If A -4521- then show that A -3 I -2 I -3 A -1

We #160;have, #160;A=4521Now,adj(A)=1 5 24and #160;A= 6 #8756;A 1= 161 5 24Now, #160;A 3I=I+3A 1LHS= #160;A 3I=4521 31001=152 2RHS=2I+3A 1=21001 3 ...

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Standard XII

Mathematics

Derivative of One Function w.r.t Another

If A =4521, t...

Question

If $A = [\begin{matrix} 4 & 5 \\ 2 & 1 \end{matrix}]$ , then show that $A - 3 I = 2 (I + 3 A^{- 1}) .$

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Solution

$We have, A = [\begin{matrix} 4 & 5 \\ 2 & 1 \end{matrix}] Now, adj (A) = [\begin{matrix} 1 & - 5 \\ - 2 & 4 \end{matrix}] and |A| = - 6 ∴ A^{- 1} = - \frac{1}{6} [\begin{matrix} 1 & - 5 \\ - 2 & 4 \end{matrix}] Now, A - 3 I = I + 3 A^{- 1} LHS = A - 3 I = [\begin{matrix} 4 & 5 \\ 2 & 1 \end{matrix}] - 3 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 5 \\ 2 & - 2 \end{matrix}] RHS = 2 (I + 3 A^{- 1}) = 2 \{[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - 3 \times \frac{1}{6} [\begin{matrix} 1 & - 5 \\ - 2 & 4 \end{matrix}]\} = 2 [\begin{matrix} 0.5 & 2.5 \\ 1 & - 1 \end{matrix}] = [\begin{matrix} 1 & 5 \\ 2 & - 2 \end{matrix}] = LHS Hence proved .$

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Similar questions

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If A=4521, then show that A-3I=2 I+3A-1.

We have, A=4521Now,adj(A)=1-5-24and A=-6∴A-1=-161-5-24Now, A-3I=I+3A-1LHS= A-3I=4521-31001=152-2RHS=2I+3A-1=21001-3×161-5-24=20.52.51-1=152-2=LHSHence proved.

If $A = [\begin{matrix} 4 & 5 \\ 2 & 1 \end{matrix}]$ , then show that $A - 3 I = 2 (I + 3 A^{- 1}) .$