Regularity and invertibility of two parameterized matrices?

Question

$$ C= \begin{bmatrix} 1+a & 2 & 3 & 4 & 5 \\ 1 & 2+a & 3 & 4 & 5 \\ 1 & 2 & 3+a & 4 & 5 \\ 1 & 2 & 3 & 4+a & 5 \\ 1 & 2 & 3 & 4 & 5+a \\ \end{bmatrix} $$

For which values of $a$ is the matrix $C$ regular?

$$ A= \begin{bmatrix} 3a & a \\ -a & 1 \\ \end{bmatrix} $$

Determine for which values of $a$ is there an inverse $A^{-1}$ and then solve it?

Please help....

Answer 1

I) We need evaluate determinant. Determinant is a multilinear function of the columns. So, $$ \require{cancel} \det C= \begin{vmatrix} 1+a & 2 & 3 & 4 & 5 \\ 1 & 2+a & 3 & 4 & 5 \\ 1 & 2 & 3+a & 4 & 5 \\ 1 & 2 & 3 & 4+a & 5 \\ 1 & 2 & 3 & 4 & 5+a \\ \end{vmatrix} = \\= \begin{vmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 2+a & 3 & 4 & 5 \\ 1 & 2 & 3+a & 4 & 5 \\ 1 & 2 & 3 & 4+a & 5 \\ 1 & 2 & 3 & 4 & 5+a \\ \end{vmatrix} + \begin{vmatrix} a & 2 & 3 & 4 & 5 \\ 0 & 2+a & 3 & 4 & 5 \\ 0 & 2 & 3+a & 4 & 5 \\ 0 & 2 & 3 & 4+a & 5 \\ 0 & 2 & 3 & 4 & 5+a \\ \end{vmatrix}=\\= \cancelto{0}{\begin{vmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3+a & 4 & 5 \\ 1 & 2 & 3 & 4+a & 5 \\ 1 & 2 & 3 & 4 & 5+a \\ \end{vmatrix}} + \begin{vmatrix} 1 & 0 & 3 & 4 & 5 \\ 1 & a & 3 & 4 & 5 \\ 1 & 0 & 3+a & 4 & 5 \\ 1 & 0 & 3 & 4+a & 5 \\ 1 & 0 & 3 & 4 & 5+a \\ \end{vmatrix} + a\begin{vmatrix} 2+a & 3 & 4 & 5 \\ 2 & 3+a & 4 & 5 \\ 2 & 3 & 4+a & 5 \\ 2 & 3 & 4 & 5+a \\ \end{vmatrix}=\\= a\begin{vmatrix} 1 & 3 & 4 & 5 \\ 1 & 3+a & 4 & 5 \\ 1 & 3 & 4+a & 5 \\ 1 & 3 & 4 & 5+a \\ \end{vmatrix} + a\begin{vmatrix} 2 & 3 & 4 & 5 \\ 2 & 3+a & 4 & 5 \\ 2 & 3 & 4+a & 5 \\ 2 & 3 & 4 & 5+a \\ \end{vmatrix} + a\begin{vmatrix} a & 3 & 4 & 5 \\ 0 & 3+a & 4 & 5 \\ 0 & 3 & 4+a & 5 \\ 0 & 3 & 4 & 5+a \\ \end{vmatrix} = \\\text{...analogous tricks...}\\= a\begin{vmatrix} 1 & 0 & 4 & 5 \\ 1 & a & 4 & 5 \\ 1 & 0 & 4+a & 5 \\ 1 & 0 & 4 & 5+a \\ \end{vmatrix} + a\begin{vmatrix} 2 & 0 & 4 & 5 \\ 2 & a & 4 & 5 \\ 2 & 0 & 4+a & 5 \\ 2 & 0 & 4 & 5+a \\ \end{vmatrix} + a^2\begin{vmatrix} 3+a & 4 & 5 \\ 3 & 4+a & 5 \\ 3 & 4 & 5+a \\ \end{vmatrix} =\\= a^2\begin{vmatrix} 1 & 4 & 5 \\ 1 & 4+a & 5 \\ 1 & 4 & 5+a \\ \end{vmatrix} + a^2\begin{vmatrix} 2 & 4 & 5 \\ 2 & 4+a & 5 \\ 2 & 4 & 5+a \\ \end{vmatrix} + a^2\begin{vmatrix} 3 & 4 & 5 \\ 3 & 4+a & 5 \\ 3 & 4 & 5+a \\ \end{vmatrix} + a^2\begin{vmatrix} a & 4 & 5 \\ 0 & 4+a & 5 \\ 0 & 4 & 5+a \\ \end{vmatrix} =\\= a^2\begin{vmatrix} 1 & 0 & 5 \\ 1 & a & 5 \\ 1 & 0 & 5+a \\ \end{vmatrix} + a^2\begin{vmatrix} 2 & 0 & 5 \\ 2 & a & 5 \\ 2 & 0 & 5+a \\ \end{vmatrix} + a^2\begin{vmatrix} 3 & 0 & 5 \\ 3 & a & 5 \\ 3 & 0 & 5+a \\ \end{vmatrix} + a^3\begin{vmatrix} 4+a & 5 \\ 4 & 5+a \\ \end{vmatrix} =\\= a^3\begin{vmatrix} 1 & 5 \\ 1 & 5+a \\ \end{vmatrix} + a^3\begin{vmatrix} 2 & 5 \\ 2 & 5+a \\ \end{vmatrix} + a^3\begin{vmatrix} 3 & 5 \\ 3 & 5+a \\ \end{vmatrix} + a^3\begin{vmatrix} 4+a & 5 \\ 4 & 5+a \\ \end{vmatrix} =\\= a^3[a + 2a + 3a + a^2 + 9a] = a^4(a + 15) $$

