Question

The matrix $\left[\begin{array}{ccc}{k}^2& 1& 4\\ k& -1& -2\\ 1& 1& 1\end{array}\right]$ is invertible for all positive constantsk.

Short answer

Therefore, the given condition is true.

Step-by-step solution

 Step 1: Matrix Definition. 

Matrix is aset of numbers arranged in rows and columns so as to form a rectangulararray.

The numbers are called the elements, or entries, of the matrix.

If there are m rows and n columns, the matrix is said to be an “ m by n ” matrix, written “$m\times n$ .”

Given. 

Given Matrix,

$\left[\begin{array}{ccc}{\mathrm{k}}^2& 1& 4\\ \mathrm{k}& -1& -2\\ 1& 1& 1\end{array}\right]$

To check whether the given condition is true or false.

We compute,

$\begin{array}{l}detA=-k^2-2+4k+4+2k^2-k\\ detA=k^2+3k+2\end{array}$

which turns out to be 0 if and only if $k=-2$or $k=-1$, so for all positive constants $k$, the matrix is regular.

Therefore, the given condition is true.