Let A be a 3 3 matrix such that A^T[[|gathered 1 0 1 gathered]]|=[[|gathered 5 2 2 gathere…

Question

Let A be a $3	imes 3$ matrix such that

${A}^{T}[[|\begin{gathered} 1 \ 0 \ 1 \end{gathered}]]|=[[|\begin{gathered} 5 \ 2 \ 2 \end{gathered}]]|,{A}^{T}[[|\begin{gathered} 0 \ 0 \ 1 \end{gathered}]]|=[[|\begin{gathered} 3 \ 1 \ 1 \end{gathered}]]|,A[[|\begin{gathered} 1 \ 0 \ 1 \end{gathered}]]|=[[|\begin{gathered} 3 \ 4 \ 4 \end{gathered}]]|A[[|\begin{gathered} 0 \ 0 \ 1 \end{gathered}]]|=[[|\begin{gathered} 1 \ 3 \ 1 \end{gathered}]]|.$

If $\det (A)=1$, then $\det (adj({A}^{2}+A))$ is equal to:

Accepted Answer

Let

$u=[[\begin{vmatrix} 1 \ 0 \ 1 \end{gathered}]]|,\, v=[[|\begin{gathered} 0 \ 0 \ 1 \end{gathered}]]|$

We are given

${A}^{T}u=[[|\begin{gathered} 5 \ 2 \ 2 \end{gathered}]]|,\, {A}^{T}v=[[|\begin{gathered} 3 \ 1 \ 1 \end{gathered}]]|$

and

$Au=[[|\begin{gathered} 3 \ 4 \ 4 \end{gathered}]]|,\, Av=[[|\begin{gathered} 1 \ 3 \ 1 \end{gathered}]]|$

We have to find

$det(adj({A}^{2}+A))$

given that

$det(A)=1$

Use the relation $A(u-v)$: First note that

$u-v=[[|\begin{gathered} 1 \ 0 \ 1 \end{gathered}]]|-[[|\begin{gathered} 0 \ 0 \ 1 \end{gathered}]]|=[[|\begin{gathered} 1 \ 0 \ 0 \end{gathered}]]|$

So,

$A[[|\begin{gathered} 1 \ 0 \ 0 \end{gathered}]]|=A(u-v)=Au-Av$

Now,

$Au-Av=[[|\begin{gathered} 3 \ 4 \ 4 \end{gathered}]]|-[[|\begin{gathered} 1 \ 3 \ 1 \end{gathered}]]|=[[|\begin{gathered} 2 \ 1 \ 3 \end{gathered}]]|$

Hence the first column of $A$ is

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

Also, since

$A[[|\begin{gathered} 0 \ 0 \ 1 \end{gathered}]]|=[[|\begin{gathered} 1 \ 3 \ 1 \end{gathered}]]|$

the third column of $A$ is

$[[|\begin{gathered} 1 \ 3 \ 1 \end{gathered}]]|$

Now use

$A[[|\begin{gathered} 1 \ 0 \ 1 \end{gathered}]]|=A[[|\begin{gathered} 1 \ 0 \ 0 \end{gathered}]]|+A[[|\begin{gathered} 0 \ 0 \ 1 \end{gathered}]]|$

which gives

$[[|\begin{gathered} 3 \ 4 \ 4 \end{gathered}]]|=[[|\begin{gathered} 2 \ 1 \ 3 \end{gathered}]]|+[[|\begin{gathered} 1 \ 3 \ 1 \end{gathered}]]|$

This is consistent.

So let

$A=[[|\begin{gathered} 2 \quad  a \quad  1 \ 1 \quad  b \quad  3 \ 3 \quad  c \quad  1 \end{gathered}]]|$

Now use $det(A)=1$.

Find the second column using determinant: Compute determinant:

$det(A)=|\begin{gathered} 2 \quad  a \quad  1 \ 1 \quad  b \quad  3 \ 3 \quad  c \quad  1 \end{vmatrix}$

Expanding along the first row,

$det(A)=2|\begin{array}{cc} b & 3 \ c & 1 \end{array}|-a|\begin{array}{cc} 1 & 3 \ 3 & 1 \end{array}|+1|\begin{array}{cc} 1 & b \ 3 & c \end{array}|$

$=2(b-3c)-a(1-9)+(c-3b)$

$=2b-6c+8a+c-3b$

$=8a-b-5c$

Given $det(A)=1$,

$8a-b-5c=1$

But we still need more information. For that use the conditions involving ${A}^{T}$.

