If A-101001ab2- then aI - bA -2 A 2 equalsa Ab -Ac a b Ad none of these

(d) none of these A=101001ab2 #8658;A2=1+ab3ab23a2ba+b+4Now,aI+bA+2A2=a000a000a+b0b00babb22b+2+2a2b62a2b46a6b2a+2b+8 #160; #160; #160; #160; #160; ...

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Byju's Answer

Standard XII

Mathematics

Property 4

If A=101001ab...

Question

If $A = [\begin{matrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2 \end{matrix}], then a I + b A + 2 A^{2}$ equals
(a) A
(b) − A
(c) ab A
(d) none of these

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Solution

(d) none of these

$A = [\begin{matrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 + a & b & 3 \\ a & b & 2 \\ 3 a & 2 b & a + b + 4 \end{matrix}] Now, a I + b A + 2 A^{2} = [\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}] + [\begin{matrix} b & 0 & b \\ 0 & 0 & b \\ a b & b^{2} & 2 b \end{matrix}] + [\begin{matrix} 2 + 2 a & 2 b & 6 \\ 2 a & 2 b & 4 \\ 6 a & 6 b & 2 a + 2 b + 8 \end{matrix}] = [\begin{matrix} 3 a + b + 2 & 2 b & b + 6 \\ 2 a & a + 2 b & b + 4 \\ a b + 6 a & b^{2} + 6 b & 3 a + 4 b + 8 \end{matrix}]$

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If A=101001ab2, then aI+bA+2 A2 equals (a) A (b) − A (c) ab A (d) none of these

(d) none of these A=101001ab2⇒A2=1+ab3ab23a2ba+b+4Now,aI+bA+2A2=a000a000a+b0b00babb22b+2+2a2b62a2b46a6b2a+2b+8 =3a+b+22bb+62aa+2bb+4ab+6ab2+6b3a+4b+8

If $A = [\begin{matrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2 \end{matrix}], then a I + b A + 2 A^{2}$ equals
(a) A
(b) − A
(c) ab A
(d) none of these