Given A - beginbmatrix 28-3 -4 - 7 lendbmatrix - compute A -1 and show that 2 A -1-9 I - A

|A| = 2 amp; 3 4 amp; 7 = 14 12 = 2 adj. A = [ 7 amp; 3; 4 amp; 2 ] A^ 1=1/|A|.adj. A=1/2[ 7 amp; 3; 4 amp; 2 ]...(i) Now, 9I ...

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Byju's Answer

Standard XII

Mathematics

Adjoint of a Matrix

Given A = beg...

Question

Given $A$ = $[\begin{matrix} 2 & - 3 - 4 & 7 \end{matrix}],$ compute $A^{- 1}$ and show that $2 A^{- 1} = 9 I - A$

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Solution

$| A | = ∣ ∣ ∣ \begin{matrix} 2 & - 3 - 4 & 7 \end{matrix} ∣ ∣ ∣ = 14 - 12 = 2 adj. A = [\begin{matrix} 7 & 3 4 & 2 \end{matrix}] A^{- 1} = \frac{1}{| A |} . adj. A = \frac{1}{2} [\begin{matrix} 7 & 3 4 & 2 \end{matrix}] . . . (i) Now, 9 I - A = 9 [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] - [\begin{matrix} 2 & - 3 - 4 & 7 \end{matrix}] = [\begin{matrix} 9 & 0 0 & 9 \end{matrix}] - [\begin{matrix} 2 & - 3 - 4 & 7 \end{matrix}] = [\begin{matrix} 9 - 2 & 0 - (- 3) 0 - (- 4) & 9 - 7 \end{matrix}] = [\begin{matrix} 7 & 3 4 & 2 \end{matrix}] = 2 \times \frac{1}{2} [\begin{matrix} 7 & 3 4 & 2 \end{matrix}] = 2 A^{- 1}$
[using $i$ ]

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Similar questions

Given A = $[\begin{matrix} 2 & - 3 - 4 & 7 \end{matrix}]$ , compute $A^{- 1}$ and show that $2 A^{- 1} = 9 I - A$ .

Q. Given

A

[\begin{matrix} 2 & - 3 - 4 & 7 \end{matrix}],

compute

A^{- 1}

and show that

2 A^{- 1} = 9 I - A

Q. Given

A = [\begin{matrix} 2 & - 3 - 4 & 7 \end{matrix}]

, compute

A^{- 1}

ans show that

2 A^{- 1} = 9 I - A

$A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & - 3 5 & 0 & 2 1 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦, B = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 1 & 2 4 & 2 & 5 2 & 0 & 3 \end{matrix} ⎤ ⎥ ⎦, C = ⎡ ⎢ ⎣ \begin{matrix} 4 & 1 & 2 0 & 3 & 2 1 & - 2 & 3 \end{matrix} ⎤ ⎥ ⎦$ compute (A+B) and (B-C). Also verify that A+(B-C)=(A+B)-C.

If $H = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 2 & 3 & 1 3 & 1 & 2 \end{matrix} ⎤ ⎥ ⎦ G = ⎡ ⎢ ⎣ \begin{matrix} 2 & 6 & - 10 0 & - 4 & 8 2 & - 8 & 4 \end{matrix} ⎤ ⎥ ⎦$

Find $3 H - \frac{1}{2} G$

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Given A = [2−3−47], compute A−1 and show that 2A−1=9I−A

|A|=∣∣∣2−3−47∣∣∣=14−12=2adj. A=[7342]A−1=1|A|.adj.A=12[7342]...(i)Now, 9I−A=9[1001]−[2−3−47]=[9009]−[2−3−47]=[9−20−(−3)0−(−4)9−7]=[7342]=2×12[7342]=2A−1 [using i]

Given $A$ = $[\begin{matrix} 2 & - 3 - 4 & 7 \end{matrix}],$ compute $A^{- 1}$ and show that $2 A^{- 1} = 9 I - A$