Let I be an identity matrix of order 2 × 2 and P = 2 - 1 5 - 3 . Then the value of n ∈ N for which P n = 5 I - 8 P is equal to ___ .

P = 2 - 1 5 - 3 5 I - 8 P = 5 0 0 5 - 16 - 8 40 - 24 = - 11 8 - 40 29 P 2 = 2 - 1 5 - 3 2 - 1 5 - 3 = - 1 1 - 5 4 P 3 = 2 - 1 5 - 3 - 1 1 - 5 4 = 3 - 2 10 - 7 ⇒ P 6 = 3 - 2 10 - 7 3 - 2 10 - 7 = - 11 8 - 40 29 = P n ⇒ n = 6

Home
PYQ PracticeFree
JEE Rank Predictor
Answer Key

Mathematics - Matrices Question with Solution | TestHub

MathematicsMatricesProduct of MatricesEasy2 minPYQ_2021

MathematicsEasynumerical

Let $I$ be an identity matrix of order $2 \times 2$ and $P = [\begin{matrix} 2 & - 1 \\ 5 & - 3 \end{matrix}]$ . Then the value of $n \in N$ for which $P^{n} = 5 I - 8 P$ is equal to ___ .

Answer:

6.00

Solution:

$P = [\begin{array}{l} 2 & - 1 \\ 5 & - 3 \end{array}]$

$5 I - 8 P = [\begin{array}{l} 5 & 0 \\ 0 & 5 \end{array}] - [\begin{matrix} 16 & - 8 \\ 40 & - 24 \end{matrix}] = [\begin{matrix} - 11 & 8 \\ - 40 & 29 \end{matrix}]$

$P^{2} = [\begin{array}{l} 2 & - 1 \\ 5 & - 3 \end{array}] [\begin{array}{l} 2 & - 1 \\ 5 & - 3 \end{array}] = [\begin{array}{l} - 1 & 1 \\ - 5 & 4 \end{array}]$

$P^{3} = [\begin{array}{l} 2 & - 1 \\ 5 & - 3 \end{array}] [\begin{array}{l} - 1 & 1 \\ - 5 & 4 \end{array}] = [\begin{matrix} 3 & - 2 \\ 10 & - 7 \end{matrix}]$

$\Rightarrow P^{6} = [\begin{matrix} 3 & - 2 \\ 10 & - 7 \end{matrix}] [\begin{matrix} 3 & - 2 \\ 10 & - 7 \end{matrix}] = [\begin{matrix} - 11 & 8 \\ - 40 & 29 \end{matrix}] = P^{n}$

$\Rightarrow n = 6$

Stream:JEESubject:MathematicsTopic:MatricesSubtopic:Product of Matrices

⏱ 2mℹ️ Source: PYQ_2021

Doubts & Discussion

Loading discussions...