3 Marks

Question 513 Marks

If $\text{A}=\begin{bmatrix}1&2\\4&1\end{bmatrix},$ then find A² + 2A + 7I.

Answer

We have, $\text{A}=\begin{bmatrix}1&2\\4&1\end{bmatrix},$
$\Rightarrow\ \text{A}^2=\begin{bmatrix}1&2\\4&1\end{bmatrix}\begin{bmatrix}1&2\\4&1\end{bmatrix}$
$\Rightarrow\ \text{A}^2=\begin{bmatrix}1+8&2+2\\4+4&8+1\end{bmatrix}=\begin{bmatrix}9&4\\8&9\end{bmatrix}$
$\therefore\ \text{A}^2+2\text{A}+7$
$=\begin{bmatrix}9&4\\8&9\end{bmatrix}+\begin{bmatrix}2&4\\8&2\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}=\begin{bmatrix}18&8\\16&18\end{bmatrix}$

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Question 523 Marks

If $\text{A}=\begin{bmatrix}0&-1&2\\4&3&-4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}4&0\\1&3\\2&6\end{bmatrix},$ then verify that $(\text{kA})'=(\text{kA}').$

Answer

We have, $\text{A}=\begin{bmatrix}0&-1&2\\4&3&-4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}4&0\\1&3\\2&6\end{bmatrix}$
We have to verify that, $(\text{kA})'=(\text{kA}')$
Now, $(\text{kA})=\begin{bmatrix}0&-\text{k}&2\text{k}\\4\text{k}&3\text{k}&-4\text{k}\end{bmatrix}$
And $(\text{kA})'=\begin{bmatrix}0&4\text{k}\\-\text{k}&3\text{k}\\2\text{k}&-4\text{k}\end{bmatrix}$
Also, $\text{kA}'=\begin{bmatrix}0&4\text{k}\\-\text{k}&3\text{k}\\2\text{k}&-4\text{k}\end{bmatrix}=(\text{kA})'$
Hence proved.

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Question 533 Marks

Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{A}-\text{B})\text{C}=\text{AC}-\text{BC}.$

Answer

We have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2

$(\text{A}-\text{B})=\begin{bmatrix}1-4&2-0\\-1-1&3-5\end{bmatrix}=\begin{bmatrix}-3&2\\-2&-2\end{bmatrix}$

$(\text{A}-\text{B})\text{C}=\begin{bmatrix}-3&2\\-2&-2\end{bmatrix}\begin{bmatrix}2&0\\1&-2\end{bmatrix}$

$=\begin{bmatrix}-4&-4\\-6&4\end{bmatrix}\ ....(\text{i})$

Now, $\text{AC}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}\begin{bmatrix}2&0\\1&-2\end{bmatrix}$

$=\begin{bmatrix}4&-4\\1&-6\end{bmatrix}\ .....(\text{ii})$

and $\text{BC}=\begin{bmatrix}4&0\\1&5\end{bmatrix}\begin{bmatrix}2&0\\1&-2\end{bmatrix}$

$=\begin{bmatrix}8&0\\7&-10\end{bmatrix}\ ....(\text{iii})$

$\therefore\ \text{AC}-\text{BC}=\begin{bmatrix}4-8&-4-0\\1-7&-6+10\end{bmatrix}$ [Using (ii) and (iii)]

$=\begin{bmatrix}-4&-4\\-6&4\end{bmatrix}$

$=(\text{A}-\text{B})\text{C}$ [Using (i)]

Hence proved.

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Question 543 Marks

If $\text{A}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix},$ then show that $\text{A}^2=\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta\end{bmatrix}.$

Answer

We have, $\text{A}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}$

$\therefore\ \text{A}^2=\text{A}.\text{A}$

$=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}.\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}$

$=\begin{bmatrix}\cos^2\theta-\sin^2\theta&\cos\theta\sin\theta+\sin\theta\cos\theta\\-\sin\theta\cos\theta-\cos\theta\sin\theta&-\sin^2\theta+\cos^2\theta\end{bmatrix}$

$=\begin{bmatrix}\cos2\theta&2\sin\theta\cos\theta\\2\sin\theta\cos\theta&\cos2\theta\end{bmatrix}$

$=\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta\end{bmatrix}$

Hence proved.

