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ICSE Class X Math 2016 Paper

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Given A=[4sin⁡30∘cos⁡0∘cos⁡0∘4sin⁡30∘]A = \begin{bmatrix} 4 \sin 30^{\circ} & \cos 0^{\circ} \\ \cos 0^{\circ} & 4 \sin 30^{\circ} \end{bmatrix} and B=[45]B = \begin{bmatrix} 4 \\ 5 \end{bmatrix} If AX=BA X = B
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Question : 28 of 57
Marks: +1, -0
Find the matrix XX.
Solution:  
Let X=[ab]X=\begin{bmatrix} a \\ b \end{bmatrix}
∴[4sin⁡30∘cos⁡0∘cos⁡0∘4sin⁡30∘][ab]\therefore \begin{bmatrix} 4\sin 30^{\circ} & \cos 0^{\circ} \\ \cos 0^{\circ} & 4\sin 30^{\circ} \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} =[45]=\begin{bmatrix} 4 \\ 5 \end{bmatrix}
⇒[4asin⁡30∘+bcos⁡0∘acos⁡0∘+4bsin⁡30∘]\Rightarrow \begin{bmatrix} 4a\sin 30^{\circ}+b\cos 0^{\circ} \\ a\cos 0^{\circ}+4b\sin 30^{\circ} \end{bmatrix} =[45]=\begin{bmatrix} 4 \\ 5 \end{bmatrix}
⇒4asin⁡30∘+bcos⁡0∘=4\Rightarrow 4a\sin 30^{\circ}+b\cos 0^{\circ}=4
⇒4a×12+b×1=4\Rightarrow 4a \times \frac{1}{2} + b \times 1 = 4
⇒2a+b=4\Rightarrow 2a+b=4 ........(1)
and acos⁡0∘+4bsin⁡30∘=5a\cos 0^{\circ}+4b\sin 30^{\circ}=5
⇒9×1+4b×12=5\Rightarrow 9 \times 1 + 4b \times \frac{1}{2} = 5
⇒a+2b=5.\Rightarrow a+2b=5. ...........(2)
Solving eqs (1) & (2), we get
a=1 and b=2a=1 \text{ and } b=2
Hence X=[12]X=\begin{bmatrix} 1 \\ 2 \end{bmatrix}
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