If \(\rm A^{-1}=\begin{bmatrix} 1& 2& 3\\ 2& 4& 3\\ 3& 1& 6\end{bmatrix}=\frac{adj(A)}{k}\), then k = ?

Concept:

For an invertible matrix A:

• A^-1 = \(\rm \frac{adj(A)}{|A|}\).
• |A-1| = |A|^-1 = \(\rm \frac{1}{|A|}\).

Calculation:

\(\rm A^{-1}=\begin{bmatrix} 1& 2& 3\\ 2& 4& 3\\ 3& 1& 6\end{bmatrix}=\frac{adj(A)}{k}\)         -----(1)

From the definition of the inverse of a matrix, 

A-1 = \(\rm \frac{adj(A)}{|A|}\)              -----(2)

Comparing equation (1) & (2), we get

k = |A|  

Using the properties of the determinant of inverse of a matrix, we have:

k = |A| = \(\rm \frac{1}{|A^{-1}|}\)        ----(3)

We know, 

A.A^-1 = I

⇒ |A.A-1| = |I| = 1

⇒ |A| |A-1| = 1

⇒ |A| = 1/ |A-1|       ....(4)

Now,

|A^-1| = 1(24 - 3) + 2(9 - 12) + 3(2 - 12) = 21 - 6 - 30 = - 15.

|A-1| = -15

Therefore, from equation (3)

k = \(\rm - \frac1{15}\).

Mistake PointsNote that, we have A^-1 matrix, not an A matrix. So to find the value of k, don't you have to use relation |A| = 1/|A^-1|