Question

if A, B and C are $n\times n$ square matrices and det(A) = 2, det(B) = 3 and det(C) = 5, then find the value of $25\times \det \left( {{A}^{2}}B{{C}^{-1}} \right)$

[a] 60
[b] 80
[c] 50
[d] 20

Answer 1

Hint: Use the fact that if A and B are two square matrices of same dimensions, then det(AB) = det(A)det(B). Use the fact that $A{{A}^{-1}}=I$. Take determinant on both sides and hence prove that $\det \left( {{A}^{-1}} \right)=\dfrac{1}{\det \left( A \right)}$. Use the above two mentioned facts and hence prove that $\det \left( {{A}^{2}}B{{C}^{-1}} \right)=\dfrac{\det {{\left( A \right)}^{2}}\det \left( B \right)}{\det \left( C \right)}$. Substitute the values of det(A), det(B) and det(C) and hence find the value of $\det \left( {{A}^{2}}B{{C}^{-1}} \right)$ and hence evaluate the given expression.

Complete step-by-step answer:
We know that if A and B are two square matrices, then det(AB)= det(A)det(B).

Now, we have ${{A}^{2}}=A\times A$

Hence, we have $\det \left( {{A}^{2}} \right)=\det \left( A \right)\det \left( A \right)=\det {{\left( A \right)}^{2}}$

Also, we know that $A{{A}^{-1}}=I$

Taking determinant on both sides, we get

$\det \left( A{{A}^{-1}} \right)=\det \left( I \right)$

Since det(I) = 1, we have

$\det \left( A \right)\det \left( {{A}^{-1}} \right)=1$

Dividing both sides by det(A) since A is non-singular, we have

$\det \left( {{A}^{-1}} \right)=\dfrac{1}{\det \left( A \right)}$

Since A, B and C are square matrices of same dimensions, we have
${{A}^{2}},B$ and ${{C}^{-1}}$ are square matrices of the same dimensions.

Since ${{A}^{2}},B$ and ${{C}^{-1}}$ are square matrices of same dimensions, we have

$\det \left( {{A}^{2}}B{{C}^{-1}} \right)=\det \left( {{A}^{2}} \right)\det \left( B \right)\det \left( {{C}^{-1}} \right)=\dfrac{\det {{\left( A \right)}^{2}}\det \left( B \right)}{\det \left( C \right)}$

Substituting the value of det(A), det(B) and det(C), we get

$\det \left( {{A}^{2}}B{{C}^{-1}} \right)=\dfrac{{{2}^{2}}\times 3}{5}=\dfrac{12}{5}$

Hence, we have

$25\times \det \left( {{A}^{2}}B{{C}^{-1}} \right)=\dfrac{12}{5}\times 25=60$

Hence option [a] is correct.


Note: [1] Students usually make mistakes in the application of the formula det(AB)= det(A)det(B). The formula is valid if and only if A and B are square matrices. Also, note that $\det \left( kA \right)\ne k\det \left( A \right)$ where k is a scalar quantity. In fact $\det \left( kA \right)={{k}^{n}}\det \left( A \right)$ if A is an $n\times n$ square matrix.