Question

Find the value of p and q such that ${{A}^{2}}+pI=qA$, where $A=\left[ \begin{matrix}
   3 & 1 \\
   7 & 5 \\
\end{matrix} \right]$.

Answer 1

Hint: First of all multiply the matrix A by itself to get the matrix ${{A}^{2}}$ then substitute the value of ${{A}^{2}}$ in the given equation and also substitute the value of “I” which is the identity matrix as $\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$ then add the left hand side of the matrices and as left hand and right hand side of the equation are equal so the matrices written on the left and right hand side is also equal and we know that when two matrices are equal so every element of the matrix is equal from that you will get equations in p and q. Solve the equations and get the value of p and q.

Complete step-by-step answer:
We have given the equation in the above problem as:
${{A}^{2}}+pI=qA$………..Eq. (1)
The matrix A is also given as:
$A=\left[ \begin{matrix}
   3 & 1 \\
   7 & 5 \\
\end{matrix} \right]$
In the equation above “I” is the identity matrix.
We know that, identity matrix is written as:
$I=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$
In the equation, ${{A}^{2}}$ is given so we are going to find ${{A}^{2}}$ matrix by multiplying A by itself.
Multiplying matrix A by itself we get,
$A.A=\left[ \begin{matrix}
   3 & 1 \\
   7 & 5 \\
\end{matrix} \right]\left[ \begin{matrix}
   3 & 1 \\
   7 & 5 \\
\end{matrix} \right]$
$\begin{align}
  & \Rightarrow {{A}^{2}}=\left[ \begin{matrix}
   3\left( 3 \right)+1\left( 7 \right) & 3\left( 1 \right)+1\left( 5 \right) \\
   7\left( 3 \right)+5\left( 7 \right) & 7\left( 1 \right)+5\left( 5 \right) \\
\end{matrix} \right] \\
 & \Rightarrow {{A}^{2}}=\left[ \begin{matrix}
   9+7 & 3+5 \\
   21+35 & 7+25 \\
\end{matrix} \right] \\
 & \Rightarrow {{A}^{2}}=\left[ \begin{matrix}
   16 & 8 \\
   56 & 32 \\
\end{matrix} \right] \\
\end{align}$
Substituting the value of A, ${{A}^{2}}$, I in eq. (1) we get,
$\left[ \begin{matrix}
   16 & 8 \\
   56 & 32 \\
\end{matrix} \right]+p\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]=q\left[ \begin{matrix}
   3 & 1 \\
   7 & 5 \\
\end{matrix} \right]$
Multiplying p by identity matrix is done by multiplying every identity element by p.
$\left[ \begin{matrix}
   16 & 8 \\
   56 & 32 \\
\end{matrix} \right]+\left[ \begin{matrix}
   p & 0 \\
   0 & p \\
\end{matrix} \right]=q\left[ \begin{matrix}
   3 & 1 \\
   7 & 5 \\
\end{matrix} \right]$……….Eq.(2)
Adding the two matrices on the left hand side of the above equation is done by adding the corresponding row and column of the first matrix with the second one.
$\left[ \begin{matrix}
   16+p & 8 \\
   56 & 32+p \\
\end{matrix} \right]=q\left[ \begin{matrix}
   3 & 1 \\
   7 & 5 \\
\end{matrix} \right]$
Multiplying q by matrix A on the right hand side of the above equation is done in the same manner as that we have done for the multiplication of p by identity.
$\left[ \begin{matrix}
   16+p & 8 \\
   56 & 32+p \\
\end{matrix} \right]=\left[ \begin{matrix}
   3q & q \\
   7q & 5q \\
\end{matrix} \right]$
The above two matrices are equal when the elements in the corresponding row and column are equal to the corresponding row and column of the other matrix.
$\begin{align}
  & 16+p=3q \\
 & 8=q \\
 & 56=7q \\
 & 32+p=5q \\
\end{align}$
From the above equation, we have got the value of q as 8. Substituting this value of q in the equation $16+p=3q$ we get,
$16+p=3\left( 8 \right)$
$\Rightarrow 16+p=24$
Subtracting 26 on both the sides of the above equation we get,
$\begin{align}
  & p=24-16 \\
 & \Rightarrow p=8 \\
\end{align}$
From the above calculations, we have got the value of p as 8 and q as 8.

Note: You can verify the values of p and q that you are getting is correct or not by substituting the value of p and q in eq. (2) and then see whether L.H.S is equal to R.H.S or not.
$\left[ \begin{matrix}
   16 & 8 \\
   56 & 32 \\
\end{matrix} \right]+\left[ \begin{matrix}
   p & 0 \\
   0 & p \\
\end{matrix} \right]=q\left[ \begin{matrix}
   3 & 1 \\
   7 & 5 \\
\end{matrix} \right]$
Substituting the value of p as -2 and q as 8 in the above equation we get,
$\left[ \begin{matrix}
   16 & 8 \\
   56 & 32 \\
\end{matrix} \right]+\left[ \begin{matrix}
   8 & 0 \\
   0 & 8 \\
\end{matrix} \right]=8\left[ \begin{matrix}
   3 & 1 \\
   7 & 5 \\
\end{matrix} \right]$
Multiplying 8 with each element of the matrix $\left[ \begin{matrix}
   3 & 1 \\
   7 & 5 \\
\end{matrix} \right]$ we get,
$\left[ \begin{matrix}
   16 & 8 \\
   56 & 32 \\
\end{matrix} \right]+\left[ \begin{matrix}
   8 & 0 \\
   0 & 8 \\
\end{matrix} \right]=\left[ \begin{matrix}
   24 & 8 \\
   56 & 40 \\
\end{matrix} \right]$
Adding the matrices present on the left hand side of the above equation we get,
$\begin{align}
  & \left[ \begin{matrix}
   16+8 & 8 \\
   56 & 32+8 \\
\end{matrix} \right]=\left[ \begin{matrix}
   24 & 8 \\
   56 & 40 \\
\end{matrix} \right] \\
 & \Rightarrow \left[ \begin{matrix}
   24 & 8 \\
   56 & 40 \\
\end{matrix} \right]=\left[ \begin{matrix}
   24 & 8 \\
   56 & 40 \\
\end{matrix} \right] \\
\end{align}$
As you can see that L.H.S is equal to R.H.S which means that the values of p and q that we have solved above are correct.