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Question 553 Marks

Answer

We have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2

$(\text{BC})=\begin{bmatrix}4&0\\1&5\end{bmatrix}\begin{bmatrix}2&0\\1&-2\end{bmatrix}$

$=\begin{bmatrix}8&0\\7&-10\end{bmatrix}$

and $\text{A}(\text{BC})=\begin{bmatrix}1&2\\-1&3\end{bmatrix}\begin{bmatrix}8&0\\7&-10\end{bmatrix}$

$=\begin{bmatrix}8+14&0-20\\-8+21&0-30\end{bmatrix}=\begin{bmatrix}22&-20\\13&-30\end{bmatrix}$

Also, $(\text{AB})=\begin{bmatrix}1&2\\-1&3\end{bmatrix}\begin{bmatrix}4&0\\1&5\end{bmatrix}$

$=\begin{bmatrix}6&10\\-1&15\end{bmatrix}$

$(\text{AB})\text{C}=\begin{bmatrix}6&10\\-1&15\end{bmatrix}\begin{bmatrix}2&0\\1&-2\end{bmatrix}$

$=\begin{bmatrix}22&-20\\13&-30\end{bmatrix}=\text{A}(\text{BC})$

Hence proved.

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Question 563 Marks

Find x, if $\begin{bmatrix}x&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix} = 0.$

Answer

Given: $\begin{bmatrix}x&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix} = 0.$
$\Rightarrow\ \begin{bmatrix} x-0-2&0-10-0&2x-5-3\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=0$
$\Rightarrow \ \begin{bmatrix} x-2&-10&2x-8\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=0$
$ \Rightarrow\ \begin{bmatrix}(x-2)x-10(4)+(2x-8)1\end{bmatrix}=0$
$\Rightarrow\ \begin{bmatrix}x^{2}-2x-40+2x-8\end{bmatrix}=0$
$\Rightarrow\ \begin{bmatrix}x^{2}-48\end{bmatrix}_{1\times1}=\left[0\right]_{1\times1}$
Equating corresponding entries, we have
$x^{2}-48=0\ \Rightarrow x^{2}=48\ \ \Rightarrow \ \ x=\pm4\sqrt{3}$

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Question 573 Marks

If $\text{A}=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix},$ and $\text{A}^{-1}=\text{A}',$ find value of $\alpha.$

Answer

We have, $\text{A}=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}$ and $\text{A}'=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}$

Also, $\text{A}^{-1}=\text{A}'$

$\Rightarrow\ \text{AA}^{-1}=\text{AA}'$

$\Rightarrow\ \text{I}=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}$

$\Rightarrow\ \begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}\cos^2\alpha+\sin^2\alpha&0\\0&\sin^2\alpha+\cos^2\alpha\end{bmatrix}$

By Using equality of matrices, we get

$\cos^2\alpha+\sin^2\alpha=1$

Which is true for all real values of $\alpha.$

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Question 583 Marks

If $\text{A}=\begin{bmatrix}0&-1&2\\4&3&-4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}4&0\\1&3\\2&6\end{bmatrix},$ then verify that $(\text{AB})'=\text{B}'\text{A}'.$

Answer

We have, $\text{A}=\begin{bmatrix}0&-1&2\\4&3&-4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}4&0\\1&3\\2&6\end{bmatrix}$

We have to verify that, $(\text{AB})'=\text{B}'\text{A}'$

$\therefore\ \text{AB}=\begin{bmatrix}3&9\\11&-15\end{bmatrix}$

$\Rightarrow\ (\text{AB})'=\begin{bmatrix}3&11\\9&-15\end{bmatrix}$

and $\text{B}'\text{A}'=\begin{bmatrix}4&1&2\\0&3&6\end{bmatrix}\begin{bmatrix}0&4\\-1&3\\2&-4\end{bmatrix}$

$=\begin{bmatrix}3&11\\9&-15\end{bmatrix}$

$=(\text{AB})'$

Hence proved.

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Question 593 Marks

If A is square matrix such that A² = A, then show that (I + A)³ = 7A + I.

Answer

We know that, A.I = I.A

So, A and I are commutative.

Therefore we can expand (I + A)³ like real number expansion.

So, (I + A)³ = I³ + 3I²A + 3IA² + A³

= I + 3IA + 3A² + AA² $($as Iⁿ = I, $\text{n}\in\text{N})$

= I + 3A + 3A + AA

= I + 3A + 3A + A²

= I + 3A + 3A + A = I + 7A

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Question 603 Marks

Express the following matrices as the sum of a symmentric and a skew symmentric matrix:

