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

Hint: 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.