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Matrix
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Matrix
All Questions (Page: 3)
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All Questions (PDF)
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Study Material (PDF)
All Questions (PDF)
English Version
Self Assesment
Sort As
Serial Number
Newly Added
Language
English
Click on each question to see answer & solution.
Question No: #21
If $ A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} $ and $ B = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix} $, then show that `(A+B)^2 \ne A^2+2AB+B^2`
Ans:
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Question No: #22
If $ A = \begin{pmatrix} 2 & 4 \\ 3 & 2 \end{pmatrix} $ and $ B = \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix} $ then show that `AB \ne BA`
Ans:
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Question No: #23
If $ A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} $ and `f(x) = x^2-2x-3`, then show that `f(A)=0`
Ans:
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Question No: #24
`A` is a square matrix such that `A^2=A`. Then find the value of `(I+A)^3-7A`
Ans: `I`
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Question No: #25
If `A` is a symmetric matrix, then `A^n` (where `n` is positive integer) is _____ matrix.
Ans: symmetric
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Question No: #26
The matrix $ \begin{bmatrix} cos\ x & sin\ x \\ -sin\ x & cos\ x \end{bmatrix} $ is - (a) symmetric & singular; (b) unit & skew-symmetric; (c) Unit & orthogonal; (d) Non-singular & orthogonal matrix.
Ans: (d) Non-singular & orthogonal matrix.
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Question No: #27
If $ A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} $, then `A^5` is -- (a) `5A`, (b) `10A`, (c) `16A`, (d) `32A`
Ans: (c) `16A`
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Question No: #28
If the matrix $ \begin{bmatrix} -x & x & 2 \\ 2 & x & -x \\ x & -2 & -x \end{bmatrix} $ is non-singular, then value of `x` is -- (a) `-2\le\x\le2`, (b) Any real number except `\pm2`, (c) `x\ge2`, (d) `x\le-2`
Ans: (b) Any real number except `\pm2`
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Question No: #29
If $ A = \begin{bmatrix} 0 & 6 \\ 0 & 0 \end{bmatrix} $ and `f(x) = 1+x+x^2+ \cdots +x^20`, then show the value of `f(A)` is -- (a) $ \begin{bmatrix} 1 & 6 \\ 0 & 1 \end{bmatrix} $ (b) $ \begin{bmatrix} 1 & 6 \\ 0 & 0 \end{bmatrix} $ (c) $ \begin{bmatrix} 1 & 6 \\ 1 & 0 \end{bmatrix} $ (d) $ \begin{bmatrix} 1 & 0 \\ 6 & 1 \end{bmatrix} $
Ans: (a)
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Question No: #30
Find a matrix `X` such that `2A+B+X = 0` where $ A = \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} $ and $ B = \begin{bmatrix} 3 & -2 \\ 1 & 5 \end{bmatrix} $
Ans: $ \begin{bmatrix} -1 & -2 \\ -7 & -13 \end{bmatrix} $
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