Linear Algebra Matrices

Question

Determining the values of a for which the Matrix A has an inverse !

A= \begin{pmatrix} 1 & a & 1 \\ 2 & a+2 & 1 \\ 1 & 2 & a \end{pmatrix}

How i solved it:

x1  + ax2     + x3  = y1
2x1 + (a+2)x2 + x3  =    y2
x1  + 2x2     + ax3 =      y3

Idont know if this is correct and i dont know what to do next?

Answer 1

We want that the only solution to the following is

$$\left(\begin{array}{ccc} 1 & a & 1 \\ 2 & a+2 & 1 \\ 1 & 2 & a\end{array}\right)\left(\begin{array}{c} x_1 \\ x_2 \\ x_3\end{array}\right) = \left(\begin{array}{c} 0 \\ 0 \\ 0\end{array}\right)$$

is the zero solution (i.e. $0 = x_1 = x_2 = x_3$). This says that $A$ is invertible. Rewriting this as an augmented matrix:

$$\left[\begin{array}{ccc|c} 1 & a & 1 & 0 \\ 2 & a+2 & 1 & 0 \\ 1 & 2 & a & 0\end{array}\right]$$

Row reducing ($R_2\to R_2-2R_1$ and $R_3\to R_3-R_1$), we get

$$\left[\begin{array}{ccc|c} 1 & a & 1 & 0 \\ 0 & 2-a & -1 & 0 \\ 0 & 2-a & a-1 & 0\end{array}\right]$$

Another row reduction ($R_3\to R_3-R_2$) gives

$$\left[\begin{array}{ccc|c} 1 & a & 1 & 0 \\ 0 & 2-a & -1 & 0 \\ 0 & 0 & a & 0\end{array}\right]$$

Since this is in upper triangular form, the only way for the solution to be trivial (meaning $x_1=x_2=x_3=0$) is if the pivots are nonzero (otherwise the matrix is not invertible). Can you take it from here?

Answer 2

Recall that a matrix is invertible if and only if its determinant is not equal to zero.

You can use Gaussian elimination to find the values of $a$ that will make $A$ invertible, that is, those values of $a$ such that $\det A\ne 0$. You can easily verify that you can reduce $A$ to the form

\begin{equation*} \begin{pmatrix} 1 & a & 1\\ 0 & 2-a & -1 \\ 0 & 0 & a \end{pmatrix} \end{equation*}

in a few easy steps with Gaussian elimination. Check it!

This reduced matrix and the original one have the same determinant. Moreover, the determinant of the reduced matrix equals the product of its diagonal entries. Therefore, we conclude that $$ \det A=(2-a)a. $$

So $A$ is invertible if and only if $a\ne 0,2$.