Answer to Question #278753 in Linear Algebra for Petra

Question

Let. (Matrix)

2             0             1

2             -2           2

0             4             1

(a) Find ^A2 and ^A3 , and verify that: ^{A 3} − ^{A 2} − 12A = −12i

holds, where I stands for the identity matrix.

(b) Find ^A−1 by multiplying the equation above on both sides by ^A−1

Answer 1

(a) Let A = $\begin{bmatrix} 2 & 0 &1\\ 2 & -2&2\\ 0&4&1 \end{bmatrix}$

So A²

= $\begin{bmatrix} 2 & 0 &1\\ 2 & -2&2\\ 0&4&1 \end{bmatrix}\begin{bmatrix} 2 & 0 &1\\ 2 & -2&2\\ 0&4&1 \end{bmatrix}$

= $\begin{bmatrix} 4+0+0 & 0+0+4 &2+0+1\\ 4-4+0 & 0+4+8&2-4+2\\ 0+8+0&0-8+4&0+8+1 \end{bmatrix}$ = $\begin{bmatrix} 4 & 4 &3\\ 0 & 12&0\\ 8&-4&9 \end{bmatrix}$

A³ = A²A

=$\begin{bmatrix} 4 & 4 &3\\ 0 & 12&0\\ 8&-4&9 \end{bmatrix}$ $\begin{bmatrix} 2 & 0 &1\\ 2 & -2&2\\ 0&4&1 \end{bmatrix}$

= $\begin{bmatrix} 8+8+0 & 0 -8+12&4+8+3\\ 0+24+0 & 0-24+0&0+24+0\\ 16-8+0&0+8+36&8-8+9 \end{bmatrix}$

= $\begin{bmatrix} 16& 4&15\\ 2 4& -24&24\\ 8&44&9 \end{bmatrix}$

Now A³ - A² =

$\begin{bmatrix} 16& 4&15\\ 2 4& -24&24\\ 8&44&9 \end{bmatrix}$ $-\begin{bmatrix} 4 & 4 &3\\ 0 & 12&0\\ 8&-4&9 \end{bmatrix}$

= $\begin{bmatrix} 12 & 0 &12\\ 24 & -36&24\\ 0&48&0 \end{bmatrix}$

12A =

$12\begin{bmatrix} 2 & 0 &1\\ 2 & -2&2\\ 0&4&1 \end{bmatrix}$ =$\begin{bmatrix} 2 4& 0 &12\\ 24 & -24&24\\ 0&48&12 \end{bmatrix}$

A³ - A² - 12A =

$\begin{bmatrix} 12 & 0 &12\\ 24 & -36&24\\ 0&48&0 \end{bmatrix}$ $-\begin{bmatrix} 2 4& 0 &12\\ 24 & -24&24\\ 0&48&12 \end{bmatrix}$

= $\begin{bmatrix} -12& 0 &0\\ 0 & -12&0\\ 0&0&-12 \end{bmatrix}$

= $-12\begin{bmatrix} 1& 0 &0\\ 0 & 1&0\\ 0&0&1 \end{bmatrix}= -12I$

Relation is verified.

b)

A³ - A² - 12A =-12I

Multiplying both sides by A^-1

A³A^-1- A²A^-1-12AA^-1=-12IA^-1

=> A²I - AI - 12I = -12A^-1

=> A² - A - 12I = -12A^-1

=> -12A^-1 = $\begin{bmatrix} 4 & 4 &3\\ 0 & 12&0\\ 8&-4&9 \end{bmatrix}$ - $\begin{bmatrix} 2 & 0 &1\\ 2 & -2&2\\ 0&4&1 \end{bmatrix}$ -$\begin{bmatrix} 12 & 0 &0\\ 0 & 12&0\\ 0&0&12 \end{bmatrix}$

= $\begin{bmatrix} 4-2-12 & 4-0-0 &3-1-0\\ 0-2-0 & 12+2-12&0-2-0\\ 8-0-0&-4-4-0&9-1-12 \end{bmatrix}$ =$\begin{bmatrix} -10 & 4 &2\\ -2 & 2&-2\\ 8&-8&-4 \end{bmatrix}$

A^-1 = $\frac{1}{6}\begin{bmatrix} 5 & -2 &-1\\ 1 & -1&1\\ -4&4&2 \end{bmatrix}$