Question

If $A$ is a square matrix of order 3 such that $\left| {Adj.A} \right| = 36$, find $\left| A \right|$.

Answer 1

Hint:In order to find $\left| A \right|$ for the value $\left| {Adj.A} \right| = 36$, we need to know about the adjoint matrix. Adjoint of a matrix is the transpose of the matrix formed by the cofactors of the given matrix. For example, if M is a matrix, then its adjoint matrix will be ${\left| {{C_m}} \right|^T}$, where ${C_m}$ is the matrix of the cofactors.

Complete step by step answer:
We are given with an equation $\left| {Adj.A} \right| = 36$, where $\left| {Adj.A} \right|$ is the adjoint of a matrix A.Now, we need to find the value of $\left| A \right|$. From the properties of the matrices, we know a formula which states that the adjoint of a matrix is equal to the determinant of the matrix raised to the power of order of matrix minus one.The formula numerically, written as:
$\left| {Adj.A} \right| = {\left| A \right|^{n - 1}}$ ……(1)
Where ‘$n$’ is the order of the matrix.

And, in our question we are given with the order of matrix as 3, so the value of ‘n’ becomes:
$n = 3$
So, we have the values $n = 3$ and $\left| {Adj.A} \right| = 36$, so substituting these two values in the equation 1 formula, we get:
$\left| {Adj.A} \right| = {\left| A \right|^{n - 1}}$
$ \Rightarrow 36 = {\left| A \right|^{3 - 1}}$
Solving the power:
$ \Rightarrow 36 = {\left| A \right|^2}$
Taking square root both the sides:
\[ \Rightarrow \sqrt {36} = \sqrt {{{\left| A \right|}^2}} \]
Since, we know that \[\sqrt {{x^2}} = x\] and \[\sqrt {36} = 6\], so according to this, we get:
\[ \Rightarrow 6 = \left| A \right|\]
\[ \therefore \left| A \right| = 6\]

Therefore, the value of $\left| A \right|$ obtained is $6$.

Note: Transpose means changing the order of columns and rows of a matrix by changing the rows to columns and columns to rows. Since, we know that square root of a number can be positive or negative but we took positive part for the matrix