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

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 .