Question

If {A^{ - 1}} = \left[ {\begin{array}{*{20}{l}}{ - 3}&0&2\\\\{ - 1}&{\dfrac{1}{2}}&{\dfrac{1}{2}}\\\2&0&{ - 1}\end{array}} \right] hence prove that ${A^3} - 6{A^2} + 7A + 2I = 0$

Answer 1

Here, we will prove the given condition. We will find the original matrix from the given inverse of a matrix and then find the square and cube of a matrix and by substituting these matrices in the equation, we will prove the given condition.

Formula Used:
We will use the following formula:
1.The inverse of a matrix is given by ${A^{ - 1}} = \dfrac{{Adj\left( A \right)}}{{\left| A \right|}}$
2.Transpose of a matrix is given by the formula ${A^T} = {\left[ {{a_{ij}}} \right]_{n \times m}}$
3.The inverse of a matrix is given by ${A^{ - 1}} = \dfrac{{Adj\left( A \right)}}{{\left| A \right|}}$

Complete step-by-step answer:
We are given the inverse of a matrix {A^{ - 1}} = \left[ {\begin{array}{*{20}{l}}{ - 3}&0&2\\\\{ - 1}&{\dfrac{1}{2}}&{\dfrac{1}{2}}\\\2&0&{ - 1}\end{array}} \right]
Now, we will find the original matrix $A$ .
We know that the inverse of an inverse matrix is an original matrix.
Now, we will find the determinant of the matrix $D$.
\Rightarrow \left| {\begin{array}{*{20}{l}}{ - 3}&0&2\\\\{ - 1}&{\dfrac{1}{2}}&{\dfrac{1}{2}}\\\2&0&{ - 1}\end{array}} \right| = - 3\left| {\begin{array}{*{20}{l}}{\dfrac{1}{2}}&{\dfrac{1}{2}}\\\0&{ - 1}\end{array}} \right| - 0\left| {\begin{array}{*{20}{l}}{ - 1}&{\dfrac{1}{2}}\\\2&{ - 1}\end{array}} \right| + 2\left| {\begin{array}{*{20}{l}}{ - 1}&{\dfrac{1}{2}}\\\2&0\end{array}} \right|
\Rightarrow \left| {\begin{array}{*{20}{l}}{ - 3}&0&2\\\\{ - 1}&{\dfrac{1}{2}}&{\dfrac{1}{2}}\\\2&0&{ - 1}\end{array}} \right| = - 3\left( {\dfrac{{ - 1}}{2}} \right) - 0 + 2\left( {\dfrac{{ - 2}}{2}} \right)
\Rightarrow \left| {\begin{array}{*{20}{l}}{ - 3}&0&2\\\\{ - 1}&{\dfrac{1}{2}}&{\dfrac{1}{2}}\\\2&0&{ - 1}\end{array}} \right| = \dfrac{3}{2} - 2
\Rightarrow \left| {\begin{array}{*{20}{l}}{ - 3}&0&2\\\\{ - 1}&{\dfrac{1}{2}}&{\dfrac{1}{2}}\\\2&0&{ - 1}\end{array}} \right| = \dfrac{3}{2} - 2 \times \dfrac{2}{2}
\Rightarrow \left| {\begin{array}{*{20}{l}}{ - 3}&0&2\\\\{ - 1}&{\dfrac{1}{2}}&{\dfrac{1}{2}}\\\2&0&{ - 1}\end{array}} \right| = \dfrac{{3 - 4}}{2}
\Rightarrow \left| {\begin{array}{*{20}{l}}{ - 3}&0&2\\\\{ - 1}&{\dfrac{1}{2}}&{\dfrac{1}{2}}\\\2&0&{ - 1}\end{array}} \right| = \dfrac{{ - 1}}{2} \ne 0
Since the determinant of the matrix is not zero, then there exist an inverse of the matrix.
Now, we will find the Adjoint of a matrix by finding the transpose of a matrix with cofactors.
