If A- -160- -160- -160-3-5-4 -160- -160- -160-2- find A2 minus- 5A minus- 14I-

Given: #160;A=3 5 42Now,A2=AA #8658;A2=3 5 423 5 42 #8658;A2=9+20 15 10 12 820+4 #8658;A2=29 25 2024A2 5A 14I #8658;A2 5A 14I=29 25 2024 53 5 42 14100 ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

If A= #160;...

Question

If $A = \"[\"\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}\"]\"$ , find A² − 5A − 14I.

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Solution

$Given : A = [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}] Now, A^{2} = A A \Rightarrow A^{2} = [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}] [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 9 + 20 & - 15 - 10 \\ - 12 - 8 & 20 + 4 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 29 & - 25 \\ - 20 & 24 \end{matrix}] A^{2} - 5 A - 14 I \Rightarrow A^{2} - 5 A - 14 I = [\begin{matrix} 29 & - 25 \\ - 20 & 24 \end{matrix}] - 5 [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}] - 14 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \Rightarrow A^{2} - 5 A - 14 I = [\begin{matrix} 29 & - 25 \\ - 20 & 24 \end{matrix}] - [\begin{matrix} 15 & - 25 \\ - 20 & 10 \end{matrix}] - [\begin{matrix} 14 & 0 \\ 0 & 14 \end{matrix}] \Rightarrow A^{2} - 5 A - 14 I = [\begin{matrix} 29 - 15 - 14 & - 25 + 25 + 0 \\ - 20 + 20 + 0 & 24 - 10 - 14 \end{matrix}] \Rightarrow A^{2} - 5 A - 14 I = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

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Similar questions

Q. If

A = [\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}]

, then find A² − 5A − 14I. Hence, obtain A³.

Q. If

A = [\begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & - 1 & 0 \end{matrix}]

, find A² − 5A + 4I and hence find a matrix X such that A² − 5A + 4I + X = 0.

Q. If

A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 1 2 & 1 & 3 1 & - 1 & 0 \end{matrix} ⎤ ⎥ ⎦

find

A^{2} - 5 A + 4 I

and hence find a matrix X such that

A^{2} - 5 A + 4 I + X = 0

Q. If

A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 1 2 & 1 & 3 1 & - 1 & 0 \end{matrix} ⎤ ⎥ ⎦

, find

A^{2} - 5 A + 4 I

and hence find a matrix

X

such that

A^{2} - 5 A + 4 I + X = 0

Q. Question 4
If

a = \frac{3 + \sqrt{5}}{2}

, then find the value of

a^{2} + \frac{1}{a^{2}}

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If A= 3-5-4 2, find A2 − 5A − 14I.

Given: A=3-5-42Now,A2=AA⇒A2=3-5-423-5-42⇒A2=9+20-15-10-12-820+4⇒A2=29-25-2024A2-5A-14I⇒A2-5A-14I=29-25-2024-53-5-42-141001⇒A2-5A-14I=29-25-2024-15-25-2010-140014⇒A2-5A-14I=29-15-14-25+25+0-20+20+024-10-14⇒A2-5A-14I=0000

If $A = \"[\"\begin{matrix} 3 & - 5 \\ - 4 & 2 \end{matrix}\"]\"$ , find A² − 5A − 14I.