# For what value of k, the matrix \[\left( \begin{matrix}

2-k & 4 \\

-5 & 1 \\

\end{matrix} \right)\] is not invertible?

Last updated date: 28th Mar 2023

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Answer

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306.6k+ views

Hint:Find the determinant of the matrix. As it’s not invertible it is equal to zero. Consider the matrix A. Find the determinant of A and equal it to zero. From that, find the value of k.

Complete step-by-step answer:

A square matrix is said to be invertible if its inverse exists and is said to be non-invertible if its determinant is equal to zero.

\[\therefore \]A square matrix that is not invertible is called singular/ degenerate. Non-square matrices \[\left( m\times n \right)\] where \[m\ne n\] do not have an inverse. In some cases there are left inverse and right inverse.

Given the matrix \[\left( \begin{matrix}

2-k & 4 \\

-5 & 1 \\

\end{matrix} \right)\]

Let’s write it as \[A=\left( \begin{matrix}

2-k & 4 \\

-5 & 1 \\

\end{matrix} \right)\]

As the matrix is non-invertible, the determinant of A is zero.

\[\left| A \right|=0\]

\[\left| \left( \begin{matrix}

2-k & 3 \\

-5 & 1 \\

\end{matrix} \right) \right|=0\]

To work out the determinant, multiply A by the determinant of the matrix that is not in a row or column and then sum up.

\[\begin{align}

& \left[ \left( 2-k \right)\left( 1 \right) \right]-\left[ \left( 3 \right)\left( -5 \right) \right] \\

& =\left( 2-k \right)-\left( -15 \right) \\

& =2-k+15=0 \\

& \therefore k=17 \\

\end{align}\]

\[\therefore \]Value of k = 17.

Note:

For the case of a non-invertible matrix, we took\[\left| A \right|=0\].

If the matrix was invertible, then \[\left| A \right|\ne 0\]but\[{{A}^{-1}}\].

Complete step-by-step answer:

A square matrix is said to be invertible if its inverse exists and is said to be non-invertible if its determinant is equal to zero.

\[\therefore \]A square matrix that is not invertible is called singular/ degenerate. Non-square matrices \[\left( m\times n \right)\] where \[m\ne n\] do not have an inverse. In some cases there are left inverse and right inverse.

Given the matrix \[\left( \begin{matrix}

2-k & 4 \\

-5 & 1 \\

\end{matrix} \right)\]

Let’s write it as \[A=\left( \begin{matrix}

2-k & 4 \\

-5 & 1 \\

\end{matrix} \right)\]

As the matrix is non-invertible, the determinant of A is zero.

\[\left| A \right|=0\]

\[\left| \left( \begin{matrix}

2-k & 3 \\

-5 & 1 \\

\end{matrix} \right) \right|=0\]

To work out the determinant, multiply A by the determinant of the matrix that is not in a row or column and then sum up.

\[\begin{align}

& \left[ \left( 2-k \right)\left( 1 \right) \right]-\left[ \left( 3 \right)\left( -5 \right) \right] \\

& =\left( 2-k \right)-\left( -15 \right) \\

& =2-k+15=0 \\

& \therefore k=17 \\

\end{align}\]

\[\therefore \]Value of k = 17.

Note:

For the case of a non-invertible matrix, we took\[\left| A \right|=0\].

If the matrix was invertible, then \[\left| A \right|\ne 0\]but\[{{A}^{-1}}\].

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