in Linear Algebra recategorized by
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2 votes
2 votes

Let $A=\begin{pmatrix} 1 & 1 & 0\\ 0 & a & b\\1 & 0 & 1 \end{pmatrix}$. Then $A^{-1}$ does not exist if $(a,b)$ is equal to

  1. $(1,-1)$
  2. $(1,0)$
  3. $(-1,-1)$
  4. $(0,1)$
in Linear Algebra recategorized by
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2 Answers

3 votes
3 votes

A matrix is not invertible if it's determinant is $0.$

$|A|= 1(a-0)-1(0-b)= a+b$

Only Option A) satisfies

edited by
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Option A satisfies the condition when we calculate the determinant we get a+b = 0 equation.

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