3 Marks Question

Question 513 Marks

Given the matrices
$\text{A}=\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4\end{bmatrix},\text{B}=\begin{bmatrix}9&7&-1\\3&5&4\\2&1&6\end{bmatrix}$ and $\text{C}=\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}$ Verify that (A + B) + C = A + (B + C).

Answer

Here,
$\text{LHS}=(\text{A}+\text{B})+\text{C}$
$=\begin{pmatrix}\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4\end{bmatrix}+\begin{bmatrix}9&7&-1\\3&5&4\\2&1&6\end{bmatrix}\end{pmatrix}+\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}$
$=\begin{pmatrix}\begin{bmatrix}2+9&1+7&1-1\\3+3&-1+5&0+4\\0+2&2+1&4+6\end{bmatrix}\end{pmatrix}+\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}$
$=\begin{bmatrix}11&8&0\\6&4&4\\2&3&10\end{bmatrix}+\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}$
$=\begin{bmatrix}11+2&8-4&0+3\\6+1&4-1&4+0\\2+9&3+4&10+5\end{bmatrix}$
$=\begin{bmatrix}13&4&3\\7&3&4\\11&7&15\end{bmatrix}$
$\text{RHS}=\text{A}+(\text{B}+\text{C})$
$=\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4\end{bmatrix}+\begin{pmatrix}\begin{bmatrix}9&7&-1\\3&5&4\\2&1&6\end{bmatrix}+\begin{bmatrix}2&-4&3\\1&-1&0\\9&4&5\end{bmatrix}\end{pmatrix}$
$=\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4\end{bmatrix}+\begin{pmatrix}\begin{bmatrix}9+2&7-4&-1+3\\3+1&5-1&4+0\\2+9&1+4&6+5\end{bmatrix}\end{pmatrix}$
$=\begin{bmatrix}2&1&1\\3&-1&0\\0&2&4 \end{bmatrix}+\begin{bmatrix}11&3&2\\4&4&4\\11&5&11\\\end{bmatrix}$
$=\begin{bmatrix}2+11&1+3&1+2\\3+4&-1+4&0+4\\0+11&2+5&4+11\\\end{bmatrix}$
$=\begin{bmatrix}13&4&3\\7&3&4\\11&7&15\\\end{bmatrix}$
$\therefore\ \text{LHS}=\text{RHS}$
Hence proved.

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Question 523 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 533 Marks

Find $A^2 - 5A + 6I$ if A = $\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}.$

Answer

$\text{A}^2-5\text{A}+6\text{I}=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}-5\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}+6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}4 + 0 + 1&0 + 0 - 1&2 + 0 + 0\\4 + 2 + 3&0 + 1 - 3&2 + 3 + 0\\2 - 2 + 0&0 - 1 - 0&1 - 3 + 0\end{bmatrix} - \begin{bmatrix}10&0&5\\10&5&15\\5&-5&0\end{bmatrix} + \begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}$
$=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-\begin{bmatrix}10&0&5\\10&5&15\\5&-5&0\end{bmatrix} + \begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix} $
$= \begin{bmatrix}5 - 10 + 6& -1-0 + 0&2 - 5 + 0\\9 - 10 + 0&-2 - 5 + 6&5-15 + 0\\0-5 + 0&-1 + 5+0&-2-0 + 6\end{bmatrix}$
$=\begin{bmatrix}1&-1&-3\\-1&-1&-10\\-5&4&4\end{bmatrix}$

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Question 543 Marks

Give an example of matrices A, B and C such that AB = AC, where A is nonzero matrix, but $\text{B}\neq\text{C}.$

Answer

Let $\text{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}2&3\\4&0\end{bmatrix}$ and $\text{C}=\begin{bmatrix}2&3\\4&4\end{bmatrix}\ [\because\ \text{B}\neq\text{C}]$
$\therefore\ \text{AB}=\begin{bmatrix}1&0\\0&0\end{bmatrix}.\begin{bmatrix}2&3\\4&0\end{bmatrix}=\begin{bmatrix}2&3\\0&0\end{bmatrix}\ ....(\text{i})$
And $\text{AC}=\begin{bmatrix}1&0\\0&0\end{bmatrix}.\begin{bmatrix}2&3\\4&0\end{bmatrix}=\begin{bmatrix}2&3\\0&0\end{bmatrix}\ ....(\text{ii})$
Thus, we see that AB = AC [using Eq. (i) and (ii)]
where, A is non-zero matrix but $\text{B}\neq\text{C}.$

