Finding the inverse of $A$

Question

Find the inverse of $$A =\left[\begin{matrix}0 & 1 & 0 & 0\\ 0& 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ a & b & c & d\end{matrix}\right]$$

My attempts:

$$A^{-1} = \frac {\operatorname{adj}A}{\det A}$$

As I can find the $\det A$ that is $\det A = -b$. Here, how can I find the inverse? Is there any easy method/ or easy procedure to find the inverse of $A$?

Here is something that will work: augment the matrix with the identity matrix $[A|I]$. Then row reduce so that $A$ becomes the identity. The matrix on the right will be the inverse of $A$. — Aaron, Jul 09 '18 at 10:16

score 3 · Accepted Answer · answered Jul 09 '18 at 10:15

I think that the formula that you mentioned is the best way for this matrix with lots of zeros. Note that you made a mistake: $\det A=-a$, not $-b$. And$$\operatorname{Adj}A=\begin{bmatrix}b & c & d & -1 \\ -a & 0 & 0 & 0 \\ 0 & -a & 0 & 0 \\ 0 & 0 & -a & 0\end{bmatrix}.$$Therefore$$A^{-1}=\begin{bmatrix}-\frac{b}{a} & -\frac{c}{a} & -\frac{d}{a} & \frac{1}{a} \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\end{bmatrix}.$$

score 3 · Answer 2 · answered Jul 09 '18 at 11:18

Let

$$\left[\begin{array}{c|ccc} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \hline a & b & c & d\end{array}\right]^{-1} := \left[\begin{array}{c|c} \mathrm p_{11}^\top & p_{12}\\ \hline \mathrm P_{21} & \mathrm p_{22}\end{array}\right]$$

Let $\mathrm r^\top := \begin{bmatrix} b & c & d\end{bmatrix}$. From

$$\left[\begin{array}{c|c} 0_3 & \mathrm I_3\\ \hline a & \mathrm r^\top\end{array}\right] \left[\begin{array}{c|c} \mathrm p_{11}^\top & p_{12}\\ \hline \mathrm P_{21} & \mathrm p_{22}\end{array}\right] = \left[\begin{array}{c|c} \mathrm I_3 & 0_3\\ \hline 0_3^\top & 1\end{array}\right]$$

we obtain $\mathrm P_{21} = \mathrm I_3$, then $\mathrm p_{22} = 0_3$, then $\mathrm p_{11}^\top = -\frac{1}{a} \mathrm r^\top$ and, lastly, $p_{12} = \frac 1a$. Thus,

$$\left[\begin{array}{c|ccc} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \hline a & b & c & d\end{array}\right]^{-1} = \color{blue}{\left[\begin{array}{ccc|c} - \frac{b}{a} & - \frac{c}{a} & - \frac{d}{a} & \frac{1}{a}\\ \hline1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\end{array}\right]}$$

score 2 · Answer 3 · answered Jul 09 '18 at 10:18

2

I suggest Gauss-Jordan method

$$[A|I]\to [I|A^{-1}]$$

Refer also to Method for Finding Matrix-Inverse Through Gauss-Jordan?

answered Jul 09 '18 at 10:18

user

154,566

1

I didn't know that that method had that name (or any other name)! – José Carlos Santos Jul 09 '18 at 10:22
1

It is the name given also by Gilbert Strang in his course if I’m not wrong ;) – user Jul 09 '18 at 10:29

score 1 · Answer 4 · answered Jul 09 '18 at 11:08

You're wrong: expanding by the first column, you obtain at once $\;\det A=-a$. We'll suppose $a\ne0$.

To find the inverse you can perform column reduction: \begin{align} \begin{bmatrix} \begin{array}{cccc|cccc} 0&1&0&0&1&0&0&0 \\ 0&0&1&0 &0&1&0&0\\ 0&0&0&1&0&0&1&0\\ a&b&c&d&0&0&0&1 \end{array} \end{bmatrix}&\xrightarrow[\scriptstyle\to\,\text{(c$_2$,c$_3$,c$_4$, c$_1$)}]{\text{(c$_1$,c$_2$,c$_3$,c$_4$)}} \begin{bmatrix} \begin{array}{cccc|cccc} 1&0&0&0&0&0&0&1 \\ 0&1&0&0 &1&0&0&0\\ 0&0&1&0&0&1&0&0\\ b&c&d&a&0&0&1&0 \end{array} \end{bmatrix}\xrightarrow{\scriptstyle\tfrac1a\text{c}_4\,\to\,\text{c}_4}\\[1ex] \begin{bmatrix} \begin{array}{cccc|cccc} 1&0&0&0&0&0&0&\frac1a \\ 0&1&0&0 &1&0&0&0\\ 0&0&1&0&0&1&0&0\\ b&c&d&1&0&0&1&0 \end{array} \end{bmatrix}&\xrightarrow[\scriptstyle\text{(c}_2-c\,\text{c}_4,\text{c}_3-d\,\text{c}_4)]{(\text{c}_1,\text{c}_2,\text{c}_3)\,\to\,(\text{c}_1-b\,\text{c}_4,\:} \begin{bmatrix} \begin{array}{cccc|cccc} \!\!1&0 &0&0&-\frac ba&-\frac ca& -\frac da&\frac1a \\ 0&1&0&0 &1&0&0&0\\ 0&0&1&0&0&1&0&0\\ 0&0&0&1&0&0&1&0 \end{array} \end{bmatrix} \end{align}

score 1 · Answer 5 · answered Jul 09 '18 at 16:16

Let

$$\mathrm M := \begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ a & b & c & d\end{bmatrix}$$

which is the companion matrix to the following (characteristic) polynomial

$$p (s) := s^4 - d s^3 - c s^2 - b s - a$$

Using the Cayley-Hamilton theorem,

$$\mathrm M^4 - d \,\mathrm M^3 - c \,\mathrm M^2 - b\,\mathrm M - a \,\mathrm I_4 = \mathrm O_4$$

Multiplying both sides by $\mathrm M^{-1}$, we obtain

$$\mathrm M^3 - d \,\mathrm M^2 - c \,\mathrm M - b\,\mathrm I_4 - a \mathrm M^{-1} = \mathrm O_4$$

Assuming that $a \neq 0$,

$$\mathrm M^{-1} = \frac{1}{a} \, \mathrm M^3 - \frac{d}{a} \,\mathrm M^2 - \frac{c}{a} \,\mathrm M - \frac{b}{a}\,\mathrm I_4$$

Using SymPy:

>>> a, b, c, d = symbols('a b c d')
>>> M = Matrix([[0,1,0,0],
                [0,0,1,0],
                [0,0,0,1],
                [a,b,c,d]])
>>> simplify((1/a)*M**3 - (d/a)*M**2 - (c/a)*M - (b/a)*eye(4))
Matrix([[-b/a, -c/a, -d/a, 1/a],
        [   1,    0,    0,   0],
        [   0,    1,    0,   0],
        [   0,    0,    1,   0]])

Finding the inverse of $A$

5 Answers5