Given A - bmatrix 280 1-1877-endbmatrix and - I - begin bmatrix 18011081endbmatrix and A 2-9 A - mI- Find m-A- -14B- 18C- -18D- 14

The correct option is A 14 Given, A^2=9A+mI ⇒[ 2 amp;0; 1 amp;7 ][ 2 amp; 0; 1 amp;7 ]=9[ 2 amp; 0; 1 amp;7 ]+m[ 1 amp; 0; ...

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Byju's Answer

Standard X

Mathematics

Equality of Matrices

Given A = bma...

Question

Given $A = [\begin{matrix} 2 & 0 - 1 & 7 \end{matrix}] a n d I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$ and $A^{2} = 9 A + m I$ . Find m.

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-18

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-14

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Solution

The correct option is D
-14

Given, $A^{2} = 9 A + m I$
$\Rightarrow [\begin{matrix} 2 & 0 - 1 & 7 \end{matrix}] [\begin{matrix} 2 & 0 - 1 & 7 \end{matrix}] = 9 [\begin{matrix} 2 & 0 - 1 & 7 \end{matrix}] + m [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 4 & 0 - 9 & 49 \end{matrix}] = [\begin{matrix} 18 & 0 - 9 & 63 \end{matrix}] + [\begin{matrix} m & 0 0 & m \end{matrix}] \Rightarrow [\begin{matrix} 4 & 0 - 9 & 49 \end{matrix}] = [\begin{matrix} 18 + m & 0 - 9 & 63 + m \end{matrix}] \Rightarrow 18 + m = 4 \Rightarrow m = - 14$

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Similar questions

Q. Given

A = [\begin{matrix} 2 & 0 - 1 & 7 \end{matrix}]

and

I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]

and

A^{2} = 9 A + m I

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and

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Given A=[20−17]and I=[1001] and A2=9A+mI. Find m.

The correct option is D -14 Given, A2=9A+mI ⇒[20−17][20−17]=9[20−17]+m[1001]⇒[40−949]=[180−963]+[m00m]⇒[40−949]=[18+m0−963+m]⇒18+m=4⇒m=−14

Given $A = [\begin{matrix} 2 & 0 - 1 & 7 \end{matrix}] a n d I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$ and $A^{2} = 9 A + m I$ . Find m.