1expand it Binomially so we I4+4A+6A2+4A3+4A4
therefore 6A2+4A3+4A4=0 solving this we will get an expression for (I-A)2 by this way you can proceed and we get the answer as (c)
If A=\begin{bmatrix} x-2 &2 &x \\ 1&2 &3 \\ x-4&-2 & -x \end{bmatrix}
and (I+A)^{4}=I + 4A,then (I-A)^{2}=?
a.0 b.\begin{bmatrix} -1 &-4 &-6 \\ -1 & -3 &-6 \\ 2& 4 & 7 \end{bmatrix}
c.\begin{bmatrix} 0 & 2 &-3 \\ -1& -1 &-3 \\ 1& 2 &4 \end{bmatrix}
d.\begin{bmatrix} 1& 2 &3 \\ 1& 2 &3 \\ -1 & -2 &-3 \end{bmatrix}how to proceed in this problem
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