Question

Let $\mathrm{M}=\left[\begin{array}{ccc}0& 1& \mathrm{a}\\ 1& 2& 3\\ 3& \mathrm{b}& 1\end{array}\right]$ and $\mathrm{adj}\left(\mathrm{M}\right)=\left[\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right]$ where $a$ and $b$ are real numbers. Which of the following options is/are correct?

• A. $\mathrm{a}+\mathrm{b}=3$
• B. $\det \left(\mathrm{adj}{\mathrm{M}}^2\right)=81$
• C. ${\left(\mathrm{adj}\mathrm{M}\right)}^{-1}+\mathrm{adj}{\mathrm{M}}^{-1}=-\mathrm{M}$
• D. If $\mathrm{M}\left[\begin{array}{l}\mathrm{\alpha }\\ \mathrm{\beta }\\ \mathrm{\gamma }\end{array}\right]=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]$ then $\mathrm{\alpha }-\mathrm{\beta }+\mathrm{\gamma }=3$

Answer 1

Answer: (A), (C), (D)

If $\mathrm{M}\left[\begin{array}{l}\mathrm{\alpha }\\ \mathrm{\beta }\\ \mathrm{\gamma }\end{array}\right]=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]$ then $\mathrm{\alpha }-\mathrm{\beta }+\mathrm{\gamma }=3$

Explanation for correct option(s):

Step 1: Given information

$\mathrm{M}=\left[\begin{array}{lll}0& 1& \mathrm{a}\\ 1& 2& 3\\ 3& \mathrm{b}& 1\end{array}\right]$

And adjoint of matrix $M$ is,

$\text{adj}\left(M\right)=\left[\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right]$

Step 2: Calculating $\left|\mathrm{M}\right|$

We also know that,

$$\begin{gathered} \left|\text{adj}\right(\mathrm{M}\left)\right|=\left|{\mathrm{M}}^2\right|=\left[\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right] \\ \Rightarrow \left|{\mathrm{M}}^2\right|=-1(6-6)+1(-10+8)-1(24-30) \\ \Rightarrow \left|{\mathrm{M}}^2\right|=4 \\ \Rightarrow \left|\mathrm{M}\right|=\pm 2 \end{gathered}$$

Step 3: Calculating $\left(\mathrm{ad}j\right(M){)}^{-1}$

We also know that,

$\mathrm{M}(adj\mathrm{M})=\left|\mathrm{M}\right|$

$\left(\text{adj}\right(M){)}^{-1}=\frac{1}{\left|\text{adj}\left(M\right)\right|}{\left[\begin{array}{ccc}0& -2& -6\\ -2& -4& -2\\ -4& -6& -2\end{array}\right]}^{\mathrm{T}}$

$\Rightarrow \left(\mathrm{adj}\left({\mathrm{M}}^{-1}\right)\right)=\frac{1}{4}\left[\begin{array}{ccc}0& -2& -4\\ -2& -4& -6\\ -6& -2& -2\end{array}\right]$

Step 4: Equating the two equations

Now

$\mathrm{M}=\left|\mathrm{M}\right|\left(\mathrm{adj}\right(M){)}^{-1}$

$\left[\begin{array}{lll}0& 1& \mathrm{a}\\ 1& 2& 3\\ 3& \mathrm{b}& 1\end{array}\right]=\frac{\left|\mathrm{M}\right|}{4}\left[\begin{array}{ccc}0& -2& -4\\ -2& -4& -6\\ -6& -2& -2\end{array}\right]$

Comparing Left Hand Side and Right Hand Side we get,

$\left|\mathrm{M}\right|=-2$

Therefore,

$\mathrm{a}=2,\mathrm{b}=1$

Step 5: Calculating $\mathrm{a}+\mathrm{b}$

Therefore,

$\mathrm{a}+\mathrm{b}=2+1$

$\Rightarrow \mathrm{a}+\mathrm{b}=3$

Thus, the value $\mathrm{a}+\mathrm{b}$ is equal $3$ .

Hence, option (A) is correct.

