If f(x)=x2+4x−5 and A=124−3 then f(A) is equal to.

Given: A=124−3∴A2=A.A=124−3124−3=1+82−64−128+9=9−4−817Now, f(x)=x2+4x−5∴f(A)=A2+4A−5=A2+4A−5I (I is a 2×2 unit matrix) =9−4−817+4124−3−51001=9−4−817+4816−12+−500−5=84780

If f(x)=x2+4x−5 and A=124−3 then f(A) is equal to.

Given: A=124−3∴A2=A.A=124−3124−3=1+82−64−128+9=9−4−817Now, f(x)=x2+4x−5∴f(A)=A2+4A−5=A2+4A−5I (I is a 2×2 unit matrix) =9−4−817+4124−3−51001=9−4−817+4816−12+−500−5=84780

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If $f (x) = x^{2} + 4 x - 5 and A = [\begin{array}{cc} 1 & 2 \\ 4 & - 3 \end{array}]$ then f(A) is equal to.

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$[\begin{array}{cc} 0 & - 4 \\ 8 & 8 \end{array}]$

$[\begin{array}{ll} 2 1 \\ 2 0 \end{array}]$

$[\begin{array}{ll} 1 1 \\ 1 0 \end{array}]$

$[\begin{array}{ll} 8 4 \\ 8 0 \end{array}]$

answer is D.

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Detailed Solution

Given: $A = [\begin{array}{cc} 1 & 2 \\ 4 & - 3 \end{array}]$

$\begin{aligned} ∴ A^{2} = A . A = [\begin{array}{cc} 1 & 2 \\ 4 & - 3 \end{array}] [\begin{array}{cc} 1 & 2 \\ 4 & - 3 \end{array}] \\ = [\begin{array}{cc} 1 + 8 & 2 - 6 \\ 4 - 12 & 8 + 9 \end{array}] = [\begin{array}{cc} 9 & - 4 \\ - 8 & 17 \end{array}] \end{aligned}$

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

$\begin{array}{l} ∴ f (A) = A^{2} + 4 A - 5 \\ = A^{2} + 4 A - 5 I (I is a 2 \times 2 unit matrix) \end{array}$

$\begin{array}{l} = [\begin{array}{cc} 9 & - 4 \\ - 8 & 17 \end{array}] + 4 [\begin{array}{cc} 1 & 2 \\ 4 & - 3 \end{array}] - 5 [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] \\ = [\begin{array}{cc} 9 & - 4 \\ - 8 & 17 \end{array}] + [\begin{array}{cc} 4 & 8 \\ 16 & - 12 \end{array}] + [\begin{array}{cc} - 5 & 0 \\ 0 & - 5 \end{array}] \\ = [\begin{array}{cc} 8 & 47 \\ 8 & 0 \end{array}] \end{array}$