\Rightarrow adj{A^{ - 1}} = {\left[ {\begin{array}{*{20}{l}}{ + \left| {\begin{array}{*{20}{l}}{\dfrac{1}{2}}&{\dfrac{1}{2}}\\\0&{ - 1}\end{array}} \right|}&{ - \left| {\begin{array}{*{20}{l}}{ - 1}&{\dfrac{1}{2}}\\\2&{ - 1}\end{array}} \right|}&{ + \left| {\begin{array}{*{20}{l}}{ - 1}&{\dfrac{1}{2}}\\\2&0\end{array}} \right|}\\\\{ - \left| {\begin{array}{*{20}{l}}0&2\\\0&{ - 1}\end{array}} \right|}&{ + \left| {\begin{array}{*{20}{l}}{ - 3}&2\\\2&{ - 1}\end{array}} \right|}&{ - \left| {\begin{array}{*{20}{l}}{ - 3}&0\\\2&0\end{array}} \right|}\\\\{ + \left| {\begin{array}{*{20}{l}}0&2\\\\{\dfrac{1}{2}}&{\dfrac{1}{2}}\end{array}} \right|}&{ - \left| {\begin{array}{*{20}{l}}{ - 3}&2\\\\{ - 1}&{\dfrac{1}{2}}\end{array}} \right|}&{ + \left| {\begin{array}{*{20}{l}}{ - 3}&0\\\\{ - 1}&{\dfrac{1}{2}}\end{array}} \right|}\end{array}} \right]^T}
\Rightarrow adj{A^{ - 1}} = {\left[ {\begin{array}{*{20}{l}}{ + \left( {\dfrac{{ - 1}}{2}} \right)}&{ - \left( {1 - \dfrac{2}{2}} \right)}&{ + \left( {\dfrac{{ - 2}}{2}} \right)}\\\0&{ + \left( {3 - 4} \right)}&0\\\\{ + \left( {\dfrac{{ - 2}}{2}} \right)}&{ - \left( {\dfrac{{ - 3}}{2} + 2} \right)}&{ + \left( {\dfrac{{ - 3}}{2}} \right)}\end{array}} \right]^T}
\Rightarrow adj{A^{ - 1}} = {\left[ {\begin{array}{*{20}{l}}{ + \left( {\dfrac{{ - 1}}{2}} \right)}&{ - \left( {1 - 1} \right)}&{ + \left( { - 1} \right)}\\\0&{ + \left( { - 1} \right)}&0\\\\{ + \left( { - 1} \right)}&{ - \left( {\dfrac{1}{2}} \right)}&{ + \left( {\dfrac{{ - 3}}{2}} \right)}\end{array}} \right]^T}
\Rightarrow adj{A^{ - 1}} = {\left[ {\begin{array}{*{20}{l}}{\left( {\dfrac{{ - 1}}{2}} \right)}&0&{\left( { - 1} \right)}\\\0&{\left( { - 1} \right)}&0\\\\{\left( { - 1} \right)}&{\left( {\dfrac{{ - 1}}{2}} \right)}&{\left( {\dfrac{{ - 3}}{2}} \right)}\end{array}} \right]^T}
Transpose of a matrix is given by the formula ${A^T} = {\left[ {{a_{ij}}} \right]_{n \times m}}$ where $A = {\left[ {{a_{ij}}} \right]_{m \times n}}$
\Rightarrow adj{A^{ - 1}} = \left[ {\begin{array}{*{20}{l}}{\left( {\dfrac{{ - 1}}{2}} \right)}&0&{\left( { - 1} \right)}\\\0&{\left( { - 1} \right)}&{\left( {\dfrac{{ - 1}}{2}} \right)}\\\\{\left( { - 1} \right)}&0&{\left( {\dfrac{{ - 3}}{2}} \right)}\end{array}} \right]
Now, we will find the inverse of a matrix.
