Question

For the matrix $A=\left[ \begin{matrix}
   3 & 2 \\
   1 & 1 \\
\end{matrix} \right]$, find the numbers a and b such that \[{{A}^{2}}+aA+bI=0\]

Answer 1

Hint: Find ${{A}^{2}}$ by multiplying the matrix A to A. multiplication of any matrix to any other matrix is done by multiplying the column of one matrix to the corresponding row elements. If two matrices are equal then all of the elements of one matrix are equal to the corresponding elements of the second matrix.

Complete step-by-step solution -
We are given matrix A as
 $A=\left[ \begin{matrix}
   3 & 2 \\
   1 & 1 \\
\end{matrix} \right]$ ………………….. (i)
And with the help of given matrix A, we need to calculate value of ‘a’, ‘b’,
${{A}^{2}}+aA+bI=0$ ……………… (ii)
It means, we need to determine ${{A}^{2}}$ from the equation (i) and hence, we have to put values of ${{A}^{2}}$, A, I in the equation (ii) to get value of ‘a’, ‘b’.
So, we have
 $A=\left[ \begin{matrix}
   3 & 2 \\
   1 & 1 \\
\end{matrix} \right]$
So, we can calculate ${{A}^{2}}$ as
${{A}^{2}}=A\times A=\left[ \begin{matrix}
   3 & 2 \\
   1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   3 & 2 \\
   1 & 1 \\
\end{matrix} \right]$
Now, we know that multiplication of two matrices is done by multiplication the column of one matrix to the row of another matrix.
So, we get
$\begin{align}
  & {{A}^{2}}=\left[ \begin{matrix}
   3 & 2 \\
   1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   3 & 2 \\
   1 & 1 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   3\times 3+2\times 1 & 3\times 2+2\times 1 \\
   1\times 3+1\times 1 & 1\times 2+1\times 1 \\
\end{matrix} \right] \\
\end{align}$
${{A}^{2}}=\left[ \begin{matrix}
   11 & 8 \\
   4 & 3 \\
\end{matrix} \right]$…………. (iii)
So, we can put the value of ${{A}^{2}}$, A ,I to the equation (ii), to get the value of a, ‘b’.
As we know I is the unit matrix in the equation (ii) of order $2\times 2$. And we also know that the unit matrix has all its elements as ‘0’ except the principal diagonal elements. The principal diagonal elements of the unit matrix is 1.
So, unit matrix of order $2\times 2$ can be given as
   $I=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$ ………………. (iv)
And 0 is the null matrix of order ‘2’ in the equation (ii). As we know, a null matrix has all its elements as ‘0’. So, we get matrix 0 as
$0=\left[ \begin{matrix}
   0 & 0 \\
   0 & 0 \\
\end{matrix} \right]$ …………………. (v)
Now, put values of ${{A}^{2}}$, A, I,0 in the equation (ii), so, we get
$\begin{align}
  & \left[ \begin{matrix}
   11 & 8 \\
   4 & 3 \\
\end{matrix} \right]+a\left[ \begin{matrix}
   3 & 2 \\
   1 & 1 \\
\end{matrix} \right]+b\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 & 0 \\
   0 & 0 \\
\end{matrix} \right] \\
 & \left[ \begin{matrix}
   11 & 8 \\
   4 & 3 \\
\end{matrix} \right]+\left[ \begin{matrix}
   3a & 2a \\
   a & a \\
\end{matrix} \right]+\left[ \begin{matrix}
   b & 0 \\
   0 & b \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 & 0 \\
   0 & 0 \\
\end{matrix} \right] \\
 & \left[ \begin{matrix}
   11+3a+b & 8+2a+0 \\
   4+a+0 & 3+a+b \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 & 0 \\
   0 & 0 \\
\end{matrix} \right] \\
\end{align}$
$\left[ \begin{matrix}
   11+3a+b & 8+2a \\
   4+a & 3+a+b \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 & 0 \\
   0 & 0 \\
\end{matrix} \right]$…………………. (vi)
Now, as we know if two matrices are equal then the corresponding elements of both the matrices will be equal.
So, we get equations from equation (vi) as
11 + 3a + b = 0 ………….(vii)
8 + 2a = 0………..(viii)
4 + a = 0……….(ix)
3 + a + b = 0…………(x)
So, we can get value of a from the equation (ix) as
a + 4 = 0
a = - 4
Now, put a = - 4 in the equation (x), so, we get
3 + (- 4) + b = 0
- 1 + b = 0
b = 1
Hence, values of a, b are – 4, 1 respectively.

Note: Another approach to solve this equation would be that we can multiply the whole expression ${{A}^{2}}+aA+bI=0\to {{A}^{-1}}$ as well. So, we get equation after the multiplication as
${{A}^{2}}+aI+b{{A}^{-1}}=0$
Now, calculate inverse of A as
${{A}^{-1}}=\dfrac{1}{\left| A \right|}{{\left[ adjA \right]}^{T}}$
So, it can be another approach.
One may go wrong if he/she do the multiplication of A, A as
$\begin{align}
  & =\left[ \begin{matrix}
   3 & 2 \\
   1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   3 & 2 \\
   1 & 1 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   3\times 3+2\times 2 & 3\times 1+2\times 1 \\
   1\times 3+1\times 2 & 1\times 1+1\times 1 \\
\end{matrix} \right] \\
\end{align}$
Which is wrong as multiplication is done using row of one matrix and row of another matrix as well. So, be clear that multiplication of two matrices is done by the column of one matrix and corresponding row of it. So, don’t be confused with it and take care of it.
Don’t expand matrices by using the expansion of determinant as matrix and determinant are two other topics. So, don’t write the matrix $\left[ \begin{matrix}
   3 & 2 \\
   1 & 1 \\
\end{matrix} \right]=3-2=1$, as matrix cannot be simplified to get only one value of it.