Chapter 3, Problem 1SE

To determine

To verify: The given statement is true or false. If $A$ is a $2\times 2$ matrix with a zero determinant, then one column of $A$ is a multiple of other.

Answer to Problem 1SE

The statement which states that ‘if $A$ is a $2\times 2$ matrix with a zero determinant, then one column of $A$ is a multiple of other’ is false.

Explanation of Solution

Given:

If $A$ is a $2\times 2$ matrix with a zero determinant, then one column of $A$ is a multiple of other.

Concept used:

The concept of determinant is used to solve the question which states that the top row components and their respective minors are taken into consideration when formulating the determinant, which is the first and easiest method. Multiply the top row's first element by its minor, then deduct the result from the second element's minor product. Up until all the items of the top row have been taken into account, alternately add and remove the product of each element of the top row with its corresponding minor.

Proof:

Let $A=\left(\begin{array}{cc}2& 3\\ 4& 6\end{array}\right)$

Now,

  $\begin{array}{l}\det A=\det \left(\begin{array}{cc}2& 3\\ 4& 6\end{array}\right)\\ \det A=\left(2\times 6\right)-\left(4\times 3\right)\\ \det A=12-12\\ \det A=0\end{array}$

Since determinant of matrix $A$ is zero but columns of $A$ are not multiple to each other. Hence, it proves that if $A$ is a $2\times 2$ matrix with a zero determinant, then one column of $A$ need not to be a multiple of other.