The inverse of a matrix is given by ${A^{ - 1}} = \dfrac{{Adj\left( A \right)}}{{\left| A \right|}}$
$\Rightarrow {\left( {{A^{ - 1}}} \right)^{ - 1}} = \dfrac{1}{{\left| {{A^{ - 1}}} \right|}}adj\left( {{A^{ - 1}}} \right)$
Thus, we get
\Rightarrow {\left( {{A^{ - 1}}} \right)^{ - 1}} = \dfrac{1}{{\dfrac{{ - 1}}{2}}}\left[ {\begin{array}{*{20}{l}}{\left( {\dfrac{{ - 1}}{2}} \right)}&0&{\left( { - 1} \right)}\\\0&{\left( { - 1} \right)}&{\left( {\dfrac{{ - 1}}{2}} \right)}\\\\{\left( { - 1} \right)}&0&{\left( {\dfrac{{ - 3}}{2}} \right)}\end{array}} \right]
\Rightarrow {\left( {{A^{ - 1}}} \right)^{ - 1}} = - 2\left[ {\begin{array}{*{20}{l}}{\left( {\dfrac{{ - 1}}{2}} \right)}&0&{\left( { - 1} \right)}\\\0&{\left( { - 1} \right)}&{\left( {\dfrac{{ - 1}}{2}} \right)}\\\\{\left( { - 1} \right)}&0&{\left( {\dfrac{{ - 3}}{2}} \right)}\end{array}} \right]
By multiplying the terms, we get
\Rightarrow {\left( {{A^{ - 1}}} \right)^{ - 1}} = \left[ {\begin{array}{*{20}{l}}{\left( {\dfrac{2}{2}} \right)}&0&{\left( 2 \right)}\\\0&{\left( 2 \right)}&{\left( {\dfrac{2}{2}} \right)}\\\\{\left( 2 \right)}&0&{\left( {\dfrac{{ - 3 \times - 2}}{2}} \right)}\end{array}} \right]
\Rightarrow {\left( {{A^{ - 1}}} \right)^{ - 1}} = \left[ {\begin{array}{*{20}{l}}1&0&2\\\0&2&1\\\2&0&3\end{array}} \right]
\Rightarrow A = \left[ {\begin{array}{*{20}{l}}1&0&2\\\0&2&1\\\2&0&3\end{array}} \right] ……………………………………………………..$\left( 1 \right)$
Now, we will find the square of the Matrix A.
$\Rightarrow {A^2} = A \times A$
\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{l}}1&0&2\\\0&2&1\\\2&0&3\end{array}} \right] \times \left[ {\begin{array}{*{20}{l}}1&0&2\\\0&2&1\\\2&0&3\end{array}} \right]
By Matrix Multiplication, we get
\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{l}}{1 + 0 + 4}&{0 + 0 + 0}&{2 + 0 + 6}\\\\{0 + 0 + 2}&{0 + 4 + 0}&{0 + 2 + 3}\\\\{2 + 0 + 6}&{0 + 0 + 0}&{4 + 0 + 9}\end{array}} \right]
\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{l}}5&0&8\\\2&4&5\\\8&0&{13}\end{array}} \right] ………………………………………………………………………………………………………..$\left( 2 \right)$
Now, we will find the cube of the matrix A
$\Rightarrow {A^3} = {A^2} \times A$
\Rightarrow {A^3} = \left[ {\begin{array}{*{20}{l}}5&0&8\\\2&4&5\\\8&0&{13}\end{array}} \right] \times \left[ {\begin{array}{*{20}{l}}1&0&2\\\0&2&1\\\2&0&3\end{array}} \right]
By Matrix Multiplication, we get
\Rightarrow {A^3} = \left[ {\begin{array}{*{20}{l}}{5 + 0 + 16}&{0 + 0 + 0}&{10 + 0 + 24}\\\\{2 + 0 + 10}&{0 + 8 + 0}&{4 + 4 + 15}\\\\{8 + 0 + 26}&{0 + 0 + 0}&{16 + 0 + 39}\end{array}} \right]
\Rightarrow {A^3} = \left[ {\begin{array}{*{20}{l}}{21}&0&{34}\\\\{12}&8&{23}\\\\{34}&0&{55}\end{array}} \right] …………………………………………$\left( 3 \right)$
Now, we will prove that ${A^3} - 6{A^2} + 7A + 2I = 0$
Now, by substituting $\left( 1 \right),\left( 2 \right),\left( 3 \right)$ , we get
