Question #278753

Let. (Matrix)

2             0             1

2             -2           2

0             4             1


(a) Find A2 and A3 , and verify that: A 3A 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

1
Expert's answer
2021-12-14T19:38:43-0500

(a) Let A = [201222041]\begin{bmatrix} 2 & 0 &1\\ 2 & -2&2\\ 0&4&1 \end{bmatrix}

So A²

= [201222041][201222041]\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}


= [4+0+00+0+42+0+144+00+4+824+20+8+008+40+8+1]\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} = [4430120849]\begin{bmatrix} 4 & 4 &3\\ 0 & 12&0\\ 8&-4&9 \end{bmatrix}

A³ = A²A

=[4430120849]\begin{bmatrix} 4 & 4 &3\\ 0 & 12&0\\ 8&-4&9 \end{bmatrix} [201222041]\begin{bmatrix} 2 & 0 &1\\ 2 & -2&2\\ 0&4&1 \end{bmatrix}

= [8+8+008+124+8+30+24+0024+00+24+0168+00+8+3688+9]\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}

= [164152424248449]\begin{bmatrix} 16& 4&15\\ 2 4& -24&24\\ 8&44&9 \end{bmatrix}

Now A³ - A² =

[164152424248449]\begin{bmatrix} 16& 4&15\\ 2 4& -24&24\\ 8&44&9 \end{bmatrix} [4430120849]-\begin{bmatrix} 4 & 4 &3\\ 0 & 12&0\\ 8&-4&9 \end{bmatrix}

= [120122436240480]\begin{bmatrix} 12 & 0 &12\\ 24 & -36&24\\ 0&48&0 \end{bmatrix}


12A =

12[201222041]12\begin{bmatrix} 2 & 0 &1\\ 2 & -2&2\\ 0&4&1 \end{bmatrix} =[2401224242404812]\begin{bmatrix} 2 4& 0 &12\\ 24 & -24&24\\ 0&48&12 \end{bmatrix}

A³ - A² - 12A =

[120122436240480]\begin{bmatrix} 12 & 0 &12\\ 24 & -36&24\\ 0&48&0 \end{bmatrix} [2401224242404812]-\begin{bmatrix} 2 4& 0 &12\\ 24 & -24&24\\ 0&48&12 \end{bmatrix}

= [120001200012]\begin{bmatrix} -12& 0 &0\\ 0 & -12&0\\ 0&0&-12 \end{bmatrix}

= 12[100010001]=12I-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 = [4430120849]\begin{bmatrix} 4 & 4 &3\\ 0 & 12&0\\ 8&-4&9 \end{bmatrix} - [201222041]\begin{bmatrix} 2 & 0 &1\\ 2 & -2&2\\ 0&4&1 \end{bmatrix} -[120001200012]\begin{bmatrix} 12 & 0 &0\\ 0 & 12&0\\ 0&0&12 \end{bmatrix}

= [421240031002012+2120208004409112]\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} =[1042222884]\begin{bmatrix} -10 & 4 &2\\ -2 & 2&-2\\ 8&-8&-4 \end{bmatrix}

A-1 = 16[521111442]\frac{1}{6}\begin{bmatrix} 5 & -2 &-1\\ 1 & -1&1\\ -4&4&2 \end{bmatrix}




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