =$$ \left(\begin{array}{c}2\\ 0\\ 2\\ 1\end{array}\right)$$

The solution $x$ is obtained by writing the coefficients of the linear combination as a column vector. This is a possible solution; there may be more.

Example 4

       $\left(\begin{array}{cccc}2& 0& 3& 1\\ 1& 3& 0& 0\\ 3& 0& 3& 0\\ 0& 2& 3& 2\end{array}\right)$$\left(\begin{array}{c}{x}_0\\ x_1\\ x_2\\ x_3\end{array}\right)$ $=$ $\left(\begin{array}{c}8\\ 2\\ 9\\ 5\end{array}\right)$

Here is one possible way of writing $y$ as a linear combination of the columns $c_i$ of $A$:

       $\left(\begin{array}{c}8\\ 2\\ 9\\ 5\end{array}\right)$ $=$ $2$ $c_0$ +$0$ $c_1$ +$1$ $c_2$ +$1$ $c_3$

       $\left(\begin{array}{c}8\\ 2\\ 9\\ 5\end{array}\right)$ $=$ $2$ $$ \left(\begin{array}{c}2\\ 1\\ 3\\ 0\end{array}\right)$$ +$0$ $$ \left(\begin{array}{c}0\\ 3\\ 0\\ 2\end{array}\right)$$ +$1$ $$ \left(\begin{array}{c}3\\ 0\\ 3\\ 3\end{array}\right)$$ +$1$ $$ \left(\begin{array}{c}1\\ 0\\ 0\\ 2\end{array}\right)$$

       $\left(\begin{array}{c}8\\ 2\\ 9\\ 5\end{array}\right)$ $=$ $x_0$ $$ \left(\begin{array}{c}2\\ 1\\ 3\\ 0\end{array}\right)$$ +$x_1$ $$ \left(\begin{array}{c}0\\ 3\\ 0\\ 2\end{array}\right)$$ +$x_2$ $$ \left(\begin{array}{c}3\\ 0\\ 3\\ 3\end{array}\right)$$ +$x_3$ $$ \left(\begin{array}{c}1\\ 0\\ 0\\ 2\end{array}\right)$$

       ⇒     $\left(\begin{array}{c}{x}_0\\ x_1\\ x_2\\ x_3\end{array}\right)$ =$$ \left(\begin{array}{c}2\\ 0\\ 1\\ 1\end{array}\right)$$

The solution $x$ is obtained by writing the coefficients of the linear combination as a column vector. This is a possible solution; there may be more.

[This content was created with SL, available on the App Store.]