If f(x) = x² + 4x - 5 and A = {bmatrix} 1 & 2 \\ 4 & -3 {bmatrix} then f ( A ) is equal to

Correct Answer is : [8480]{bmatrix} 8 & 4 \\ 8 & 0 {bmatrix}[8840]

{bmatrix} 0 & -4 \\ 8 & 8 {bmatrix}

{bmatrix} 2 & 1 \\ 2 & 0 {bmatrix}

{bmatrix} 1 & 1 \\ 1 & 0 {bmatrix}

{bmatrix} 8 & 4 \\ 8 & 0 {bmatrix}

Correct Answer is : [8480]{bmatrix} 8 & 4 \\ 8 & 0 {bmatrix}[8840]

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AP and TS EAMCET Engineering Model Paper 5

Question : 11 of 160

Marks: +1, -0

f(x) = x^{2} + 4x - 5

A = \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix}

then f ( A ) is equal to

Solution:

Since

f(x)=x^{2}+4x-5

, for the matrix

A

the constant term becomes

-5I

, so

f(A)=A^{2}+4A-5I.

Compute

A^{2}

A^{2}=\begin{bmatrix}1&2\\4&-3\end{bmatrix}\begin{bmatrix}1&2\\4&-3\end{bmatrix}=\begin{bmatrix}9&-4\\-8&17\end{bmatrix}.

Also,

4A=\begin{bmatrix}4&8\\16&-12\end{bmatrix},\qquad 5I=\begin{bmatrix}5&0\\0&5\end{bmatrix}.

Therefore,

f(A)=\begin{bmatrix}9&-4\\-8&17\end{bmatrix}+\begin{bmatrix}4&8\\16&-12\end{bmatrix}-\begin{bmatrix}5&0\\0&5\end{bmatrix}=\begin{bmatrix}8&4\\8&0\end{bmatrix}.

Thus the correct option is

\begin{bmatrix}8&4\\8&0\end{bmatrix}

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