Question

Consider the 3 × 3 matrix A =   0 1 k 2 k −6 2 7 4  . For what values of k is matrix A invertible? (a) k ∈ R (b) all real k except 2 and 5 (c) k ≥ 0 (d) k = 0 (e) no value of k makes A invertible

Answer 1

Answer:

b)

Step-by-step explanation:

A is invertible if and only if det(A)≠0. Let's compute the determinant of A and find the values k for which it is nonzero.

Using Sarrus's rule, we obtain that

$\det A=0(k)(4)+1(-6)(2)+k(2)(7)-2(k)(k)-7(-6)(0)-4(2)(1)=-12+14k-2k^2-8=-2k^2+14k-20=-2(k^2-7k+10)=2(k-5)(k-2)$

Note that the determinant is a quadratic equation on k, which can be factored as above.

Now the determinant is only zero if k=5 or k=2 (the zeroes of the quadratic polynomial). Therefore, if k≠2,5 the determinant is nonzero so A is invertible.

Answer 2

Answer:

All real k except 2 and 5.

Step-by-step explanation:

For a matrix to be invertible the determinant of it should not be equal to zero.

A matrix P is invertible if |P| ≠ 0

The matrix given here is:

$\left[\begin{array}{ccc}0&1&k\\2&k&-6\\2&7&4\end{array}\right]$

For the above matrix to be invertible:

$0\times(4k+42)-1\times(8+12)+k\times(14-2k)\neq 0\\-20+14k-2k^{2}\neq 0\\k^{2}-7k+10\neq 0\\(k-2)(k-5)\neq 0\\k\neq 2,k\neq 5$

Matrix is invertible for all real k except 2 and 5.