Question

Let A be a square matrix of order 3 such that

$Tr.(A^{-1}) = 3$ and $det(A^{-1}) = \frac{1}{5}$.

If $(A^{-1}B^{-1}A)^{-1} = 2(adj A)$, then

• A. det (B) = 5000
• B. det (B) = 200
• C. Tr.(B) = 30
• D. Tr.(B) = 15

Answer 1

Answer: (B), (C)

det (B) = 200, Tr.(B) = 30

Answer 2

We are given $Tr(A^{-1}) = 3$ and $det(A^{-1}) = \frac{1}{5}$.

From the property $det(A^{-1}) = \frac{1}{det(A)}$, we have $det(A) = \frac{1}{det(A^{-1})} = \frac{1}{1/5} = 5$.

We are given the equation $(A^{-1}B^{-1}A)^{-1} = 2(adj A)$.

Using the property $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$, we have $(A^{-1}B^{-1}A)^{-1} = A^{-1}(B^{-1})^{-1}(A^{-1})^{-1} = A^{-1}BA$.

Using the property $adj P = det(P) P^{-1}$, we have $adj A = det(A) A^{-1}$. Substituting $det(A) = 5$, we get $adj A = 5 A^{-1}$.

So, the given equation becomes: $A^{-1}BA = 2(5 A^{-1})$ $A^{-1}BA = 10 A^{-1}$

To solve for B, we can pre-multiply both sides by A: $A(A^{-1}BA) = A(10 A^{-1})$ $(AA^{-1})BA = 10 (AA^{-1})$ Since A is invertible (as $det(A) = 5 \neq 0$), $AA^{-1} = I_3$, where $I_3$ is the identity matrix of order 3. $I_3 BA = 10 I_3$ $BA = 10 I_3$

From $BA = 10 I_3$, we can find the determinant of B. Taking the determinant of both sides: $det(BA) = det(10 I_3)$ Using the properties $det(XY) = det(X)det(Y)$ and $det(cI_n) = c^n det(I_n)$ for a scalar c and identity matrix $I_n$: $det(B)det(A) = 10^3 det(I_3)$ $det(B) \times 5 = 1000 \times 1$ $det(B) = \frac{1000}{5} = 200$.

Now, let's find the trace of B. From $BA = 10 I_3$, we can post-multiply by $A^{-1}$ (since A is invertible): $BAA^{-1} = 10 I_3 A^{-1}$ $B I_3 = 10 A^{-1}$ $B = 10 A^{-1}$

Now, take the trace of B: $Tr(B) = Tr(10 A^{-1})$ Using the property $Tr(cA) = c Tr(A)$: $Tr(B) = 10 Tr(A^{-1})$ We are given that $Tr(A^{-1}) = 3$. $Tr(B) = 10 \times 3 = 30$.

So, we have found that $det(B) = 200$ and $Tr(B) = 30$.