For a square matrix \( \mathrm{A}=\begin{bmatrix}2 & 1 & 3\\0 & -1 & 4\\1 & 2 & 0\end{bmatrix} \), which of the following statements is/are TRUE ?
A. det (A) = 9.
B. \( \mathrm{A_{21}}\,.\,\mathrm{A_{33}}=-12 \) , where \( \mathrm{A_{ij}} \) is the cofactor of \( \mathrm{a_{ij}} \)
C. det (A) = \( \mathrm{a_{11}}\,\mathrm{A_{11}}+\mathrm{a_{12}}\,\mathrm{A_{12}}+\mathrm{a_{13}}\,\mathrm{A_{31}} \).
D. \( |\mathrm{A}\mathrm{A}'|\ne|\mathrm{A}|^2 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → B only
The cofactors needed are
\( \displaystyle A_{11}=(-1)(0)-(4)(2)=-8 \)
\( \displaystyle A_{12}=-\left[(0)(0)-(4)(1)\right]=4 \)
\( \displaystyle A_{13}=(0)(2)-(-1)(1)=1 \)
\( \displaystyle A_{21}=-\left[(1)(0)-(3)(2)\right]=6 \)
\( \displaystyle A_{31}=(1)(4)-(3)(-1)=7 \)
\( \displaystyle A_{33}=(2)(-1)-(1)(0)=-2 \)
Expanding along the first row, each element is multiplied by the cofactor of the same element:
\( \displaystyle \det A=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13} \)
\( \displaystyle =2(-8)+1(4)+3(1) \)
\( \displaystyle =-9 \)
So det A = −9, \( A_{21}=6 \), \( A_{31}=7 \) and \( A_{33}=-2 \).
Checking the statements
(A) det (A) = 9. — Incorrect. det A is −9, not 9.
(B) \( \mathrm{A_{21}}\,.\,\mathrm{A_{33}}=-12 \), where \( \mathrm{A_{ij}} \) is the cofactor of \( \mathrm{a_{ij}} \) — Correct. \( 6\times(-2)=-12 \).
(C) det (A) = \( \mathrm{a_{11}}\,\mathrm{A_{11}}+\mathrm{a_{12}}\,\mathrm{A_{12}}+\mathrm{a_{13}}\,\mathrm{A_{31}} \). — Incorrect. The last term needs \( A_{13} \), not \( A_{31} \): \( 2(-8)+1(4)+3(7)=9\ne -9 \).
(D) \( |\mathrm{A}\mathrm{A}'|\ne|\mathrm{A}|^2 \) — Incorrect. \( |AA'|=|A|\,|A'|=|A|^2 \), since a matrix and its transpose have the same determinant.