If A -23-10- show that A2-2 A-3 I2-0

Given: #160;A=23 10Now,A2=AA #8658;A2=23 1023 10 #8658;A2=4 36+0 2+0 3+0 #8658;A2=16 2 3A2 2A+3I2 #8658; #160;A2 2A+3I2=16 2 3 223 10+31001 #8658; ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

If A =23-10, ...

Question

If $A = [\begin{matrix} 2 & 3 \\ - 1 & 0 \end{matrix}]$ , show that A² − 2A + 3I₂ = O

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Solution

$Given : A = [\begin{matrix} 2 & 3 \\ - 1 & 0 \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 2 & 3 \\ - 1 & 0 \end{matrix}] [\begin{matrix} 2 & 3 \\ - 1 & 0 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 4 - 3 & 6 + 0 \\ - 2 + 0 & - 3 + 0 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 & 6 \\ - 2 & - 3 \end{matrix}] A^{2} - 2 A + 3 I_{2} \Rightarrow A^{2} - 2 A + 3 I_{2} = [\begin{matrix} 1 & 6 \\ - 2 & - 3 \end{matrix}] - 2 [\begin{matrix} 2 & 3 \\ - 1 & 0 \end{matrix}] + 3 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A^{2} - 2 A + 3 I_{2} = [\begin{matrix} 1 & 6 \\ - 2 & - 3 \end{matrix}] - [\begin{matrix} 4 & 6 \\ - 2 & 0 \end{matrix}] + [\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}] \Rightarrow A^{2} - 2 A + 3 I_{2} = [\begin{matrix} 1 - 4 + 3 & 6 - 6 + 0 \\ - 2 + 2 + 0 & - 3 + 0 + 3 \end{matrix}] \Rightarrow A^{2} - 2 A + 3 I_{2} = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = 0 Hence proved .$

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If A= 23-10, show that A2 − 2A + 3I2 = O

Given: A=23-10Now,A2=AA⇒A2=23-1023-10⇒A2=4-36+0-2+0-3+0⇒A2=16-2-3A2-2A+3I2⇒ A2-2A+3I2=16-2-3-223-10+31001⇒ A2-2A+3I2=16-2-3-46-20+3003⇒ A2-2A+3I2=1-4+36-6+0-2+2+0-3+0+3⇒ A2-2A+3I2=0000=0Hence proved.

If $A = [\begin{matrix} 2 & 3 \\ - 1 & 0 \end{matrix}]$ , show that A² − 2A + 3I₂ = O