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in Matrices & determinants by (476 points)
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Consider a matrix  A = \(\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}\). If |2 adj(3 adj(4A-1))| = 2a.3b, then the value of a - 2b is 

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A = \(\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}\)

|A| = -1 + 1 + 1 + 1 + 1 + 1 = 4

Now, 

|2 adj (3adj(4A-1))| = 23|adj(3adj (4A-1))|

(\(\because\) |KA| = Kn|A|)

 = 23|3adj(4A-1)|2 (\(\because\) |adj A| = |A|n-1)

 = 23(33|adj (4A-1)|)2 (\(\because\) |KA| = Kn|A|)

= 23.36 |adj(4A-1)|4

= 23.36(43. (A-1))4 (\(\because\) |KA| = Kn|A|)

 23.36(43. \(\frac14\)) (\(\because\) |A-1| = \(\frac1{|A|}=\frac14\))

= 23.36(42)4

= 23.36.48

= 23.36.216

= 219.36

\(\therefore\) a = 19, b = 6

\(\therefore\) a - 2b = 19 - 12 = 7

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