mjhjmtu

I'm given $$beginbmatrix3&x\-2&-3\endbmatrix$$
and am asked to find x such that it's inverse would equal itself. To attempt this I first tried to put the question into an augmented matrix and got this:

$$beginbmatrix1&x/3&1/3&0\0&(x/3)-(3/2)&1/3&1/2\ endbmatrix$$

I found that my answer was wrong so I tried:

$$beginbmatrix3&x\-2&-3\endbmatrix$$
times
$$beginbmatrixx_1&x_2\x_3&x_4\endbmatrix$$ to try and solve for x but found similar dissatisfactory results.

The answer is listed as x = 4; how might I go about solving this? Did I just make a mistake with my methods or is this the entirely wrong way about?

Hint: what is $$ pmatrix3 & xcr -2 & -3^2 $$
and what do you need it to be?

The determinant of the matrix is
$$
beginvmatrix
3 & x \ -2 & -3
endvmatrix=2x-9.
$$
Assuming this matrix has an inverse, the determinant of that inverse will be the reciprocal of the determinant. So, we must have
$$
9+2x=frac12x-9Rightarrow2x-9=pm1.
$$
Now, $2x-9=1$ when $x=5$, and $2x-9=-1$ when $x=4$. So, these are the only two possible values. From here, you can check these two values to see what works.

If we have a $ 2 times 2 $ matrix $A$ then the inverse can be given by

$$ A^-1 = frac1ad-bcbeginbmatrix d & -b \ -c & a endbmatrix tag1 $$

Then with your matrix $A$

$$A = beginbmatrix 3 & x \ -2 & -3 endbmatrix tag2 $$

$$ A^-1 = frac1-9+2xbeginbmatrix -3 & -x \ 2 & 3 endbmatrix tag3 $$

You need to find where they are the same. I won't do it all I guess.

$A^-1 = A$ is equivalent to $A^2 = I$ or $A^2-I = 0$.

Hence the polynomial $lambda^2-1 = (lambda-1)(lambda+1)$ must annihilate $A$. Clearly $A ne pm I$ so the characteristic polynomial of $A$ must be equal to $lambda^2-1$.

We have

$$det(A - lambda I) = beginvmatrix 3-lambda & x \ -2 & -3-lambdaendvmatrix = lambda^2 - 9 + 2x$$

Hence $2x-9 = -1$ which gives $x = 4$.

Both methods you attempted must give correct answer.

Method 1. Finding the inverse from the augmented matrix:
$$ left[
beginarraycc
1&0&3&x\
0&1&-2&-3
endarray
right] stackrelR_1/3=
left[beginarraycc
frac13&0&1&frac x3\
0&1&-2&-3
endarray
right] stackrel2R_1+R_2to R_2=\
left[beginarraycc
frac13&0&1&frac x3\
frac23&1&0&frac 2x3-3
endarray
right] stackrelfrac32x-9cdot R_2=
left[beginarraycc
frac13&0&1&frac x3\
frac22x-9&frac32x-9&0&1
endarray
right] stackrel-fracx3cdot R_2+R_1to R_1=\
left[beginarraycc
frac13-frac2x3(2x-9)&-fracx2x-9&1&0\
frac22x-9&frac32x-9&0&1
endarray
right].$$
So, it must be:
$$left[beginarraycc
frac13-frac2x3(2x-9)&-fracx2x-9\
frac22x-9&frac32x-9
endarrayright]=
left[beginarraycc
3&x\
-2&-3
endarrayright] Rightarrow x=4.$$
Method 2. Multiply by its inverse (by condition it must be equal to the original matrix and remember the result is an identity matrix: $Acdot A^-1=I$):
$$left[beginarraycc
3&x\
-2&-3
endarrayright]
left[beginarraycc
3&x\
-2&-3
endarrayright]=
left[beginarraycc
1&0\
0&1
endarrayright] Rightarrow \
left[beginarraycc
9-2x&0\
0&-2x+9
endarrayright]=
left[beginarraycc
1&0\
0&1
endarrayright] Rightarrow x=4.$$