Question

Find A, if \[A\left[ \begin{matrix}
   1 & 2 & 3 \\
   2 & 5 & 7 \\
   2 & 0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   3 & -2 & -1 \\
   -4 & 1 & -1 \\
   2 & 0 & 1 \\
\end{matrix} \right]\]

Answer 1

Hint: Multiply the inverse of the matrix given on LHS, both the sides. Then, find the inverse of the matrix which was present earlier in LHS following every step. Now, multiply the inverse matrix with the matrix present on RHS earlier.

Complete step-by-step answer:
Given in the question:
\[A\left[ \begin{matrix}
   1 & 2 & 3 \\
   2 & 5 & 7 \\
   2 & 0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   3 & -2 & -1 \\
   -4 & 1 & -1 \\
   2 & 0 & 1 \\
\end{matrix} \right]\]
Suppose, \[\left[ \begin{matrix}
   1 & 2 & 3 \\
   2 & 5 & 7 \\
   2 & 0 & 1 \\
\end{matrix} \right]=B\]
B is a $ 3\times 3 $ read as 3 by 3 matrix, which means B matrix contains 3 rows and 3 columns.
\[\left. \underbrace{\left[ \begin{matrix}
   1 & 2 & 3 \\
   2 & 5 & 7 \\
   2 & 0 & 1 \\
\end{matrix} \right]}_{\text{Column}} \right\}\text{Rows}\left( 3\times 3 \right)\]
Given here, \[A\times B=\left[ \begin{matrix}
   3 & -2 & -1 \\
   -4 & 1 & -1 \\
   2 & 0 & 1 \\
\end{matrix} \right]\]
Since, we do not know the matrix A, we must have to solve matrix B with the final product matrix present in the RHS.
To move matrix B from $ \text{LHS}\to \text{RHS} $ we multiply B matrix with its own inverse $ {{B}^{-1}} $ which yield identity matrix.
\[\text{Identity matrix I = }\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]\]
Any matrix when multiplied with its own inverse always results in identity matrix this is predefined matrix property.
So, \[B\times {{B}^{-1}}=I\]
Now, in the question, we multiply $ {{B}^{-1}} $ on both LHS as well as RHS. So, that things overall remain unchanged.
\[\begin{align}
  & A\times \left( B\times {{B}^{-1}} \right)=\left[ \begin{matrix}
   3 & -2 & -1 \\
   -4 & 1 & -1 \\
   2 & 0 & 1 \\
\end{matrix} \right]\times {{B}^{-1}} \\
 & \Rightarrow A\times I=\left[ \begin{matrix}
   3 & -2 & -1 \\
   -4 & 1 & -1 \\
   2 & 0 & 1 \\
\end{matrix} \right]\times {{B}^{-1}} \\
\end{align}\]
Now, we will find $ {{B}^{-1}} $ there are general rules to find inverse of my matrix.
Step 1:- Find minors of each element of the matrix.
For example, \[\left[ \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right]\]
Minor of $ {{1}^{st}} $ element $ {{a}_{1}} $ can be found out by eliminating the row and the column containing it:
\[\text{Minor of }{{\text{a}}_{\text{1}}}=\left( \begin{matrix}
   {{b}_{2}} & {{b}_{3}} \\
   {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right)\text{ can be solved as }{{\text{b}}_{2}}{{c}_{3}}-{{b}_{3}}{{c}_{2}}\]
Therefore, for matrix B minor of each element are:
\[\begin{align}
  & \Rightarrow \left( \begin{matrix}
   \left( \begin{matrix}
   5 & 7 \\
   0 & 1 \\
\end{matrix} \right) & \left( \begin{matrix}
   2 & 7 \\
   2 & 1 \\
\end{matrix} \right) & \left( \begin{matrix}
   2 & 5 \\
   2 & 0 \\
\end{matrix} \right) \\
   \left( \begin{matrix}
   2 & 3 \\
   0 & 1 \\
\end{matrix} \right) & \left( \begin{matrix}
   1 & 3 \\
   2 & 1 \\
\end{matrix} \right) & \left( \begin{matrix}
   1 & 2 \\
   2 & 0 \\
\end{matrix} \right) \\
   \left( \begin{matrix}
   2 & 3 \\
   5 & 7 \\
\end{matrix} \right) & \left( \begin{matrix}
   1 & 3 \\
   2 & 7 \\
\end{matrix} \right) & \left( \begin{matrix}
   1 & 2 \\
   2 & 5 \\
\end{matrix} \right) \\
\end{matrix} \right) \\
 & \Rightarrow \left( \begin{matrix}
   5 & -12 & -10 \\
   2 & -5 & -4 \\
   -1 & 1 & 1 \\
\end{matrix} \right) \\
\end{align}\]
Step 2:- Matrix of co-factors.
Co factors are just associated with +ve, -ve signs alternatively.
\[\Rightarrow \left( \begin{matrix}
   5\left( + \right) & -12\left( - \right) & -10\left( + \right) \\
   2\left( - \right) & -5\left( + \right) & -4\left( - \right) \\
   -1\left( + \right) & 1\left( - \right) & 1\left( + \right) \\
\end{matrix} \right)=\left( \begin{matrix}
   5 & 12 & -10 \\
   -2 & -5 & 4 \\
   -1 & -1 & 1 \\
\end{matrix} \right)\]
Step 3:- Adjugate: transverse all the elements diagonally.
\[\Rightarrow \left( \begin{matrix}
   5 & 12 & -10 \\
   -2 & -5 & 4 \\
