Let A= 2 −2 −4−1 3 4 1 −2 xand A2 = A. If r is the rank of A, then r+x=

A2= 2 −2 −4−2 3 4 1 −2 x 2 −2 −4−1 3 4 1 −2 x A2 2 2 −16−4x−1 3 16+4x4+x −8−2x −12+x2A2=A∴4+x=1⇒x=−3A2 −2 −4−1 3 41 −2 −4|A|=2(−9+8)+2(3−4)−4(2−3) =−2−2+4=0Rank A=2⇒r+x=2−3=−1

Let A= 2 −2 −4−1 3 4 1 −2 xand A2 = A. If r is the rank of A, then r+x=

A2= 2 −2 −4−2 3 4 1 −2 x 2 −2 −4−1 3 4 1 −2 x A2 2 2 −16−4x−1 3 16+4x4+x −8−2x −12+x2A2=A∴4+x=1⇒x=−3A2 −2 −4−1 3 41 −2 −4|A|=2(−9+8)+2(3−4)−4(2−3) =−2−2+4=0Rank A=2⇒r+x=2−3=−1

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Let $A = [\begin{array}{l} 2 - 2 - 4 \\ - 1 3 4 \\ 1 - 2 x \end{array}]$ and A²= A. If r is the rank of A, then $r + x =$

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Detailed Solution

$A^{2} = [\begin{array}{l} 2 - 2 - 4 \\ - 2 3 4 \\ 1 - 2 x \end{array}] [\begin{array}{l} 2 - 2 - 4 \\ - 1 3 4 \\ 1 - 2 x \end{array}]$
$^{A^{2} [\begin{array}{l} 2 2 - 16 - 4 x \\ - 1 3 16 + 4 x \\ 4 + x - 8 - 2 x - 12 + x^{2} \end{array}]}$
$\begin{array}{l} A^{2} = A \\ ∴ 4 + x = 1 \Rightarrow x = - 3 \\ A [\begin{array}{l} 2 - 2 - 4 \\ - 1 3 4 \\ 1 - 2 - 4 \end{array}] \end{array}$
$\begin{array}{l} | A | = 2 (- 9 + 8) + 2 (3 - 4) - 4 (2 - 3) \\ = - 2 - 2 + 4 = 0 \\ R a n k A = 2 \Rightarrow r + x = 2 - 3 = - 1 \end{array}$