Question

answer for this question with a correct solution
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Answer 1

Step-by-step explanation:

Given:

$A=\left[\begin{array}{ccc}1&1&3\\5&2&6\\-2&-1&-3\end{array}\right]$

To find: A³

Solution:

It is question of matrix multiplication.

Matrix Multiplication: It is only possible when columns of first matrix and rows of second matrix are equal.

Matrix Multiplication not always commutative.

$A^2=\left[\begin{array}{ccc}1&1&3\\5&2&6\\-2&-1&-3\end{array}\right]\times\left[\begin{array}{ccc}1&1&3\\5&2&6\\-2&-1&-3\end{array}\right]$

$A^2=\left[\begin{array}{ccc}1+5-6&1+2-3&3+6-9\\5+10-12&5+4-6&15+12-18\\-2-5+6&-2-2+3&-6-6+9\end{array}\right]\\\\A^2=\left[\begin{array}{ccc}0&0&0\\3&3&9\\-1&-1&-3\end{array}\right]$

Again multiply A² with A to get A³.

$A^3=\left[\begin{array}{ccc}0&0&0\\3&3&9\\-1&-1&-3\end{array}\right]\times\left[\begin{array}{ccc}1&1&3\\5&2&6\\-2&-1&-3\end{array}\right]$

$A^3=\left[\begin{array}{ccc}0&0&0\\3+15-18&3+6-9&9+18-27\\-1-5+6&-1-2+3&-3-6+9\end{array}\right]\\\\A^3=\left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right]$

Final answer:

A³ is a Null matrix.

Hope it helps you.

To learn more on brainly:

Find the inverse of the matrix ( if exists ) and state the reason if it doesn't exists.

[tex]\left[ \begin{array} {ccc}...

https://brainly.in/question/45794130

Answer 2

$\large\underline{\sf{Solution-}}$

Given matrix is

$\rm :\longmapsto\: \begin{gathered}\sf A=\left[\begin{array}{ccc}1&1&3\\5&2&6\\ - 2&-1& - 3\end{array}\right]\end{gathered}$

Consider,

$\red{\rm :\longmapsto\: {A}^{2}}$

$\rm \:  =  \: A \times A$

$\\ \rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}1&1&3\\5&2&6\\ - 2&-1& - 3\end{array}\right]\end{gathered}\begin{gathered}\sf \left[\begin{array}{ccc}1&1&3\\5&2&6\\ - 2&-1& - 3\end{array}\right]\end{gathered}$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}1 + 5 - 6&1 + 2 - 3&3 + 6 - 9\\5 + 10 - 12&5 + 4 - 6&15 + 12 - 18\\ - 2 - 5 + 6&-2 - 2 + 3& - 6 - 6 + 9\end{array}\right]\end{gathered}$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\3&3&9\\ - 1&-1& - 3\end{array}\right]\end{gathered}$

$\\ \bf\implies \: {A}^{2} = \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\3&3&9\\ - 1&-1& - 3\end{array}\right]\end{gathered}$

Now, Consider

$\red{\rm :\longmapsto\: {A}^{3}}$

$\rm \:  =  \: {A}^{2} \times A$

$\\ \rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\3&3&9\\ - 1&-1& - 3\end{array}\right]\end{gathered} \times \begin{gathered}\sf \left[\begin{array}{ccc}1&1&3\\5&2&6\\ - 2&-1& - 3\end{array}\right]\end{gathered}$

$\\ \rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\3 + 15 - 18&3 + 6 - 9&9 + 18 - 27\\ - 1 - 5 + 6&-1 - 2 + 3& - 3 - 6 + 9\end{array}\right]\end{gathered}$

$\\ \rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\0&0&0\\ 0&0&0\end{array}\right]\end{gathered}$

$\\ \bf\implies \: {A}^{3}   =  \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\0&0&0\\ 0&0&0\end{array}\right]\end{gathered}$

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Learn More :-

Matrix multiplication is defined when number of columns of pre multiplier is equal to the number of rows of post multiplier.

• Matrix multiplication may or may not be Commutative.

• Matrix multiplication is Associative. i.e (AB)C = A(BC)

• Matrix multiplication is Distributive. i.e. A ( B + C ) = AB + AC