Question

Given \(A = \left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 6}\\7&{21}\\{ - 3}&{ - 9}\end{array}} \right]\), find one nontrivial solution of \(A{\bf{x}} = 0\) by inspection. [Hint:Think of the equation \(A{\bf{x}} = 0\) written as a vector equation.]

Short answer

The solution is \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}3\\{ - 1}\end{array}} \right]\).

Step-by-step solution

Write the condition for the product of a vector and a matrix

It is known that the representation of a column of a matrix is \(A = \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{ \cdot \cdot \cdot }&{{a_n}}\end{array}} \right]\), and the representation of a vector is \[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\ \vdots \\{{x_n}}\end{array}} \right]\].

According to the definition, the entries in vector x reflect the values in a linear combination of matrix A columns.

Also, the product by using the row-vector rule is defined as shown below:

\[\begin{array}{c}A{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{ \cdot \cdot \cdot }&{{a_n}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\ \vdots \\{{x_n}}\end{array}} \right]\\ = {x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n}\end{array}\]

The number of columns in matrix \(A\)should be equal to the number of entries in vector x so that \[A{\bf{x}}\] can be defined.

Write vector x

Consider matrix \(A = \left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 6}\\7&{21}\\{ - 3}&{ - 9}\end{array}} \right]\). Since there are two columns, so the number of entries should be 2. Thus, vector \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\).

Write the matrix equation

Consider the equation \(A{\bf{x}} = 0\)as shown below:

\(\begin{array}{c}A{\bf{x}} = 0\\\left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 6}\\7&{21}\\{ - 3}&{ - 9}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = 0\end{array}\)

Thus, \(\left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 6}\\7&{21}\\{ - 3}&{ - 9}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = 0\).

Obtain one nontrivial solution

Solve the equation \(\left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 6}\\7&{21}\\{ - 3}&{ - 9}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = 0\) by inspection.

\({x_1}\left[ {\begin{array}{*{20}{c}}{ - 2}\\7\\{ - 3}\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 6}\\{21}\\{ - 9}\end{array}} \right] = 0\)

From the two columns \(\left[ {\begin{array}{*{20}{c}}{ - 2}\\7\\{ - 3}\end{array}} \right]\) and \(\left[ {\begin{array}{*{20}{c}}{ - 6}\\{21}\\{ - 9}\end{array}} \right]\), it is observed that column \(\left[ {\begin{array}{*{20}{c}}{ - 6}\\{21}\\{ - 9}\end{array}} \right]\) is three times column \(\left[ {\begin{array}{*{20}{c}}{ - 2}\\7\\{ - 3}\end{array}} \right]\).

By inspection, the values of \({x_1}\) and \({x_2}\) are 3 and \( - 1\), respectively. Thus, \({\bf{x}} = \left[ {\begin{array}{*{20}{c}}3\\{ - 1}\end{array}} \right]\).

Therefore, one nontrivial solution is 3 and \( - 1\).