Let A=[[|gathered 1 2 7 4 -2 8 3 8 -7 gathered]]| and (A- I)=0, where is a real number.

Question

Let $A=[[|\begin{gathered} 1 \quad  2 \quad  7 \ 4 \quad  -2 \quad  8 \ 3 \quad  8 \quad  -7 \end{gathered}]]|$ and $\det (A-\alpha I)=0$, where $\alpha $ is a real number. If the largest possible value of $\alpha $ is $p$, then the circle $(x-p{)}^{2}+(y-2p{)}^{2}=320$, intersects the co-ordinate axes at

Accepted Answer

We need the largest real value of $\alpha $ such that

$det(A-\alpha I)=0$

That means $\alpha $ is an eigenvalue of the matrix

$A=[[\begin{vmatrix} 1 \quad  2 \quad  7 \ 4 \quad  -2 \quad  8 \ 3 \quad  8 \quad  -7 \end{gathered}]]|.$

So first we find the eigenvalues of $A$.

Form $A-\alpha I$: $A-\alpha I=[[|\begin{gathered} 1-\alpha  \quad  2 \quad  7 \ 4 \quad  -2-\alpha  \quad  8 \ 3 \quad  8 \quad  -7-\alpha \end{gathered}]]|$

Now,

$det(A-\alpha I)=|\begin{gathered} 1-\alpha  \quad  2 \quad  7 \ 4 \quad  -2-\alpha  \quad  8 \ 3 \quad  8 \quad  -7-\alpha \end{vmatrix}=0$

Expand the determinant: Expanding along the first row:

$det(A-\alpha I)=(1-\alpha )|\begin{array}{cc} -2-\alpha  & 8 \ 8 & -7-\alpha \end{array}|-2|\begin{array}{cc} 4 & 8 \ 3 & -7-\alpha \end{array}|+7|\begin{array}{cc} 4 & -2-\alpha \ 3 & 8 \end{array}|$

Now calculate each minor:

$|\begin{array}{cc} -2-\alpha  & 8 \ 8 & -7-\alpha \end{array}|=(-2-\alpha )(-7-\alpha )-64$

$=(\alpha +2)(\alpha +7)-64={\alpha }^{2}+9\alpha +14-64={\alpha }^{2}+9\alpha -50$

Next,

$|\begin{array}{cc} 4 & 8 \ 3 & -7-\alpha \end{array}|=4(-7-\alpha )-24=-28-4\alpha -24=-52-4\alpha $

And,

$|\begin{array}{cc} 4 & -2-\alpha \ 3 & 8 \end{array}|=32-3(-2-\alpha )=32+6+3\alpha =38+3\alpha $

So,

$det(A-\alpha I)=(1-\alpha )({\alpha }^{2}+9\alpha -50)-2(-52-4\alpha )+7(38+3\alpha )$

Now simplify:

$(1-\alpha )({\alpha }^{2}+9\alpha -50)={\alpha }^{2}+9\alpha -50-{\alpha }^{3}-9{\alpha }^{2}+50\alpha $

$=-{\alpha }^{3}-8{\alpha }^{2}+59\alpha -50$

Also,

$-2(-52-4\alpha )=104+8\alpha $

and

$7(38+3\alpha )=266+21\alpha $

Therefore,

$det(A-\alpha I)=-{\alpha }^{3}-8{\alpha }^{2}+59\alpha -50+104+8\alpha +266+21\alpha $

$=-{\alpha }^{3}-8{\alpha }^{2}+88\alpha +320$

So the equation becomes

$-{\alpha }^{3}-8{\alpha }^{2}+88\alpha +320=0$

Multiplying by $-1$:

${\alpha }^{3}+8{\alpha }^{2}-88\alpha -320=0$

Factorise: Try integer roots. On putting $\alpha =8$:

${8}^{3}+8({8}^{2})-88(8)-320=512+512-704-320=0$

So $(\alpha -8)$ is a factor.

Divide:

${\alpha }^{3}+8{\alpha }^{2}-88\alpha -320=(\alpha -8)({\alpha }^{2}+16\alpha +40)$

Now solve

${\alpha }^{2}+16\alpha +40=0$

$\alpha =\frac{-16\pm \sqrt{256-160}}{2}=\frac{-16\pm \sqrt{96}}{2}=\frac{-16\pm 4\sqrt{6}}{2}=-8\pm 2\sqrt{6}$

Hence the eigenvalues are

$8,\, -8+2\sqrt{6},\, -8-2\sqrt{6}$

So the largest possible value is

$p=8$

Put $p=8$ in the circle: Given circle:

$(x-p{)}^{2}+(y-2p{)}^{2}=320$

Substitute $p=8$:

$(x-8{)}^{2}+(y-16{)}^{2}=320$

Its centre is $(8,16)$ and radius is

$r=\sqrt{320}=8\sqrt{5}$

Find intersections with the coordinate axes: On the $x$-axis
For points on the $x$-axis, $y=0$.

$(x-8{)}^{2}+(0-16{)}^{2}=320$

$(x-8{)}^{2}+256=320$

$(x-8{)}^{2}=64$

$x-8=\pm 8$

So,

$x=16\, \, x=0$

Thus the circle meets the $x$-axis at

$(16,0),(0,0)$

On the $y$-axis
For points on the $y$-axis, $x=0$.

$(0-8{)}^{2}+(y-16{)}^{2}=320$

$64+(y-16{)}^{2}=320$

$(y-16{)}^{2}=256$

$y-16=\pm 16$

So,

$y=32\, \, y=0$

Thus the circle meets the $y$-axis at

$(0,32),(0,0)$

Count distinct intersection points: The distinct points are

$(16,0),(0,0),(0,32)$

So the circle intersects the coordinate axes at

$3$

Hence, the correct option is:

Solutions · 1