Let A = <math><mrow><mrow><mo>|</mo><mrow><mtable><mtr columnalign="center"><mtd columnalign="center"><mn>1</mn></mtd><mtd columnalign="center"><mi>a</mi></mtd><mtd columnalign="center"><mi>a</mi></mtd></mtr><mtr columnalign="center"><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>1</mn></mtd><mtd columnalign="center"><mi>b</mi></mtd></mtr><mtr columnalign="center"><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>1</mn></mtd></mtr></mtable></mrow><mo>|</mo></mrow></mrow></math> , a, b <math><mrow><mo>∈</mo></mrow></math> R. If for some n <math><mrow><mo>∈</mo></mrow></math> N, An = <math><mrow><mrow><mo>|</mo><mrow><mtable><mtr columnalign="center"><mtd columnalign="center"><mn>1</mn></mtd><mtd columnalign="center"><mn>4</mn><mn>8</mn></mtd><mtd columnalign="center"><mn>2</mn><mn>1</mn><mn>6</mn><mn>0</mn></mtd></mtr><mtr columnalign="center"><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>1</mn></mtd><mtd columnalign="center"><mn>9</mn><mn>6</mn></mtd></mtr><mtr columnalign="center"><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>0</mn></mtd><mtd columnalign="center"><mn>1</mn></mtd></mtr></mtable></mrow><mo>|</mo></mrow></mrow></math> then n + a + b is equal to

A=[100010001]+[0aa00b000]=l+BB2=[0aa00b000]×[0aa00b000]=[00ab000000]B3 = 0∴An=(1+B)n=nC0l+nC1B+nC2B2+nC3B3+....On comparing we get na = 48, nb = 96 and na + n(n−1)2ab=2160⇒a=4,n=12 and b=8⇒n+a+b=24

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Maths Class 12th Maths Matrices

Let A = $| \begin{matrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{matrix} |$ , a, b $\in$ R. If for some n $\in$ N, Aⁿ = $| \begin{matrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{matrix} |$ then n + a + b is equal to

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Payal Gupta | Contributor-Level 10

2 months ago

$A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] + [\begin{matrix} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{matrix}] = l + B$

$B^{2} = [\begin{matrix} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{matrix}] \times [\begin{matrix} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 & a b \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

B³ = 0

$∴ A^{n} = {(1 + B)}^{n} =^{n} C_{0} l +^{n} C_{1} B +^{n} C_{2} B^{2} +^{n} C_{3} B^{3} + . . . .$

On comparing we get na = 48, nb = 96 and na + $\frac{n (n - 1)}{2} a b = 2160$

$\Rightarrow a = 4, n = 12 a n d b = 8$

$\Rightarrow n + a + b = 24$

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Let A = $| \begin{matrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{matrix} |$ , a, b $\in$ R. If for some n $\in$ N, Aⁿ = $| \begin{matrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{matrix} |$ then n + a + b is equal to

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Let A = |1aa01b001| , a, b ∈ R. If for some n ∈ N, An = |14821600196001| then n + a + b is equal to

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Let A = $| \begin{matrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{matrix} |$ , a, b $\in$ R. If for some n $\in$ N, Aⁿ = $| \begin{matrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{matrix} |$ then n + a + b is equal to