Let A=[[|gathered 1 1 2 -2 0 1 1 3 5 gathered]]|.

Question

Let $A=[[|\begin{gathered} 1 \quad  1 \quad  2 \ -2 \quad  0 \quad  1 \ 1 \quad  3 \quad  5 \end{gathered}]]|$. Then the sum of all elements of the matrix $adj(adj(2(adjA{)}^{-1}))$ is equal to:

Accepted Answer

Given

$A=[[|\begin{gathered} 1 \quad  1 \quad  2 \ -2 \quad  0 \quad  1 \ 1 \quad  3 \quad  5 \end{gathered}]]|$

We have to find the sum of all elements of

$adj(adj(2(adjA{)}^{-1}))$

$detA=1|\begin{array}{cc} 0 & 1 \ 3 & 5 \end{array}|-1|\begin{array}{cc} -2 & 1 \ 1 & 5 \end{array}|+2|\begin{array}{cc} -2 & 0 \ 1 & 3 \end{array}|$

$=1(0\cdot 5-1\cdot 3)-1((-2)\cdot 5-1\cdot 1)+2((-2)\cdot 3-0\cdot 1)$

$=-3-(-11)+2(-6)$

$=-3+11-12=-4$

So,

$detA=-4$

For an $n	imes n$ matrix,

$det(adjA)=(detA{)}^{n-1}$

Since $A$ is $3	imes 3$,

$det(adjA)=(detA{)}^{2}=(-4{)}^{2}=16$

Now let

$B=2(adjA{)}^{-1}$

We need $adj(adjB)$.

For a $3	imes 3$ matrix, the property is:

$adj(adjB)=(detB)\, B$

Since

$B=2(adjA{)}^{-1}$

and for a $3	imes 3$ matrix,

$det(2M)={2}^{3}det(M)=8det(M)$

therefore,

$detB=8det((adjA{)}^{-1})$

Also,

$det((adjA{)}^{-1})=\frac{1}{det(adjA)}=\frac{1}{16}$

Hence,

$detB=8\cdot \frac{1}{16}=\frac{1}{2}$

So,

$adj(adjB)=(detB)B=\frac{1}{2}\, B$

But

$B=2(adjA{)}^{-1}$

Thus,

$adj(adjB)=\frac{1}{2}\cdot 2(adjA{)}^{-1}=(adjA{)}^{-1}$

So the required matrix is simply

$(adjA{)}^{-1}$

We know,

$adjA=(detA){A}^{-1}$

Hence,

$(adjA{)}^{-1}=\frac{1}{detA}A$

Since $detA=-4$,

$(adjA{)}^{-1}=\frac{1}{-4}A=-\frac{1}{4}A$

Therefore the required matrix is

$-\frac{1}{4}A$

First find the sum of elements of $A$:

$1+1+2-2+0+1+1+3+5=12$

Therefore sum of elements of $-\frac{1}{4}A$ is

$-\frac{1}{4}\cdot 12=-3$

Final Answer
$-3$

So, the correct option is D.

Solutions · 1