If A=≤ft[{array}{rr} -1 & 2 \\ 3 & 1 {array}] then find f(A) where f(x)=x²-2x+3 ?

Correct Answer is : [12−4−68]≤ft[{array}{rr} 12 & -4 \\ -6 & 8 {array}][12−6−48]

≤ft[{array}{rr} 12 & 4 \\ 6 & 8 {array}]

≤ft[{array}{rr} 12 & 4 \\ -6 & 8 {array}]

≤ft[{array}{rr} 12 & -4 \\ -6 & 8 {array}]

≤ft[{array}{rr} 12 & -4 \\ 6 & 8 {array}]

Correct Answer is : [12−4−68]≤ft[{array}{rr} 12 & -4 \\ -6 & 8 {array}][12−6−48]

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UPSC NDA Math Model Paper 8

Question : 114 of 120

Marks: +1, -0

A=\left[\begin{array}{rr} -1 & 2 \\ 3 & 1 \end{array}\right]

f(A)

f(x)=x^{2}-2x+3 ?

Solution:

Given:

A=\left[\begin{array}{rr} -1 & 2 \\ 3 & 1 \end{array}\right]

f(x)=x^{2}-2x+3

Here, we have to find the value of f(A)

\because f(x)=x^{2}-2x+3

\Rightarrow f(A)=A^{2}-2\cdot A+3\cdot I

where I is an identity matrix of order 2.

\Rightarrow A^{2}=\left[\begin{array}{rr} 7 & 0 \\ 0 & 7 \end{array}\right], 2\cdot A=\left[\begin{array}{rr} -2 & 4 \\ 6 & 2 \end{array}\right]

and

3\cdot I=\left[\begin{array}{rr} 3 & 0 \\ 0 & 3 \end{array}\right]

\Rightarrow A^{2}-2\cdot A+3\cdot I=\left[\begin{array}{rr} 7 & 0 \\ 0 & 7 \end{array}\right]-\left[\begin{array}{rr} -2 & 4 \\ 6 & 2 \end{array}\right]

+\left[\begin{array}{rr} 3 & 0 \\ 0 & 3 \end{array}\right]

\Rightarrow A^{2}-2\cdot A+3\cdot I=\left[\begin{array}{rr} 12 & -4 \\ -6 & 8 \end{array}\right]

Hence,

f(A)=\left[\begin{array}{rr} 12 & -4 \\ -6 & 8 \end{array}\right]

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