Question

Solve this and I'll give you brainliest ​

Answer 1

Answer:

let's try

Step-by-step explanation:

answer is 1

I think so dont know why

Answer 2

Given Question

$\rm :\longmapsto\:If \: A = \bigg[ \begin{matrix}3&1 \\ - 1&2 \end{matrix} \bigg] \: and \: {A}^{2} - 5A + 7I = 0 \: then \: I =$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}3&1 \\ - 1&2 \end{matrix} \bigg]$

and

$\rm :\longmapsto\: {A}^{2} - 5A + 7I = 0$

$\red{ \sf \: Pre-multiply \: by \: {A}^{ - 1} \: on \: both \: sides}$

So, we get

$\rm :\longmapsto\: {A}^{ - 1} {A}^{2} - 5 {A}^{ - 1} A + 7 {A}^{ - 1} I = 0$

$\rm :\longmapsto\: ({A}^{ - 1}A)A - 5 I + 7 {A}^{ - 1} = 0$

$\rm :\longmapsto\: IA - 5 I + 7 {A}^{ - 1} = 0$

$\rm :\longmapsto\: A - 5 I + 7 {A}^{ - 1} = 0$

$\rm :\longmapsto\: A + 7 {A}^{ - 1} = 5I$

$\rm :\longmapsto\:I = \dfrac{1}{5}(A + 7 {A}^{ - 1})$

$\rm :\longmapsto\:I = \dfrac{1}{5}A + \dfrac{7}{5} {A}^{ - 1}$

Verification

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}3&1 \\ - 1&2 \end{matrix} \bigg]$

So,

$\rm :\longmapsto\: |A| = 6 - ( - 1) = 7$

Now, Evaluation of Cofactors

$\rm :\longmapsto\:A_{11} = {( - 1)}^{1 + 1}2 = 2$

$\rm :\longmapsto\:A_{12} = {( - 1)}^{1 + 2}( - 1)= 1$

$\rm :\longmapsto\:A_{21} = {( - 1)}^{2 + 1} 1= - 1$

$\rm :\longmapsto\:A_{22} = {( - 1)}^{2 + 2} 3= 3$

So,

$\rm :\longmapsto\:adjA = \bigg[ \begin{matrix}2& - 1 \\1&3 \end{matrix} \bigg]$

So,

$\rm :\longmapsto\: {A}^{ - 1} =\dfrac{1}{ |A| } adjA = \dfrac{1}{7} \bigg[ \begin{matrix}2& - 1 \\1&3 \end{matrix} \bigg]$

Now, Consider

$\rm :\longmapsto\:\dfrac{1}{5}A + \dfrac{7}{5} {A}^{ - 1}$

$\rm \:  =  \: \dfrac{1}{5}\bigg[ \begin{matrix}3& 1 \\ - 1&2 \end{matrix} \bigg] + \dfrac{7}{5} \times \dfrac{1}{7}\bigg[ \begin{matrix}2& - 1 \\ 1&3 \end{matrix} \bigg]$

$\rm \:  =  \: \dfrac{1}{5}\bigg[ \begin{matrix}3& 1 \\ - 1&2 \end{matrix} \bigg] + \dfrac{1}{5}\bigg[ \begin{matrix}2& - 1 \\ 1&3 \end{matrix} \bigg]$

$\rm \:  =  \: \dfrac{1}{5}\bigg(\bigg[ \begin{matrix}3&1 \\ - 1&2 \end{matrix} \bigg] + \bigg[ \begin{matrix}2& - 1 \\ 1&3 \end{matrix} \bigg]\bigg)$

$\rm \:  =  \: \dfrac{1}{5}\bigg[ \begin{matrix}5&0 \\ 0&5 \end{matrix} \bigg]$

$\rm \:  =  \: \bigg[ \begin{matrix}1&0 \\ 0&1 \end{matrix} \bigg]$

$\rm \:  =  \: I$

Hence, Verified

Hence, Option (d) is correct.