Matrices

NCERT Solution

Exercise 2 Part 3

Question 4: If

A = \begin{bmatrix}1 & 2 & -3\\5 & 0 & 2\\1 & -1 & 1\end{bmatrix}
B = \begin{bmatrix}3 & -1 & 2\\4 & 2 & 5\\2 & 0 & 3\end{bmatrix}
C = \begin{bmatrix}4 & 1 & 2\\0 & 3 & 2\\1 & -2 & 3\end{bmatrix}

Then compute (A + B) and (A – C). Also verify that A + (B - C) = (A + B) - C

Solution:

A + B = \begin{bmatrix}4 & 1 & -1\\9 & 2 & 7\\3 & -1 & 4\end{bmatrix}
B - C = \begin{bmatrix}-1 & -2 & 0\\4 & -1 & 3\\1 & 2 & 0\end{bmatrix}
Now, A + (B – C) = \begin{bmatrix}0 & 0 & -3\\9 & -1 & 5\\2 & 1 & 1\end{bmatrix}
And (A+B) – C = \begin{bmatrix}0 & 0 & -2\\9 & -1 & 5\\2 & 1 & 1\end{bmatrix}

Thus, A + (B - C) = (A + B) - C


Question 5: If

A = \begin{bmatrix}2/3 & 1 & 5/3\\1/3 & 2/3 & 4/3\\7/3 & 2 & 2/3\end{bmatrix}
B = \begin{bmatrix}2/5 & 3/5 & 1\\1/5 & 2/5 & 4/5\\7/5 & 6/5 & 2/5\end{bmatrix}

Compute 3A – 5B

Solution:

3A = \begin{bmatrix}2 & 3 & 5\\1 & 2 & 4\\7 & 6 & 2\end{bmatrix}
5B = \begin{bmatrix}2 & 3 & 5\\1 & 2 & 4\\7 & 6 & 2\end{bmatrix}
Hence, 3A – B = \begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

Question 6: Simplify

`text(cos)θ` \begin{bmatrix}cosθ & sinθ\\-sinθ & cosθ\end{bmatrix} `+text(sin)θ` \begin{bmatrix}cosθ & -cosθ\\cosθ & sinθ\end{bmatrix}

Solution:

= \begin{bmatrix}cos^2θ & sinθcosθ\\-sinθcosθ & cos^2θ\end{bmatrix} + \begin{bmatrix}sin^2θ & -sinθcosθ\\sinθcosθ & sin^2θ\end{bmatrix}
= \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}


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