Hence, if $a\ne0$ and $a\ne-15$ matrix is regular.

II) $$ \det A = a^2 + 3a $$ If $a\ne0$ and $a\ne-3$, matrix is regular and $$ A^{-1} = \frac{1}{a(a+3)}\begin{pmatrix} 1 & -a\\ a & 3a \end{pmatrix} $$

Answer 2

Actually, there is an easier way to find the values for the matrix $C$ for which it is invertible:

Look at the following representation $$ \begin{pmatrix} 1+a & 2 & 3 & 4 & 5 \\ 1 & 2+a & 3 & 4 & 5 \\ 1 & 2 & 3+a & 4 & 5 \\ 1 & 2 & 3 & 4+a & 5 \\ 1 & 2 & 3 & 4 & 5+a \\ \end{pmatrix}= \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \\ \end{pmatrix}+ a \operatorname{Id}_{5\times5}=B+a \operatorname{Id}_{5\times5} $$ which is the eigenvalue problem for matrix B in $\lambda=-a$. That for $\lambda=0$ we have an eigenvalue is clear, the dimension of the eigenspace is $4$, which is clear because with Gauss we get $$ \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \\ \end{pmatrix}\to \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{pmatrix}\ $$ so in fact we are looking for one other eigenvalue $\lambda_2$ with eigenspace dimension $1$, so that everything adds up to $5$ eventually.

If we look hard enough, we can easily figure out it has to be $\lambda=15\Leftrightarrow a=-15$. For this to be seen, observe that $$ \begin{pmatrix} 1+a \\ 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}+ \begin{pmatrix} 2 \\ 2+a \\ 2 \\ 2 \\ 2 \end{pmatrix}+ \begin{pmatrix} 3 \\ 3 \\ 3+a \\ 3 \\ 3 \end{pmatrix}+ \begin{pmatrix} 4 \\ 4 \\ 4 \\ 4+a \\ 4 \end{pmatrix}= \begin{pmatrix} 10+a \\ 10+a \\ 10+a \\ 10+a \\ 10 \end{pmatrix}\equiv(-1)\begin{pmatrix} 5 \\ 5 \\ 5 \\ 5 \\ 5+a \end{pmatrix} $$ which is true for $a=-15$. Since now we have 5 dimensional eigenspace we are finished and conclude that for $a_1=0$ and $a_2=-15$ the matrix $C$ is not invertible.

For the second part just use Cramer's rule which will give you $$ A^{-1}=\frac 1{a(3+a)}\begin{pmatrix} 1 & -a \\ a & 3a \end{pmatrix} $$ which indeed is defined for all $a\in\mathbb{R}\backslash\{0,-3\}$