[Math] Find values so matrix not invertible

linear algebra

$$
\begin{pmatrix}
2 & 4 & k \\
1 & 3 & 2 \\
3 & k & 9 \\
\end{pmatrix}
$$

For what values of $k$ is the above matrix not invertible.

Need help. Don't know where to start really. Thanks.`

Best Answer

Hint: A matrix is not invertible iff it's determinant is zero

Solve for the value of $k$ that makes the determinant $0$

Related Solutions

[Math] What values of $a$ make this matrix not invertible

Calculate the determinant of A $$\det(A)=a(a^2-1)-1(a-1)+1(1-a)=a^3-a-a+1+1-a=a^3-3a+2$$ The matrix is not invertible when det(A) is equal to zero.

You can solve or guess the solution the obvious solutions which are $a=1$ (double root) and $a=-2$. As stated in the comments, the value $a=1$ is obvious already before that point, since it is the value that makes all columns of the matrix equal (and thus linearly dependent).

So, for every other value of $a$, i.e for $a\in \mathbb{R}-\{1,-2\}$ the matrix is invertible, which means that in this case the column vectors are linearly independent and therefore they span $\mathbb{R}^3$.

[Math] Determining the values of $\lambda$ for which the matrix is invertible

The key principle here is that a matrix is invertible if an only if its determinant is non-zero.

We apply this principle here as follows:

Set

$A(\lambda) = \begin{bmatrix} \lambda &-1 & 0 \\ -1 & \lambda & -1 \\ 0 & -1 & \lambda\end{bmatrix}; \tag{1}$

then we have

$\det A(\lambda) = \lambda^3 - 2\lambda; \tag{2}$

thus $\det A(\lambda) = 0$ precisely when

$0 = \lambda^3 - 2 \lambda = \lambda(\lambda^2 - 2); \tag{3}$

the roots of this equation are easily seen to be

$\lambda = 0, \pm \sqrt 2; \tag{4}$

$A(\lambda)$ is thus invertible for all other values of $\lambda$.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Best Answer

Related Solutions

[Math] What values of $a$ make this matrix not invertible

[Math] Determining the values of $\lambda$ for which the matrix is invertible

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