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

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

If $A = [\begin{array}{cc} 3 & - 5 \\ - 4 & 2 \end{array}]$ , then find A² – 5A – 14I. Hence, obtain A³.

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$A = [\begin{array}{cc} 3 & - 5 \\ - 4 & 2 \end{array}] A^{2} = [\begin{array}{cc} 3 & - 5 \\ - 4 & 2 \end{array}] [\begin{array}{cc} 3 & - 5 \\ - 4 & 2 \end{array}]$

= [\begin{array}{cc} 9 + 20 & - 15 - 10 \\ - 12 - 8 & 20 + 4 \end{array}]

= [\begin{array}{cc} 29 & - 25 \\ - 20 & 24 \end{array}] A^{2} - 5 A - 14 I = [\begin{array}{cc} 29 & - 25 \\ - 20 & 24 \end{array}] - 5 [\begin{array}{cc} 3 & - 5 \\ - 4 & 2 \end{array}] - 14 [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}]

= [\begin{array}{cc} 29 - 15 - 14 & - 25 + 25 - 0 \\ - 20 + 20 - 0 & 24 - 10 - 14 \end{array}]

= [\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}]

Therefore, A² – 5A – 14I = 0 ......................(1)

PerMultiplying the (1) by A, we get,

A(A² – 5A – 14I) = A. 0

⇒ A³ – 5A² – 14A = 0

⇒ A³ = 5A² + 14A

A^{3} = 5 [\begin{array}{cc} 29 & - 25 \\ - 20 & 24 \end{array}] + 14 [\begin{array}{cc} 3 & - 5 \\ - 4 & 2 \end{array}]

= [\begin{array}{cc} 145 + 42 & - 125 - 70 \\ - 100 - 56 & 120 + 28 \end{array}]

= [\begin{array}{cc} 187 & - 195 \\ - 156 & 148 \end{array}]

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