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3&{ - 3} \\

{ - 3}&3

\end{array}} \right)\] and \[{A^2} = \lambda A\], then write the value of \[\lambda \].

Answer
Verified

Hint- We cannot multiply any matrices. The conditions of the multiplication are:

The number of columns of the first matrix must equal the number of rows of the second matrix.

And the result will have the same number of rows as the first matrix and the same number of columns as the second matrix.

In general, to multiply \[m \times n\] a matrix by a \[n \times p\] matrix, the \[n\]’s must be the same, and the result is a \[m \times p\] matrix.

Multiply a number with a scalar number:

To multiply a number by a number is easy. Every element will be multiplied by the given scalar number.

Multiply a matrix by a matrix:

It is called the dot product. In this process, each element of the first row of the first matrix will be multiplied by the respective elements of the first column of the second matrix.

__Complete step by step answer:__

It is given that the matrix is,

\[A = \left( {\begin{array}{*{20}{c}}

3&{ - 3} \\

{ - 3}&3

\end{array}} \right)\]

Here, the given matrix \[A\] is of order \[2 \times 2\]. So, the matrix \[{A^2}\]will be also \[2 \times 2\] order.

So,

\[{A^2} = \left( {\begin{array}{*{20}{c}}

3&{ - 3} \\

{ - 3}&3

\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}

3&{ - 3} \\

{ - 3}&3

\end{array}} \right)\]

As per the process, we will multiply each element of the first row of the first matrix by the respective elements of the first column of the second matrix.

\[{A^2} = \left( {\begin{array}{*{20}{c}}

{9 + 9}&{ - 9 - 9} \\

{ - 9 - 9}&{9 + 9}

\end{array}} \right)\]

Simplifying we get,

\[{A^2} = \left( {\begin{array}{*{20}{c}}

{18}&{ - 18} \\

{ - 18}&{18}

\end{array}} \right)\]

Now we will represent the matrix \[{A^2}\]in terms of \[A\].

So, we can write that,

\[{A^2} = \left( {\begin{array}{*{20}{c}}

{18}&{ - 18} \\

{ - 18}&{18}

\end{array}} \right) = 6\left( {\begin{array}{*{20}{c}}

3&{ - 3} \\

{ - 3}&3

\end{array}} \right)\]

So, we can write as,

\[{A^2} = 6A\]…. (1)

Again, it is given that, \[{A^2} = \lambda A\]…. (2)

Comparing equation (1) and (2) we get,

\[\lambda = 6\]

Hence, the value of \[\lambda = 6\].

Note – For the multiplication of matrices if the number of columns of the first matrix is not equal to the number of rows of the second matrix, the multiplication is not possible.

The number of columns of the first matrix must equal the number of rows of the second matrix.

And the result will have the same number of rows as the first matrix and the same number of columns as the second matrix.

In general, to multiply \[m \times n\] a matrix by a \[n \times p\] matrix, the \[n\]’s must be the same, and the result is a \[m \times p\] matrix.

Multiply a number with a scalar number:

To multiply a number by a number is easy. Every element will be multiplied by the given scalar number.

Multiply a matrix by a matrix:

It is called the dot product. In this process, each element of the first row of the first matrix will be multiplied by the respective elements of the first column of the second matrix.

It is given that the matrix is,

\[A = \left( {\begin{array}{*{20}{c}}

3&{ - 3} \\

{ - 3}&3

\end{array}} \right)\]

Here, the given matrix \[A\] is of order \[2 \times 2\]. So, the matrix \[{A^2}\]will be also \[2 \times 2\] order.

So,

\[{A^2} = \left( {\begin{array}{*{20}{c}}

3&{ - 3} \\

{ - 3}&3

\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}

3&{ - 3} \\

{ - 3}&3

\end{array}} \right)\]

As per the process, we will multiply each element of the first row of the first matrix by the respective elements of the first column of the second matrix.

\[{A^2} = \left( {\begin{array}{*{20}{c}}

{9 + 9}&{ - 9 - 9} \\

{ - 9 - 9}&{9 + 9}

\end{array}} \right)\]

Simplifying we get,

\[{A^2} = \left( {\begin{array}{*{20}{c}}

{18}&{ - 18} \\

{ - 18}&{18}

\end{array}} \right)\]

Now we will represent the matrix \[{A^2}\]in terms of \[A\].

So, we can write that,

\[{A^2} = \left( {\begin{array}{*{20}{c}}

{18}&{ - 18} \\

{ - 18}&{18}

\end{array}} \right) = 6\left( {\begin{array}{*{20}{c}}

3&{ - 3} \\

{ - 3}&3

\end{array}} \right)\]

So, we can write as,

\[{A^2} = 6A\]…. (1)

Again, it is given that, \[{A^2} = \lambda A\]…. (2)

Comparing equation (1) and (2) we get,

\[\lambda = 6\]

Hence, the value of \[\lambda = 6\].

Note – For the multiplication of matrices if the number of columns of the first matrix is not equal to the number of rows of the second matrix, the multiplication is not possible.

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