If $A = \begin{bmatrix} a & b & c \ x & y & \mathcal{z} \ p & q & r \end{bmatrix}$, $B = \begin{bmatrix} q & - b & y \ - p & a & - x \ r & - c & \mathcal{z} \end{bmatrix}$ and if $A$ is invertible, Then which of the following is not true?

Question

Dr. Abubacker · Accepted Answer

(a)
$|B| = \left| \begin{matrix}
q & - b & y \
 - p & a & - x \
r & - c & \mathcal{z}
\end{matrix} ight|$(Multiplying $R_{2}$ by $- 1$)
$= - \left| \begin{matrix}
q & - b & y \
p & - a & x \
r & - c & \mathcal{z}
\end{matrix} ight|$(Multiplying $C_{2}$ by $- 1$)
$= \left| \begin{matrix}
q & b & y \
p & a & x \
r & c & \mathcal{z}
\end{matrix} ight|$(Changing $R_{1}$ with $R_{2}$)
$= - \left| \begin{matrix}
q & b & y \
p & a & x \
r & c & \mathcal{z}
\end{matrix} ight|$
$= - \left| \begin{matrix}
q & b & y \
p & a & x \
r & c & \mathcal{z}
\end{matrix} ight|$
Hence $|A| = - |B|,$ obviously when $|A| 
eq 0,\ |B| 
eq 0$. Also, $|adj\ B| = |B|^{2}$
$= \left( - |A| ight)^{2} = |A|^{2}$

Q. If $A = \begin{bmatrix} a & b & c \\ x & y & \mathcal{z} \\ p & q & r \end{bmatrix}$, $B = \begin{bmatrix} q & - b & y \\ - p & a & - x \\ r & - c & \mathcal{z} \end{bmatrix}$ and if $A$ is invertible, Then which of the following is not true?

Explanation

Reviewed by Dr. Abubacker

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