\[A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix} and \text{ I} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]

Given : `A =[[3 1],[-1 2]]` Now, `A^2`=AA `⇒A^2= [[3 1],[-1 2]] [[3 1],[-1 2]]` `⇒A^2=[[9-1 3+2],[-3-2 -1+4]]` `⇒A^2=[[8 5],[-5 3]]` `A^2=5A+λI` ⇒`[[8 5],[-5 3]]=5[[3 1],[-1 2]]+λ[[1 0],[0 1]]` ⇒`[[8 5],[-5 3]]` =`[[15 5],[-5 10]]+[[λ 0],[0 λ]]` ⇒`[[8 5],[-5 3]]`=`[[15+λ 5+0],[-5+0 10+λ]]` ⇒`[[8 5],[-5 3]]``[[15+λ 5+0],[-5+0 10+λ]]` The corresponding elements of two equal matrices are equal. &there4; 8=15+λ ⇒8−15=λ ⇒−7=λ &there4; λ=−7

If A= `[[3 1],[-1 2]]` And I= `[[1 0],[0 1]]`,Then Find λ So That A2 = 5a + λI. - Mathematics

प्रश्न

$A = [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] a n d I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

योग

उत्तर

Given : $A = [\begin{matrix} 3 1 \\ - 1 2 \end{matrix}]$

Now,

$A^{2}$ =AA

$\Rightarrow A^{2} = [\begin{matrix} 3 1 \\ - 1 2 \end{matrix}] [\begin{matrix} 3 1 \\ - 1 2 \end{matrix}]$

$\Rightarrow A^{2} = [\begin{matrix} 9 - 1 3 + 2 \\ - 3 - 2 - 1 + 4 \end{matrix}]$

$\Rightarrow A^{2} = [\begin{matrix} 8 5 \\ - 5 3 \end{matrix}]$

$A^{2} = 5 A + λ I$

⇒ $[\begin{matrix} 8 5 \\ - 5 3 \end{matrix}] = 5 [\begin{matrix} 3 1 \\ - 1 2 \end{matrix}] + λ [\begin{matrix} 1 0 \\ 0 1 \end{matrix}]$

⇒ $[\begin{matrix} 8 5 \\ - 5 3 \end{matrix}]$ = $[\begin{matrix} 15 5 \\ - 5 10 \end{matrix}] + [\begin{matrix} λ 0 \\ 0 λ \end{matrix}]$

⇒ $[\begin{matrix} 8 5 \\ - 5 3 \end{matrix}]$ = $[\begin{matrix} 15 + λ 5 + 0 \\ - 5 + 0 10 + λ \end{matrix}]$

⇒ $[\begin{matrix} 8 5 \\ - 5 3 \end{matrix}]$ $[\begin{matrix} 15 + λ 5 + 0 \\ - 5 + 0 10 + λ \end{matrix}]$

The corresponding elements of two equal matrices are equal.

∴ 8=15+λ

⇒8−15=λ

⇒−7=λ

∴ λ=−7

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अध्याय 5: Algebra of Matrices - Exercise 5.3 [पृष्ठ ४३]

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अध्याय 5 Algebra of Matrices
Exercise 5.3 | Q 28 | पृष्ठ ४३

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If A= $[\begin{matrix} 31 \\ - 12 \end{matrix}]$ And I= $[\begin{matrix} 10 \\ 01 \end{matrix}]$ ,Then Find λ So That A2 = 5a + λI. - Mathematics

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