Question

For what value of K, the matrix \[\left( {\begin{array}{\times {20}{l}}
  k&2 \\
  3&4
\end{array}} \right)\]has no inverse.

Answer 1

Hint:
We are asked the value of K in the above matrix \[\left( {\begin{array}{\times {20}{l}}
  k&2 \\
  3&4
\end{array}} \right)\]and we are given with the condition that the inverse of the given matrix \[\left( {\begin{array}{\times {20}{l}}
  k&2 \\
  3&4
\end{array}} \right)\] in the question does not exist .So using that information we can easily solve this question. The above given matrix is \[2 \times 2\] square matrix (which means it has \[2\]rows and \[2\]columns as we can see clearly).The inverse for such kind of matrix A is simply calculated by firstly calculating the determinant of the square matrix that is \[\det A\] and if the \[0\] then the inverse does not exist so by using this condition we can get the answer to this question.

Complete step by step solution:
Here in this question we are given with the matrix \[\left( {\begin{array}{\times {20}{l}}
  k&2 \\
  3&4
\end{array}} \right)\]and we have to find the value of K for this matrix does not having an inverse
So for inverse to exist or not we have to check a condition and that is the determinant of the given matrix whose inverse has to be found should not be \[0\]that is \[\det A \ne 0\]for the inverse to exist
And determinant of the given matrix is calculated by solving the matrix that is in the case of a \[2\times 2\]square matrix of the form \[\left( {\begin{array}{\times {20}{l}}
  a&b \\
  c&d
\end{array}} \right)\]is equal to \[ad - bc\]
So we shall find the determinant of the given matrix and then equating the resultant equation to the \[0\]and hence we then solve for K and get our required answer
For the determinant of the matrix \[\left( {\begin{array}{\times {20}{l}}
  k&2 \\
  3&4
\end{array}} \right)\]is \[4k - 6\]
Now equating the resultant equation that is \[4k - 6\]equal to \[0\]for solving for K
\[4k - 6 = 0\]
\[4k = 6\]
\[k = \dfrac{6}{4}\]
\[k = \dfrac{3}{2}\]

Hence the \[k = \dfrac{3}{2}\] which is our desired answer that we want

Note:
In the above question we have seen how to solve for getting the determinant of a \[2\times 2\] square matrix of the form \[\left( {\begin{array}{\times {20}{l}}
  a&b \\
  c&d
\end{array}} \right)\]is equal to \[ad - bc\]but for the \[3\times 3\] square matrix The same determinant can be calculated by solving the given matrix along the row or the column .