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I'm given $$\begin{bmatrix}3&x\\-2&-3\\\end{bmatrix}$$ 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:

$$\begin{bmatrix}1&x/3&1/3&0\\0&(x/3)-(3/2)&1/3&1/2\\ \end{bmatrix}$$

I found that my answer was wrong so I tried:

$$\begin{bmatrix}3&x\\-2&-3\\\end{bmatrix}$$ times $$\begin{bmatrix}x_1&x_2\\x_3&x_4\\\end{bmatrix}$$ 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?

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5 Answers 5

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Hint: what is $$ \pmatrix{3 & x\cr -2 & -3}^2 $$ and what do you need it to be?

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    $\begingroup$ \begin{bmatrix}9-2x&3x-3x\\-6+6&-2x+9\\\end{bmatrix} I can see how this makes it more clear that x is 4 by relating A(A^-1) = I_n. I am surprised I didn't think of just doing A^2 = I_n; thank you for your guidance. $\endgroup$
    – Charlatan
    Commented Sep 13, 2018 at 3:20
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The determinant of the matrix is $$ \begin{vmatrix} 3 & x \\ -2 & -3 \end{vmatrix}=2x-9. $$ Assuming this matrix has an inverse, the determinant of that inverse will be the reciprocal of the determinant. So, we must have $$ 9+2x=\frac{1}{2x-9}\Rightarrow2x-9=\pm1. $$ 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.

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

$$ A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \tag{1} $$

Then with your matrix $A$

$$A = \begin{bmatrix} 3 & x \\ -2 & -3 \end{bmatrix} \tag{2} $$

$$ A^{-1} = \frac{1}{-9+2x}\begin{bmatrix} -3 & -x \\ 2 & 3 \end{bmatrix} \tag{3} $$

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

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  • $\begingroup$ I didn't even think of trying this method but thank you for pointing this out. $\endgroup$
    – Charlatan
    Commented Sep 13, 2018 at 3:26
  • $\begingroup$ you're welcome. $\endgroup$
    – user3417
    Commented Sep 13, 2018 at 3:30
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$A^{-1} = A$ 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

$$\det(A - \lambda I) = \begin{vmatrix} 3-\lambda & x \\ -2 & -3-\lambda\end{vmatrix} = \lambda^2 - 9 + 2x$$

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

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Both methods you attempted must give correct answer.

Method 1. Finding the inverse from the augmented matrix: $$ \left[ \begin{array}{cc|cc} 1&0&3&x\\ 0&1&-2&-3 \end{array} \right] \stackrel{R_1/3}= \left[\begin{array}{cc|cc} \frac13&0&1&\frac x3\\ 0&1&-2&-3 \end{array} \right] \stackrel{2R_1+R_2\to R_2}=\\ \left[\begin{array}{cc|cc} \frac13&0&1&\frac x3\\ \frac23&1&0&\frac {2x}3-3 \end{array} \right] \stackrel{\frac{3}{2x-9}\cdot R_2}= \left[\begin{array}{cc|cc} \frac13&0&1&\frac x3\\ \frac{2}{2x-9}&\frac{3}{2x-9}&0&1 \end{array} \right] \stackrel{-\frac{x}{3}\cdot R_2+R_1\to R_1}=\\ \left[\begin{array}{cc|cc} \frac13-\frac{2x}{3(2x-9)}&-\frac{x}{2x-9}&1&0\\ \frac{2}{2x-9}&\frac{3}{2x-9}&0&1 \end{array} \right].$$ So, it must be: $$\left[\begin{array}{cc} \frac13-\frac{2x}{3(2x-9)}&-\frac{x}{2x-9}\\ \frac{2}{2x-9}&\frac{3}{2x-9} \end{array}\right]= \left[\begin{array}{cc} 3&x\\ -2&-3 \end{array}\right] \Rightarrow x=4.$$ Method 2. Multiply by its inverse (by condition it must be equal to the original matrix and remember the result is an identity matrix: $A\cdot A^{-1}=I$): $$\left[\begin{array}{cc} 3&x\\ -2&-3 \end{array}\right] \left[\begin{array}{cc} 3&x\\ -2&-3 \end{array}\right]= \left[\begin{array}{cc} 1&0\\ 0&1 \end{array}\right] \Rightarrow \\ \left[\begin{array}{cc} 9-2x&0\\ 0&-2x+9 \end{array}\right]= \left[\begin{array}{cc} 1&0\\ 0&1 \end{array}\right] \Rightarrow x=4.$$

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