$\Rightarrow {A^3} - 6{A^2} + 7A + 2I = 0$
\Rightarrow \left[ {\begin{array}{*{20}{l}}{21}&0&{34}\\\\{12}&8&{23}\\\\{34}&0&{55}\end{array}} \right] - 6\left[ {\begin{array}{*{20}{l}}5&0&8\\\2&4&5\\\8&0&{13}\end{array}} \right] + 7\left[ {\begin{array}{*{20}{l}}1&0&2\\\0&2&1\\\2&0&3\end{array}} \right] + 2\left[ {\begin{array}{*{20}{l}}1&0&0\\\0&1&0\\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}{21}&0&{34}\\\\{12}&8&{23}\\\\{34}&0&{55}\end{array}} \right] + \left[ {\begin{array}{*{20}{l}}{ - 30}&0&{ - 48}\\\\{ - 12}&{ - 24}&{ - 30}\\\\{ - 48}&0&{ - 78}\end{array}} \right] + \left[ {\begin{array}{*{20}{l}}7&0&{14}\\\0&{14}&7\\\\{14}&0&{21}\end{array}} \right] + \left[ {\begin{array}{*{20}{l}}2&0&0\\\0&2&0\\\0&0&2\end{array}} \right]
By adding, we get
\Rightarrow \left[ {\begin{array}{*{20}{l}}{21}&0&{40}\\\\{12}&8&{23}\\\\{34}&0&{55}\end{array}} \right] - 6\left[ {\begin{array}{*{20}{l}}5&0&8\\\2&4&5\\\8&0&{13}\end{array}} \right] + 7\left[ {\begin{array}{*{20}{l}}1&0&2\\\0&2&1\\\2&0&3\end{array}} \right] + 2\left[ {\begin{array}{*{20}{l}}1&0&0\\\0&1&0\\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}{21 - 30 + 7 + 1}&{0 + 0 + 0 + 0}&{34 - 48 + 14 + 0}\\\\{12 - 12 + 0 + 0}&{8 - 24 + 14 + 2}&{23 - 30 + 7 + 0}\\\\{34 - 48 + 14 + 0}&{0 + 0 + 0 + 0}&{55 - 78 + 21 + 2}\end{array}} \right]
\Rightarrow \left[ {\begin{array}{*{20}{l}}{21}&0&{40}\\\\{12}&8&{23}\\\\{34}&0&{55}\end{array}} \right] - 6\left[ {\begin{array}{*{20}{l}}5&0&8\\\2&4&5\\\8&0&{13}\end{array}} \right] + 7\left[ {\begin{array}{*{20}{l}}1&0&2\\\0&2&1\\\2&0&3\end{array}} \right] + 2\left[ {\begin{array}{*{20}{l}}1&0&0\\\0&1&0\\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}{30 - 30}&0&{48 - 48}\\\\{12 - 12}&{24 - 24}&{30 - 30}\\\\{48 - 48}&0&{78 - 78}\end{array}} \right]
\Rightarrow \left[ {\begin{array}{*{20}{l}}{21}&0&{40}\\\\{12}&8&{23}\\\\{34}&0&{55}\end{array}} \right] - 6\left[ {\begin{array}{*{20}{l}}5&0&8\\\2&4&5\\\8&0&{13}\end{array}} \right] + 7\left[ {\begin{array}{*{20}{l}}1&0&2\\\0&2&1\\\2&0&3\end{array}} \right] + 2\left[ {\begin{array}{*{20}{l}}1&0&0\\\0&1&0\\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}0&0&0\\\0&0&0\\\0&0&0\end{array}} \right]
Therefore, ${A^3} - 6{A^2} + 7A + 2I = 0$ .

Note: We know that the matrix multiplication is possible only if the number of columns in one matrix is equal to the number of rows in another matrix. $I$ is the Identity matrix which should always be equal to the dimensions of the other matrix in the given condition. Identity matrix is a diagonal matrix with its diagonal as 1. If all the elements in a matrix are zero, then the matrix is said to be a null matrix. So, the given ${A^3} - 6{A^2} + 7A + 2I$ is a null matrix. We should remember that the square of the matrix is defined as the product of itself two times and not the square of the elements in the matrix.