
If$A$ is $n\times n$ matrix, then $adj\,(adj.A)=$.
A. $|A{{|}^{n-1}}A|$
B. $|A{{|}^{n-2}}A|$
C. $|A{{|}^{n}}n|$
D. None of these.
Answer
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Hint: The matrix of order $n\times n$ can be defined as a square matrix in which equal number of rows and columns are present.
To find $adj\,(adj.A)$, we will use the inverse matrix formula and matrix property $|adj.A|=\,|A{{|}^{n-1}}$ and \[A.{{A}^{-1}}=I\].
Formula Used: \[{{A}^{-1}}=\dfrac{1}{|A|}\,adj(A)\]
\[A.{{A}^{-1}}=I\], where $I$ is identity matrix.
$|adj.A|=\,|A{{|}^{n-1}}$
Complete step by step solution: We are given that $A$ is a square matrix, and we have to find the value of $adj\,(adj.A)$.
To derive the value of $adj\,(adj.A)$ we will use the formula of inverse matrix \[{{A}^{-1}}=\dfrac{1}{|A|}\,adj(A)\] and property \[A.{{A}^{-1}}=I\].
\[{{A}^{-1}}=\dfrac{1}{|A|}\,adj(A)\]………..\[\left( 1 \right)\]
Using property \[A.{{A}^{-1}}=I\], we can derive the value of \[{{A}^{-1}}\],
\[{{A}^{-1}}=\dfrac{I}{A}\]………………..\[\left( 2 \right)\]
Comparing equation \[\left( 1 \right)\]and \[\left( 2 \right)\]we get,
\[\dfrac{I}{A}=\dfrac{adj(A)}{|A|}\,\]
Now We will find the value of determinant.
\[|A|I=A(adj.A)\,\]……\[\left( 3 \right)\]
We will replace $A$ with \[(adj.A)\,\]in the equation.
Then,
\[adj.A(adj(adj.A))\,=|adj.A|I\]
Now using property $|adj.A|=\,|A{{|}^{n-1}}$we can write,
\[adj.A(adj(adj.A))\,=|A{{|}^{n-1}}I\]
We will multiply the equation on both side by $A$,
\[A\left( adj.A(adj(adj.A)) \right)\,=A.|A{{|}^{n-1}}I\]
\[(A\,adj.A)(adj(adj.A))\,=|A{{|}^{n-1}}I.A\]
As we know that $I.A=A$ so,
\[(A\,adj.A)(adj(adj.A))\,=|A{{|}^{n-1}}A\]
From equation \[\left( 3 \right)\]we will substitute the value \[|A|I=A(adj.A)\,\],
\[|A|I.(adj(adj.A))\,=|A{{|}^{n-1}}A\]
\[(adj(adj.A))\,=\dfrac{|A{{|}^{n-1}}A}{|A|.I}\]
\[(adj(adj.A))\,=|A{{|}^{n-2}}.A\]
Now as we have to find the value of $adj\,(adj.A)$, we can say that it will be $A\,(adj.A)=|A|I$. The value of $adj\,(adj.A)$is \[(adj(adj.A))\,=|A{{|}^{n-2}}.A\] where$A$ is square matrix
Option ‘B’ is correct
Note:
The adjoint of a square matrix can be defined as the transpose of the co-factor of its matrix. The transpose of the matrix means interchanging the elements of rows with columns and elements of columns with rows.
When a matrix is multiplied by its adjoint then the product formed is identity matrix multiplied by the determinant of that matrix. If $A$ is the matrix then the relation will be written as$A\,(adj.A)=\,(adj.A).A=|A|I$.
To find $adj\,(adj.A)$, we will use the inverse matrix formula and matrix property $|adj.A|=\,|A{{|}^{n-1}}$ and \[A.{{A}^{-1}}=I\].
Formula Used: \[{{A}^{-1}}=\dfrac{1}{|A|}\,adj(A)\]
\[A.{{A}^{-1}}=I\], where $I$ is identity matrix.
$|adj.A|=\,|A{{|}^{n-1}}$
Complete step by step solution: We are given that $A$ is a square matrix, and we have to find the value of $adj\,(adj.A)$.
To derive the value of $adj\,(adj.A)$ we will use the formula of inverse matrix \[{{A}^{-1}}=\dfrac{1}{|A|}\,adj(A)\] and property \[A.{{A}^{-1}}=I\].
\[{{A}^{-1}}=\dfrac{1}{|A|}\,adj(A)\]………..\[\left( 1 \right)\]
Using property \[A.{{A}^{-1}}=I\], we can derive the value of \[{{A}^{-1}}\],
\[{{A}^{-1}}=\dfrac{I}{A}\]………………..\[\left( 2 \right)\]
Comparing equation \[\left( 1 \right)\]and \[\left( 2 \right)\]we get,
\[\dfrac{I}{A}=\dfrac{adj(A)}{|A|}\,\]
Now We will find the value of determinant.
\[|A|I=A(adj.A)\,\]……\[\left( 3 \right)\]
We will replace $A$ with \[(adj.A)\,\]in the equation.
Then,
\[adj.A(adj(adj.A))\,=|adj.A|I\]
Now using property $|adj.A|=\,|A{{|}^{n-1}}$we can write,
\[adj.A(adj(adj.A))\,=|A{{|}^{n-1}}I\]
We will multiply the equation on both side by $A$,
\[A\left( adj.A(adj(adj.A)) \right)\,=A.|A{{|}^{n-1}}I\]
\[(A\,adj.A)(adj(adj.A))\,=|A{{|}^{n-1}}I.A\]
As we know that $I.A=A$ so,
\[(A\,adj.A)(adj(adj.A))\,=|A{{|}^{n-1}}A\]
From equation \[\left( 3 \right)\]we will substitute the value \[|A|I=A(adj.A)\,\],
\[|A|I.(adj(adj.A))\,=|A{{|}^{n-1}}A\]
\[(adj(adj.A))\,=\dfrac{|A{{|}^{n-1}}A}{|A|.I}\]
\[(adj(adj.A))\,=|A{{|}^{n-2}}.A\]
Now as we have to find the value of $adj\,(adj.A)$, we can say that it will be $A\,(adj.A)=|A|I$. The value of $adj\,(adj.A)$is \[(adj(adj.A))\,=|A{{|}^{n-2}}.A\] where$A$ is square matrix
Option ‘B’ is correct
Note:
The adjoint of a square matrix can be defined as the transpose of the co-factor of its matrix. The transpose of the matrix means interchanging the elements of rows with columns and elements of columns with rows.
When a matrix is multiplied by its adjoint then the product formed is identity matrix multiplied by the determinant of that matrix. If $A$ is the matrix then the relation will be written as$A\,(adj.A)=\,(adj.A).A=|A|I$.
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