A and B be 3 × 3 matrices such that AB + A + B = 0 , then

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3 and 4 .Determinants and Matrices

normal

$A$ and $B$ be $3 \times 3$ matrices such that $AB + A + B = 0$ , then

$(A + B)^2 = A^2 + 2AB + B^2$

$|A| = |B|$

$A^2 = B^2$

None

Solution

$\mathrm{AB}+\mathrm{A}+\mathrm{B}+\mathrm{I}=\mathrm{I}$

$\Rightarrow \mathrm{A}(\mathrm{B}+\mathrm{I})+\mathrm{I}(\mathrm{B}+\mathrm{I}) \Rightarrow(\mathrm{A}+\mathrm{I})(\mathrm{B}+\mathrm{I})=\mathrm{I}$

so $(\mathrm{B}+\mathrm{I}) $ and $(\mathrm{A}+\mathrm{I})$ are inverse of each other

$({\text{B}} + {\text{I}})({\text{A}} + {\text{I}}) = ({\text{A}} + {\text{I}})({\text{B}} + {\text{I}}) \Rightarrow \boxed{{\text{AB}} = {\text{BA}}}$

so $(\mathrm{A}+\mathrm{B})^{2}=\mathrm{A}^{2}+2 \mathrm{AB}+\mathrm{B}^{2}$

Standard 12

Mathematics

If $A + B = \left[ {\begin{array}{{20}{c}}1&0\\1&1\end{array}} \right]$and $A – 2B = \left[ {\begin{array}{{20}{c}}{ – 1}&1\\0&{ – 1}\end{array}} \right]\,,$ then $A=$

easy

Find the matrix $X$ so that $X\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{rrr}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$

hard

If $A = \left( {\begin{array}{*{20}{c}}i&1\\0&i\end{array}} \right)$, then ${A^4}$ equals

easy

For how many value $(s)$ of $x$ in the closed interval $[ – 4,\,\, – 1]$ is the matrix $\left[ {\begin{array}{*{20}{c}}3&{ – 1 + x}&2\\3&{ – 1}&{x + 2}\\{x + 3}&{ – 1}&2\end{array}} \right]$ singular

medium

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