   -1 & -1 & 1 \\
\end{matrix} \right)=\left( \begin{matrix}
   5 & -2 & -1 \\
   12 & -5 & -1 \\
   -10 & 4 & 1 \\
\end{matrix} \right)\]
Step 4:- Multiply by 1/determinant.
Determinant = total number of matrix.
\[D={{a}_{1}}\times \left( \text{minors of }{{\text{a}}_{\text{1}}} \right)-{{a}_{2}}\times \left( \text{minors of }{{\text{a}}_{2}} \right)+{{a}_{3}}\times \left( \text{minors of }{{\text{a}}_{3}} \right)\]
D = is given for the original matrix B.
\[B=\left[ \begin{matrix}
   1 & 2 & 3 \\
   2 & 5 & 7 \\
   2 & 0 & 1 \\
\end{matrix} \right]\]
So,
\[\begin{align}
  & D=1\times \left( \begin{matrix}
   5 & 7 \\
   0 & 1 \\
\end{matrix} \right)-2\times \left( \begin{matrix}
   2 & 7 \\
   2 & 1 \\
\end{matrix} \right)+3\times \left( \begin{matrix}
   2 & 5 \\
   2 & 0 \\
\end{matrix} \right) \\
 & \Rightarrow 1\left( 5 \right)-2\left( 12 \right)+3\left( -10 \right) \\
 & \Rightarrow 5+24-30 \\
 & \Rightarrow -1 \\
\end{align}\]
Now,
\[\begin{align}
  & {{B}^{-1}}=\dfrac{1}{D}\times \left( \text{Matrix obtained in step 3} \right) \\
 & \Rightarrow \dfrac{1}{D}\times \left( \begin{matrix}
   5 & -2 & -1 \\
   12 & -5 & -1 \\
   -10 & 4 & 1 \\
\end{matrix} \right) \\
 & \Rightarrow -1\left( {} \right)=\left( \begin{matrix}
   -5 & 2 & 1 \\
   -12 & 5 & 1 \\
   10 & -4 & -1 \\
\end{matrix} \right) \\
\end{align}\]
So, we have,
\[AI=\left[ \begin{matrix}
   3 & -2 & -1 \\
   -4 & 1 & -1 \\
   2 & 0 & 1 \\
\end{matrix} \right]\times \left( \begin{matrix}
   -5 & 2 & 1 \\
   -12 & 5 & 1 \\
   10 & -4 & -1 \\
\end{matrix} \right)\]
We now have to multiply both the matrices on RHS.
Matrices can only be multiplied if and only if:
Number of rows of first matrices = Number of columns of second matrices.
Here, both matrices to be multiplied are $ \left( 3\times 3 \right) $
Therefore, can be multiplied:
\[\Rightarrow \left[ \begin{matrix}
   3 & -2 & -1 \\
   -4 & 1 & -1 \\
   2 & 0 & 1 \\
\end{matrix} \right]\times \left( \begin{matrix}
   -5 & 2 & 1 \\
   -12 & 5 & 1 \\
   10 & -4 & -1 \\
\end{matrix} \right)\]
Elements of rows of first matrix will be multiplied by the elements of columns of second matrix:
\[\begin{align}
  & \Rightarrow \left( \begin{matrix}
   \left( 3\times -5 \right)+\left( -2\times -12 \right)+\left( -1\times 10 \right) & \left( 3\times 2 \right)+\left( -2\times 5 \right)+\left( -1\times -4 \right) & \left( 3\times 1 \right)+\left( -2\times 1 \right)+\left( -1\times 1 \right) \\
   \left( -4\times 5 \right)+\left( 1\times -12 \right)+\left( -1\times 10 \right) & \left( -4\times 2 \right)+\left( 1\times 5 \right)+\left( -1\times 4 \right) & \left( -4\times 1 \right)+\left( 1\times 1 \right)+\left( -1\times 1 \right) \\
   \left( 2\times -5 \right)+\left( 0\times -12 \right)+\left( 1\times 10 \right) & \left( 2\times 2 \right)+\left( 0\times 5 \right)+\left( 1\times -4 \right) & \left( 2\times 1 \right)+\left( 0\times 1 \right)+\left( 1\times -1 \right) \\
\end{matrix} \right) \\
 & \Rightarrow \left( \begin{matrix}
   -1 & 0 & 2 \\
   -2 & 1 & -5 \\
   0 & 0 & 1 \\
\end{matrix} \right) \\
\end{align}\]
Thus, \[AI=\left( \begin{matrix}
   -1 & 0 & 2 \\
   -2 & 1 & -5 \\
   0 & 0 & 1 \\
\end{matrix} \right)\]
According to the matrix properties AI=A.
Whenever an identity matrix:
\[\text{Identity matrix I = }\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]\] is multiplied with any other matrix, leaving the product as the matrix which was multiplied itself.
\[\therefore \text{ A}\times \text{I=A}\]
Therefore, required matrix A is \[A=\left( \begin{matrix}
   -1 & 0 & 2 \\
   -2 & 1 & -5 \\
   0 & 0 & 1 \\
\end{matrix} \right)\]

Note: You should be knowing all the properties of matrices that is, how to find the inverse of a matrix, how to multiply two matrices. You must be very careful while doing the calculation, because the calculation part is complex here. It is important to note that we cannot divide the matrix and get A directly, so we must assume the matrix as B and then apply the concept of inverse of a matrix, identity matrix and solve such questions. Usually students make mistakes while writing the cofactor matrix by interchanging the sign of terms, so be very careful while solving that portion.