$\begin{bmatrix}3&5\\1&-1\end{bmatrix}$

Answer

$\text{Let}\text{A}=\begin{bmatrix}3&5\\1&-1\end{bmatrix},\text{then}\ \text{A}'=\begin{bmatrix}3&1\\5&-1\end{bmatrix}$
Now, $\text{A+ A}'=\begin{bmatrix}3&5\\1&-1\end{bmatrix}+\begin{bmatrix}3&1\\5&-1\end{bmatrix}=\begin{bmatrix}6&6\\6&-2\end{bmatrix}$
Let $\text{P}=\frac{1}{2}\text{(A+A}')=\frac{1}{2}\begin{bmatrix}6&6\\6&-2\end{bmatrix}=\begin{bmatrix}3&3\\3&-1\end{bmatrix}$
Now, $\text{P}'=\begin{bmatrix}3&3\\3&-1\end{bmatrix}=\text{P}$
Thus $\text{P}=\frac{1}{2}(\text{A}+\text{A}')$ is a symmentric matrix.
Now, $\text{A}-\text{A}'=\begin{bmatrix}3&5\\1&-1\end{bmatrix}-\begin{bmatrix}3&1\\5&-1\end{bmatrix}=\begin{bmatrix}0&4\\-4&0\end{bmatrix}$
Let $\text{Q}=\frac{1}{2}\text{(A} -\text{A}')=\frac{1}{2}\begin{bmatrix}0&4\\-4&0\end{bmatrix}=\begin{bmatrix}0&2\\-2&0\end{bmatrix}$
Now, $\text{Q}'=\begin{bmatrix}0&2\\-2&0\end{bmatrix}=-\text{Q}$
Thus, $\text{Q}=\frac{1}{2}\text{(A} - \text{A}')$ is a skew - symmentric matrix.
Representing A as the sum of P and Q:
$\text{P + Q}=\begin{bmatrix}3&3\\3&-1\end{bmatrix}+\begin{bmatrix}0&2\\-2&0\end{bmatrix}=\begin{bmatrix}3&5\\1&-1\end{bmatrix}=\text{A}$

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Question 613 Marks

If $\text{A}=\begin{bmatrix}1&2\\0&3 \end{bmatrix}$ is written as B + C, where B is a symmetric matrix and C is a skew- symmetric matrix, then B is equal to.

Answer

Given: $\text{A}=\begin{bmatrix}1&2\\0&3 \end{bmatrix}$
$\Rightarrow\text{A}^{\text{T}}=\begin{bmatrix}1&0\\2&3 \end{bmatrix}$
Let $\text{B}=\frac{1}{2}(\text{A+A}^{\text{T}})=\frac{1}{2}\bigg(\begin{bmatrix}1&2\\0&3 \end{bmatrix}+\begin{bmatrix}1&0\\2&3 \end{bmatrix}\bigg)$
$=\frac{1}{2}\begin{bmatrix}1+1&2+0\\0+2&3+3 \end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix} 2&2\\2&6\end{bmatrix}$
$=\begin{bmatrix} 1&1\\1&3\end{bmatrix}$
Now,
$\text{B}^{\text{T}}=\begin{bmatrix}1&1\\1&3 \end{bmatrix}=\text{B}$
Therefore, B is symmetric matrix.
Let $\text{C}=\frac{1}{2}(\text{A}-\text{A}^{\text{T}})=\frac{1}{2}\bigg(\begin{bmatrix}1&2\\0&3 \end{bmatrix}-\begin{bmatrix}1&0\\2&3 \end{bmatrix}\bigg)$
$=\frac{1}{2}\begin{bmatrix}1-1&2-0\\0-2&3-3 \end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}0&2\\-2&2 \end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}0&2\\-1&0 \end{bmatrix}$
$\therefore\text{C}^{\text{T}}=\begin{bmatrix}0&1\\-1&0 \end{bmatrix}^{\text{T}}=\begin{bmatrix}0&-1\\1&0 \end{bmatrix}=-\begin{bmatrix}0&1\\-1&0 \end{bmatrix}=\text{C}$
So, C is a skew-symmetric matrix.
Now,
$\text{B + C}=\begin{bmatrix}1&1\\1&3 \end{bmatrix}+\begin{bmatrix}0&1\\-1&0 \end{bmatrix}=\begin{bmatrix}1+0&1+1\\1-1&3+0 \end{bmatrix}=\begin{bmatrix}1&2\\0&3 \end{bmatrix}=\text{A}$
$\therefore\text{B}=\begin{bmatrix}1&1\\1&3 \end{bmatrix}$

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Question 623 Marks

If $\text{x}\begin{bmatrix}2\\3 \end{bmatrix}+\text{y}\begin{bmatrix}-1\\1 \end{bmatrix}=\begin{bmatrix}10\\5 \end{bmatrix},$ find the value of x.

Answer

$\text{x}\begin{bmatrix}2\\3 \end{bmatrix}+\text{y}\begin{bmatrix}-1\\1 \end{bmatrix}=\begin{bmatrix}10\\5 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{x}-\text{y}\\3\text{x + y} \end{bmatrix}=\begin{bmatrix}10\\5 \end{bmatrix}$
Corresponding elements of equal matrices are equal.
⇒ 2x - y = 10 and 3x + y = 5
⇒ y = 2x - 10 and 3x + (2x - 10) = 5
⇒ y = 2x - 10 and 5x = 15
⇒ y = 2x - 10 and x = 3
⇒ y = 2(3) - 10 and x = 3
⇒ y = -4 and x = 3
$\therefore$ x = 3 and y = -4
Hence, the value of x is 3.

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Maths 3 Marks