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Mathematics Question on Matrices

If f(x)=x2+4x−5f (x) = x^2 + 4x - 5 and A=[12 4−3]A = \begin{bmatrix}1&2\\\ 4&-3\end{bmatrix} then f (A) is equal to

A

[0\-4 88]\begin{bmatrix} 0 & \- 4 \\\ 8 & 8 \end{bmatrix}

B

[21 20]\begin{bmatrix} 2 & 1 \\\ 2 & 0 \end{bmatrix}

C

[11 10]\begin{bmatrix} 1 & 1 \\\ 1 & 0 \end{bmatrix}

D

[84 80]\begin{bmatrix} 8 & 4 \\\ 8 & 0 \end{bmatrix}

Answer

[84 80]\begin{bmatrix} 8 & 4 \\\ 8 & 0 \end{bmatrix}

Explanation

Solution

Given : A=[12 4−3]A = \begin{bmatrix}1&2\\\ 4&-3\end{bmatrix} ∴A2=A.A=[12 4−3][12 4−3]\therefore A^{2} = A.A = \begin{bmatrix}1&2\\\ 4&-3\end{bmatrix}\begin{bmatrix}1&2\\\ 4&-3\end{bmatrix} =[1+82−6 4−128+9]=[9−4 −817] = \begin{bmatrix}1+8&2-6\\\ 4-12&8+9\end{bmatrix} = \begin{bmatrix}9&-4\\\ -8&17\end{bmatrix} Now, f(x)=x2+4x−5 f \left(x\right) = x^{2} + 4x - 5 ∴f(A)=A2+4A−5 \therefore f \left(A\right) = A^{2} + 4A - 5 =A2+4A−5I= A^{2} + 4A - 5 I (I is a 2×22 \times 2unit matrix) =[9−4 −817]+4[12 4−3]−5[10 01]= \begin{bmatrix}9&-4\\\ -8&17\end{bmatrix} + 4 \begin{bmatrix}1&2\\\ 4&-3\end{bmatrix}- 5 \begin{bmatrix}1&0\\\ 0&1\end{bmatrix} =[9−4 −817]+[48 16−12]+[−50 0−5] = \begin{bmatrix}9&-4\\\ -8&17\end{bmatrix} + \begin{bmatrix}4&8\\\ 16&-12\end{bmatrix}+ \begin{bmatrix}-5&0\\\ 0&-5\end{bmatrix} =[84 80]= \begin{bmatrix}8&4\\\ 8&0\end{bmatrix}