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ICSE Class X Math 2014 Paper

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Question : 37 of 52
Marks: +1, -0
Let A=[210−2],B=[41−3−2]A=\begin{bmatrix}2 & 1 \\ 0 & -2\end{bmatrix}, B=\begin{bmatrix}4 & 1 \\ -3 & -2\end{bmatrix} and C=[−32−14]C=\begin{bmatrix}-3 & 2 \\ -1 & 4\end{bmatrix} Find A2+AC−5BA^2+A C-5 B.
Solution:  
Given:
A=[210−2],B  =[41−3−2],A=\begin{bmatrix}2 & 1 \\ 0 & -2\end{bmatrix}, {B}\;=\begin{bmatrix}4 & 1 \\ -3 & -2\end{bmatrix}, C=[−32−14]{C}=\begin{bmatrix}-3 & 2 \\ -1 & 4\end{bmatrix}
A2=A⋅A  =[210−2][210−2]{A}^2= {A} \cdot {A}\;=\begin{bmatrix}2 & 1 \\ 0 & -2\end{bmatrix}\begin{bmatrix}2 & 1 \\ 0 & -2\end{bmatrix}
  =[4+02−20+00+4]\;=\begin{bmatrix}4+0 & 2-2 \\ 0+0 & 0+4\end{bmatrix}
  =[4004]\;=\begin{bmatrix}4 & 0 \\ 0 & 4\end{bmatrix}
AC  =[210−2][−32−14]{AC}\;=\begin{bmatrix}2 & 1 \\ 0 & -2\end{bmatrix}\begin{bmatrix}-3 & 2 \\ -1 & 4\end{bmatrix}
  =[−6−14+40+20−8]=[−782−8]\;=\begin{bmatrix}-6-1 & 4+4 \\ 0+2 & 0-8\end{bmatrix}=\begin{bmatrix}-7 & 8 \\ 2 & -8\end{bmatrix}
and 5B  =5[41−3−2]5 B\;=5\begin{bmatrix}4 & 1 \\ -3 & -2\end{bmatrix}
  =[205−15−10]\;=\begin{bmatrix}20 & 5 \\ -15 & -10\end{bmatrix}
Now,
  A2+AC−5B=[4004]\;A^2+A C-5 B=\begin{bmatrix}4 & 0 \\ 0 & 4\end{bmatrix} +[−782−8]+\begin{bmatrix}-7 & 8 \\ 2 & -8\end{bmatrix}     −[205−15−10]\;\; -\begin{bmatrix}20 & 5 \\ -15 & -10\end{bmatrix}
    =[4−7−200+8−50+2+154−8+10]\;\; =\begin{bmatrix}4-7-20 & 0+8-5 \\ 0+2+15 & 4-8+10\end{bmatrix}
    =[−233176]\;\; =\begin{bmatrix}-23 & 3 \\ 17 & 6\end{bmatrix}
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