Step 6: Calculating $(adj\mathrm{M}{)}^{-1}+ad{\mathrm{j}}^{-1}=-\mathrm{M}$

We know that,

$(adj\mathrm{M}{)}^{-1}+4{\mathrm{d}}_{\mathrm{j}}\left({\mathrm{M}}^{-1}\right)=2(adj\mathrm{M}{)}^{-1}$

$\Rightarrow (\mathrm{adjM}{)}^{-1}+\mathrm{adj}\left({\mathrm{M}}^{-1}\right)=2\left(\frac{\mathrm{M}}{\left|\mathrm{M}\right|}\right)\left[\because adj\left({\mathrm{M}}^{-1}\right)=\left(\mathrm{adj}-\mathrm{M}\right)\right]$

$\Rightarrow (\mathrm{adjM}{)}^{-1}+\mathrm{adj}\left({\mathrm{M}}^{-1}\right)=\frac{-2M}{2}=-M\left((\mathrm{adjM}{)}^{-1}=\frac{\mathrm{M}}{\left|\mathrm{M}\right|}\right)$

Therefore,

$(\mathrm{adjM}{)}^{-2}+\mathrm{adj}\left({\mathrm{M}}^{-1}\right)=-\mathrm{M}$

Therefore,

$\left(\mathrm{adjM}{)}^{-1}\right)+{\mathrm{adjM}}^{-1}=-\mathrm{M}$

Therefore, option (C) is correct.

Step 7: Checking for Option (D)

If, $\mathrm{M}\left[\begin{array}{l}\mathrm{\alpha }\\ \mathrm{\beta }\\ \mathrm{\gamma }\end{array}\right]=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]$ then $\mathrm{\alpha }-\mathrm{\beta }+\mathrm{\gamma }=3$

$\mathrm{M}=\left[\begin{array}{lll}0& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]$

Now if,

$\mathrm{M}\left[\begin{array}{l}\mathrm{\alpha }\\ \mathrm{\beta }\\ \mathrm{\gamma }\end{array}\right]=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]$

$\Rightarrow \left[\begin{array}{lll}0& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]\left[\begin{array}{l}\mathrm{\alpha }\\ \mathrm{\beta }\\ \mathrm{\gamma }\end{array}\right]=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]$

$$\begin{gathered} \Rightarrow \mathrm{\beta }+2\mathrm{\gamma }=1-11 \\ \mathrm{\alpha }+2\mathrm{\beta }+3\mathrm{\gamma }=2-\left(2\right) \\ 3\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=3-\left(3\right) \end{gathered}$$

By Solving the above three Equation we get,

$\mathrm{\alpha }=1,\mathrm{\beta }=-1$ and $\mathrm{\gamma }=1$

Therefore,

$$\begin{gathered} \mathrm{\alpha }-\mathrm{\beta }+\mathrm{\gamma }=1-(-1)+1 \\ \Rightarrow \mathrm{\alpha }-\mathrm{\beta }+\mathrm{\gamma }=1+1+1 \\ \Rightarrow \mathrm{\alpha }-\mathrm{\beta }+\mathrm{\gamma }=3 \end{gathered}$$

Therefore, $\mathrm{\alpha }-\mathrm{\beta }+\mathrm{\gamma }=3.$

Hence, Option (D) is correct.

Explanation for incorrect option:

Step 8: Calculating $\det \left({\mathrm{adjM}}^2\right)$

We know that,

$\left.\Rightarrow \det ({\mathrm{adjM}}^2\right)={\left|{\mathrm{M}}^2\right|}^2$

$\Rightarrow \det \left(\mathrm{ad}\dot{\mathrm{j}}{\mathrm{M}}^2\right)={\left|\mathrm{M}\right|}^4$

$\Rightarrow \det \left({\mathrm{adjM}}^2\right)=16\left[\because |\mathrm{m}{|}^2=4\right]$

Therefore,

$\det \left({\mathrm{adjM}}^2\right)$ is not equal to $81$.

Thus, option (B) is incorrect.

Hence, Option (A), (C) and (D) are the correct answers.