Question

If \[A=\left[ \begin{matrix}
   2 & 3 \\
   -1 & 2 \\
\end{matrix} \right]\] and \[f\left( x \right)={{x}^{2}}-4x+7\], show that \[f\left( A \right)=0\]. Use this to find \[{{A}^{3}}\] and \[{{A}^{5}}\].

Answer 1

Hint: This question is from the topic of matrix. We will first find out the value of \[{{A}^{2}}\]. After that, we will put the value of x as the matrix A in the equation \[f\left( x \right)={{x}^{2}}-4x+7\] and prove that the value of \[f\left( A \right)\] is zero. And, after that we will find the value of \[{{A}^{3}}\] and \[{{A}^{5}}\] by using \[A=\left[ \begin{matrix}
   2 & 3 \\
   -1 & 2 \\
\end{matrix} \right]\].

Complete step by step answer:
Let us solve this question.
In this question, we have given that \[A=\left[ \begin{matrix}
   2 & 3 \\
   -1 & 2 \\
\end{matrix} \right]\], and a function of x is given as \[f\left( x \right)={{x}^{2}}-4x+7\]. This question has asked us to prove that \[f\left( A \right)=0\] from the given function \[f\left( x \right)={{x}^{2}}-4x+7\] and it is asked to find the value of \[{{A}^{3}}\] and \[{{A}^{5}}\].
Let us first find the value \[{{A}^{2}}\].
The term \[{{A}^{2}}\] can also be written as
\[{{A}^{2}}=A\times A\]
After putting the value of A as \[\left[ \begin{matrix}
   2 & 3 \\
   -1 & 2 \\
\end{matrix} \right]\], we can write the above equation as
\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}
   2 & 3 \\
   -1 & 2 \\
\end{matrix} \right]\times \left[ \begin{matrix}
   2 & 3 \\
   -1 & 2 \\
\end{matrix} \right]\]
Here, in right hand side of the equation, the number of columns of the first matrix is equal to number of rows of the second matrix. So, we can multiply them.
After the multiplying the matrices, we can write
\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}
   2\times 2+3\times \left( -1 \right) & 2\times 3+3\times 2 \\
   \left( -1 \right)\times 2+2\times \left( -1 \right) & \left( -1 \right)\times 3+2\times 2 \\
\end{matrix} \right]\]
The above multiplied matrix can also be written as
\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}
   4-3 & 6+6 \\
   -2-2 & -3+4 \\
\end{matrix} \right]\]
The above equation can also be written as
\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}
   1 & 12 \\
   -4 & 1 \\
\end{matrix} \right]\]
A function of x is given in the question which is \[f\left( x \right)={{x}^{2}}-4x+7\].
In this function, we will put the value of x as the matrix A.
After putting the value of x as the matrix A, we get
\[f\left( A \right)={{A}^{2}}-4A+7\]
As we have found above that \[{{A}^{2}}=\left[ \begin{matrix}
   1 & 12 \\
   -4 & 1 \\
\end{matrix} \right]\] and it is given in the question that \[A=\left[ \begin{matrix}
   2 & 3 \\
   -1 & 2 \\
\end{matrix} \right]\], so using these values in the above equation, we get
\[\Rightarrow f\left( A \right)=\left[ \begin{matrix}
   1 & 12 \\
   -4 & 1 \\
\end{matrix} \right]-4\left[ \begin{matrix}
   2 & 3 \\
   -1 & 2 \\
\end{matrix} \right]+7\]
As we know that the identity matrix for \[2\times 2\] is \[{{I}_{2\times 2}}=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\], and the value of identity matrix is 1.
So, we can write
\[\Rightarrow f\left( A \right)=\left[ \begin{matrix}
   1 & 12 \\
   -4 & 1 \\
\end{matrix} \right]-4\left[ \begin{matrix}
   2 & 3 \\
   -1 & 2 \\
\end{matrix} \right]+7\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\]
After multiplying 4 to all the elements in the second matrix and 7 to all the elements in 3rd matrix, we get
\[\Rightarrow f\left( A \right)=\left[ \begin{matrix}
   1 & 12 \\
   -4 & 1 \\
\end{matrix} \right]-\left[ \begin{matrix}
   8 & 12 \\
   -4 & 8 \\
\end{matrix} \right]+\left[ \begin{matrix}
   7 & 0 \\
   0 & 7 \\
\end{matrix} \right]\]
Now, after adding the matrices, we can write the above equation as
\[\Rightarrow f\left( A \right)=\left[ \begin{matrix}
   1-8+7 & 12-12+0 \\
   -4+4+0 & 1-8+7 \\
\end{matrix} \right]\]
The above equation can also be written as
\[\Rightarrow f\left( A \right)=\left[ \begin{matrix}
   0 & 0 \\
   0 & 0 \\
\end{matrix} \right]\]
Hence, the term \[f\left( A \right)\] is the matrix who’s all the elements are zero.
So, we can say that \[f\left( A \right)=0\]
Now, we will find out the value of \[{{A}^{3}}\] and \[{{A}^{5}}\].
\[{{A}^{3}}={{A}^{2}}A=\left[ \begin{matrix}
   1 & 12 \\
   -4 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   2 & 3 \\
   -1 & 2 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1\times 2+12\times \left( -1 \right) & 1\times 3+12\times 2 \\
   \left( -4 \right)\times 2+1\times \left( -1 \right) & \left( -4 \right)\times 3+1\times 2 \\
\end{matrix} \right]\]
The above equation can also be written as
\[\Rightarrow {{A}^{3}}=\left[ \begin{matrix}
   -10 & 27 \\
   -9 & -10 \\
\end{matrix} \right]\]
For the value of \[{{A}^{5}}\], we will write
\[{{A}^{5}}={{A}^{3}}{{A}^{2}}=\left[ \begin{matrix}
   1 & 12 \\
   -4 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   -10 & 27 \\
   -9 & -10 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1\times \left( -10 \right)+12\times \left( -9 \right) & 1\times 27+12\times \left( -10 \right) \\
   \left( -4 \right)\times \left( -10 \right)+1\times \left( -9 \right) & \left( -4 \right)\times 27+1\times \left( -10 \right) \\
\end{matrix} \right]\]
The above equation can also be written as
\[{{A}^{5}}=\left[ \begin{matrix}
   -118 & -93 \\
   31 & -118 \\
\end{matrix} \right]\]
Hence, we have got the value of \[{{A}^{5}}\] as \[\left[ \begin{matrix}
   -118 & -93 \\
   31 & -118 \\
\end{matrix} \right]\] and \[{{A}^{3}}\] as \[\left[ \begin{matrix}
   -10 & 27 \\
   -9 & -10 \\
\end{matrix} \right]\]

Note: As this question is from the topic of matrix, so we should have a better knowledge in that topic. We should know how to multiply two matrices. Always remember that whenever we are going to multiply the two matrices, then the number of columns of the first matrix should be equal to the number of rows of the second matrix. We should know about the identity matrix. Identity matrix is the square matrix of the order \[n\times n\] and all elements are zero except the diagonal element. The diagonal elements of the identity matrix are always 1.