Question

In Exercises 29–32, (a) does the equation \(A{\bf{x}} = 0\) have a nontrivial solution and (b) does the equation \(A{\bf{x}} = {\bf{b}}\) have at least one solution for every possible b?

32. A is a \(2 \times 4\) matrix with two pivot positions.

Short answer

(a) The system of equations has a nontrivial solution.

(b) The matrix equation has a solution for every possible b.

Step-by-step solution

(a) Step 1: Write the condition for the matrix according to the pivot positions

It is given that the order of the matrix Ais \(2 \times 4\). This means that there are 2 rows and 4 columns. Also, it is given that there are two pivot positions.

Since there are four columns, there must be 4 variables. Due to the pivot positions, out of four variables, two are free, and two are basic.

Construct a matrix according to the condition (a)

Matrix Aof \(2 \times 4\) order in the augmented form \(\left[ {\begin{array}{*{20}{c}}A&0\end{array}} \right]\) is written as shown below:

It is observed that there are two free variables in \(A{\bf{x}} = 0\).

Thus, the system of equations has a nontrivial solution.

(b) Step 3: Construct a matrix according to the condition (b)

Here, each row consists of pivot positions.

Consider vector \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{{b_1}}\\{{b_2}}\\{{b_3}}\end{array}} \right]\).

Matrix Aof \(2 \times 4\) order in the augmented form \(\left[ {\begin{array}{*{20}{c}}A&{\bf{b}}\end{array}} \right]\) is written as shown below:

It is observed that there are possible values of b in \(A{\bf{x}} = {\bf{b}}\). So, in \({\mathbb{R}^2}\), the matrix equation \(A{\bf{x}} = {\bf{b}}\) has a solution for every possible b.

Thus, the system of equations has a solution for every possible value b.