Use transpose conditions: Since

${A}^{T}[[\begin{vmatrix} 0 \ 0 \ 1 \end{gathered}]]|=[[|\begin{gathered} 3 \ 1 \ 1 \end{gathered}]]|$

this means the third column of ${A}^{T}$ is $[[|\begin{gathered} 3 \ 1 \ 1 \end{gathered}]]|$, so the third row of $A$ is

$[\begin{gathered} 3 \quad  1 \quad  1 \end{gathered}]$

Thus,

$c=1$

Now $A$ becomes

$A=[[|\begin{gathered} 2 \quad  a \quad  1 \ 1 \quad  b \quad  3 \ 3 \quad  1 \quad  1 \end{gathered}]]|$

Next, use

${A}^{T}[[|\begin{gathered} 1 \ 0 \ 1 \end{gathered}]]|=[[|\begin{gathered} 5 \ 2 \ 2 \end{gathered}]]|$

But

${A}^{T}[[|\begin{gathered} 1 \ 0 \ 1 \end{gathered}]]|={A}^{T}[[|\begin{gathered} 1 \ 0 \ 0 \end{gathered}]]|+{A}^{T}[[|\begin{gathered} 0 \ 0 \ 1 \end{gathered}]]|$

Now ${A}^{T}[[|\begin{gathered} 1 \ 0 \ 0 \end{gathered}]]|$ is the first column of ${A}^{T}$, i.e. the first row of $A$, which is

$[[|\begin{gathered} 2 \ a \ 1 \end{gathered}]]|$

And

${A}^{T}[[|\begin{gathered} 0 \ 0 \ 1 \end{gathered}]]|=[[|\begin{gathered} 3 \ 1 \ 1 \end{gathered}]]|$

Therefore,

$[[|\begin{gathered} 2 \ a \ 1 \end{gathered}]]|+[[|\begin{gathered} 3 \ 1 \ 1 \end{gathered}]]|=[[|\begin{gathered} 5 \ 2 \ 2 \end{gathered}]]|$

So,

$[[|\begin{gathered} 5 \ a+1 \ 2 \end{gathered}]]|=[[|\begin{gathered} 5 \ 2 \ 2 \end{gathered}]]|$

Hence,

$a=1$

Now use $det(A)=1$:

$8a-b-5c=1$

with $a=1,c=1$,

$8-b-5=1$

$3-b=1$

$b=2$

Thus,

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

Find ${A}^{2}+A$: First compute ${A}^{2}$:

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

Multiplying row by column:

${A}^{2}=[[|\begin{gathered} 8 \quad  5 \quad  6 \ 13 \quad  8 \quad  10 \ 10 \quad  6 \quad  7 \end{gathered}]]|$

So,

${A}^{2}+A=[[|\begin{gathered} 8 \quad  5 \quad  6 \ 13 \quad  8 \quad  10 \ 10 \quad  6 \quad  7 \end{gathered}]]|+[[|\begin{gathered} 2 \quad  1 \quad  1 \ 1 \quad  2 \quad  3 \ 3 \quad  1 \quad  1 \end{gathered}]]|=[[|\begin{gathered} 10 \quad  6 \quad  7 \ 14 \quad  10 \quad  13 \ 13 \quad  7 \quad  8 \end{gathered}]]|$

Use determinant property: For a $3	imes 3$ matrix $M$,

$det(adjM)=det(M{)}^{3-1}=det(M{)}^{2}$

So we first find

$det({A}^{2}+A)$

Now,

${A}^{2}+A=A(A+I)$

Hence,

$det({A}^{2}+A)=det(A)det(A+I)$

Given $det(A)=1$, so

$det({A}^{2}+A)=det(A+I)$

Now

$A+I=[[|\begin{gathered} 3 \quad  1 \quad  1 \ 1 \quad  3 \quad  3 \ 3 \quad  1 \quad  2 \end{gathered}]]|$

Its determinant is

$det(A+I)=|\begin{gathered} 3 \quad  1 \quad  1 \ 1 \quad  3 \quad  3 \ 3 \quad  1 \quad  2 \end{vmatrix}$

Expanding along the first row:

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

$=3(6-3)-(2-9)+(1-9)$

$=3\cdot 3-(-7)-8$

$=9+7-8=8$

So,

$det({A}^{2}+A)=8$

Therefore,

$det(adj({A}^{2}+A))=det({A}^{2}+A{)}^{2}={8}^{2}=64$

Final Answer
$64$

So, the correct option is D.

Solutions · 1