Let A = 1 - 1 0 0 1 - 1 0 0 1 and B = 7 A 20 - 20 A 7 + 2 I , where I is an identity matrix of order 3 × 3 . If B = b i j , then b 13 is equal to

Let A = 1 - 1 0 0 1 - 1 0 0 1 = I + C where I = 1 0 0 0 1 0 0 0 1 , C = 0 - 1 0 0 0 - 1 0 0 0 Then, C 2 = 0 0 1 0 0 0 0 0 0 , C 3 = 0 0 0 0 0 0 0 0 0 = C 4 = C 5 = … … . Now, B = 7 A 20 - 20 A 7 + 2 I = 7 ( I + C ) 20 - 20 ( I + C ) 7 + 2 I = 7 I + 20 C + C 2 20 C 2 + . . . - 20 I + 7 C + C 2 7 C 2 + . . + 2 I = – 11 I + 910 C 2 = - 11 0 910 0 - 11 0 0 0 - 11 So, b 13 = 910

Mathematics - Matrices Question with Solution | TestHub

MathematicsMatricesProduct of MatricesHard2 minPYQ_2021

MathematicsHardnumerical

Let $A = [\begin{matrix} \begin{matrix} 1 & - 1 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{matrix} \end{matrix}]$ and $B = 7 A^{20} - 20 A^{7} + 2 I$ ,
where $I$ is an identity matrix of order $3 \times 3 .$ If $B = [b_{i j}]$ , then $b_{13}$ is equal to

Answer:

910.00

Solution:

Let $A = [\begin{matrix} \begin{matrix} 1 & - 1 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{matrix} \end{matrix}] = I + C$

where $I = [\begin{matrix} \begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \end{matrix}], C = [\begin{matrix} \begin{matrix} 0 & - 1 & 0 \\ 0 & 0 & - 1 \\ 0 & 0 & 0 \end{matrix} \end{matrix}]$

Then,

$C^{2} = \begin{matrix} [\begin{matrix} \begin{array}{l} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \end{matrix}] \end{matrix}$ ,

$C^{3} = [\begin{matrix} \begin{array}{l} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \end{matrix}] = C^{4} = C^{5} = \dots \dots .$

Now,

$B = 7 A^{20} - 20 A^{7} + 2 I$

$= 7 (I + C)^{20} - 20 (I + C)^{7} + 2 I$

$= 7 (I + 20 C + {}^{20}C_{2} C^{2} + . . .) - 20 (I + 7 C + {}^{7}C_{2} C^{2} + . .) + 2 I$

$= - 11 I + 910 C^{2}$

$= [\begin{matrix} - 11 & 0 & 910 \\ 0 & - 11 & 0 \\ 0 & 0 & - 11 \end{matrix}]$

So,

$b_{13} = 910$

Stream:JEESubject:MathematicsTopic:MatricesSubtopic:Product of Matrices

⏱ 2mℹ️ Source: PYQ_2021

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