Math, asked by yakannatokala9, 10 days ago

answer for this question with a correct solution

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Answered by hukam0685
3

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

Answered by mathdude500
4

\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

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