Vtyjkyuk

For what values ​​of "$a$" will we have the $dim(textim (M))$ of the matrix smaller than $3$?

$$beginalign
M=
beginpmatrix
a & 2 & 3 \
a & a & 4 \
a & a & a \
endpmatrix
endalign
$$

Hint:

For $$M=beginpmatrixa&2&3\a&a&4\a&a&aendpmatrix$$ to give $dim(textim(M))=3$, you need $$M:Bbb R^3rightarrow Bbb R^3$$ to be an isomorphism. If this is not the case, then the dimension of the image is less than 3.

@AllanOliveira With $a=0$ you get $$beginpmatrix0&2&3\0&0&4\0&0&0endpmatrixbeginpmatrixx\y\zendpmatrix=ybeginpmatrix2\0\0endpmatrix+zbeginpmatrix3\4\0endpmatrix$$ surely this dimension is $2$ ($<3$).

Since OP already accepted the answer, I'll expand it. The map $M:Bbb R^3rightarrowBbb R^3$ is an isomorphism precisely when the matrix $M$ is invertible, this happens if and only if $det(M)neq 0$, so we need to find the values of $a$ for which $det(M)=0$. Computing the determinant we get $$det(M)=a(a^2-6a+8)=a(a-2)(a-4)$$ Therefore, $dim(textim(M))<3$ when $a=0,2,4$.

If a square matrix is rank-deficient, its rows/columns are linearly dependent, which in turn means that its determinant vanishes. Compute the determinant of your matrix and set it to zero to get a condition on $a$. There are some obvious elementary row operations that you can perform on the matrix to make it upper-triangular, which will make the computation of its determinant very simple.