Step 1

Answer(1):

The given statement is:

If the number of equations in a linear system exceeds the number of unknowns, then the system must be inconsistent - thus no solution.

The given statement is false because it's not compulsory that, if If the number of equations in a linear system exceeds the number of unknowns, then the system must be inconsistent, it can be consistent.

For example:

\(\begin{cases} x=8\\2x=16 \end{cases}\)

In this system, the number of equations is two and the number of variables is one, that is the number of equations is exceeded by the number of variables.

Thus, this system is consistent.

Step 2

Answer(2):

The given statement is:

If B has a column with zeros, then AB will also have a column with zeros, if this product is defined.

The given statement is false because it's not compulsory that, If B has a column with zeros, then AB will also have a column with zeros if this product is defined.

For example:

Consider, \(A=\begin{bmatrix}1 & 0 \\2 & 1 \end{bmatrix} , B=\begin{bmatrix}3 & 0 \\2 & 0 \end{bmatrix}\) are two matrices, then the product of these matrices is given as:

\(AB=\begin{bmatrix}(1\times3+0\times2) & (2\times3+1\times2) \\(1\times0+0\times0) & (2\times0+1\times0) \end{bmatrix}\)

\(AB=\begin{bmatrix}3 & 8 \\0 & 0 \end{bmatrix}\)

Clearly, their product is defined, and not any column is zero.

Therefore, the given statement is false.

Step 3

Answer(4):

The given statement is:

Suppose A is an n x n matrix and assume \(A^2 = O\), where O is the zero matrices. Then A = O.

The given statement is false because it's not compulsory that, If A is an n x n matrix and assume \(A^2 = O\), where O is the zero matrices. Then A = O.

For example:

Consider, \(A=\begin{bmatrix}0 & 0 \\1 & 0 \end{bmatrix}\) is a non-zero matrix but its square is zero.

Therefore, the given statement is false.

Step 4

Hence, all false statements are 1, 2 and 4.

Answer(1):

The given statement is:

If the number of equations in a linear system exceeds the number of unknowns, then the system must be inconsistent - thus no solution.

The given statement is false because it's not compulsory that, if If the number of equations in a linear system exceeds the number of unknowns, then the system must be inconsistent, it can be consistent.

For example:

\(\begin{cases} x=8\\2x=16 \end{cases}\)

In this system, the number of equations is two and the number of variables is one, that is the number of equations is exceeded by the number of variables.

Thus, this system is consistent.

Step 2

Answer(2):

The given statement is:

If B has a column with zeros, then AB will also have a column with zeros, if this product is defined.

The given statement is false because it's not compulsory that, If B has a column with zeros, then AB will also have a column with zeros if this product is defined.

For example:

Consider, \(A=\begin{bmatrix}1 & 0 \\2 & 1 \end{bmatrix} , B=\begin{bmatrix}3 & 0 \\2 & 0 \end{bmatrix}\) are two matrices, then the product of these matrices is given as:

\(AB=\begin{bmatrix}(1\times3+0\times2) & (2\times3+1\times2) \\(1\times0+0\times0) & (2\times0+1\times0) \end{bmatrix}\)

\(AB=\begin{bmatrix}3 & 8 \\0 & 0 \end{bmatrix}\)

Clearly, their product is defined, and not any column is zero.

Therefore, the given statement is false.

Step 3

Answer(4):

The given statement is:

Suppose A is an n x n matrix and assume \(A^2 = O\), where O is the zero matrices. Then A = O.

The given statement is false because it's not compulsory that, If A is an n x n matrix and assume \(A^2 = O\), where O is the zero matrices. Then A = O.

For example:

Consider, \(A=\begin{bmatrix}0 & 0 \\1 & 0 \end{bmatrix}\) is a non-zero matrix but its square is zero.

Therefore, the given statement is false.

Step 4

Hence, all false statements are 1, 2 and 4.