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Question 553 Marks

If A and B are square matrices of the same order such that $AB = BA,$ then prove by induction that $AB’’ = B’’A.$ Further prove that $(AB)’’ = A’’B’’$ for all $n \Rightarrow N.$

Answer

Given: $AB = BA ...(i)$
Let $p(n): AB^n = B^nA ...(ii)$
For $n = 1, p(n):$ becomes $AB = BA$
$\therefore p(1)$ is true for $n = 1.$
For $n = k, p(k): AB^k = B^kA$
Multiplying both sides by $B, AB^kB = B^kAB \Rightarrow AB^{k+1} = B^kAB$
$\Rightarrow AB^{k+1} = B^{k+1}A [$From eq. $(i)]$
$\therefore p(k +1)$ is also true.
Therefore, $p(n)$ is true for all $\text{n}\in\text{N}$ by P.M.I.

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Question 563 Marks

$\text{A}=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$, then verify that A'A = I

Answer

$ \text{A}=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$\therefore\ \text{A}'=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}$
$\text{A}'\text{A}=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$=\begin{bmatrix}(\sin\alpha)(\sin\alpha)+(-\cos\alpha)(-\cos\alpha)&(\sin\alpha)(\cos\alpha)+(-\cos\alpha)(\sin\alpha)\$\cos\alpha)(\sin\alpha)+(\sin\alpha)(-\cos\alpha)&(\cos\alpha)(\cos\alpha)+(\sin\alpha)(\sin\alpha)\end{bmatrix}$
$=\begin{bmatrix}\sin^2\alpha+\cos^2\alpha&\sin\alpha\cos\alpha-\sin\alpha\cos\alpha\\\sin\alpha\cos\alpha-\sin\alpha\cos\alpha&\cos^2\alpha+\sin^2\alpha\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}=\text{I}$
Hence, we have verified that A'A = I

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Question 573 Marks

Let $\text{A} = \begin{bmatrix}0 & 1\\0 & 0 \end{bmatrix},$show that $(aI + bA)^n = a^nI + na^{n-1}$ bA where I is the identity matrix of order $2$ and $ \text{n} \in \text{N}.$

Answer

Using Mathematical Induction, we see the result is true for $n = 1,$ for
$(aI + bA)^n = a^nI + na^{n - 1} bA$
Given: $p(k)$ is true, i.e. $(aI + bA)^k = a^kI + ka^{k - 1}bA$
To prove: $(aI + bA)^{k + 1} = a^{k + 1}I + (k + 1)a^kbA$
Proof: L.H.S. $= (aI + bA)^{k + 1} = (aI + bA)^k (aI + bA)$
$= (a^kI + ka^{k - 1} bA)(aI + bA)$
$= a^{k + 1} I \times I + ka^kbAI + a^kbAI + ka^{k - 1}b^2A.A$
$= a^{k + 1} I \times I + ka^kbAI + a^kbAI + ka^{k - 1}b^2.0$
$= a^{k +1}I + (k +1) a^kbA =$ R.H.S.
Thus, $p(k + 1)$ is true, therefore, $p(n)$ is true.

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Question 583 Marks

Find the matrix A such that
$\begin{bmatrix}2&1&3\end{bmatrix}\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=\text{A}$

Answer

Let $\text{A}=[\text{x}]$
$ \Rightarrow\begin{bmatrix}2&1&3\end{bmatrix}\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=\text{A}$
$ \Rightarrow\begin{bmatrix}2&1&3\end{bmatrix}\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=[\text{x}]$
$ \Rightarrow\begin{bmatrix}-2-1+0&0+1+3&-2+0+3\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=[\text{x}]$
$ \Rightarrow\begin{bmatrix}-3&4&1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}=[\text{x}]$
$ \Rightarrow\begin{bmatrix}-3+0-1\end{bmatrix}=[\text{x}]$
$\Rightarrow\begin{bmatrix}-4\end{bmatrix}=[\text{x}]$
The corresponding elements of two equal matrices are equal.
$\therefore\ \text{x}=-4$
$\therefore\ \text{A}=[-4]$

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Question 593 Marks

Show that $\text{AB}\neq\text{BA}$ in the following cases:
$\text{A}=\begin{bmatrix}5&-1\\6&7\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&1\\3&4\end{bmatrix}$

Answer

$\text{AB}=\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}10-3&5-4\\12+21&6+28\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}7&1\\33&34\end{bmatrix}\ \dots(1)$
Also,
$\text{BA}=\begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}10+6&-2+7\\15+24&-3+28\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}16&5\\39&25\end{bmatrix}\ \dots(2)$
$\therefore\ \text{AB}\neq\text{BA}$ From eqs. (1) and (2)

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Question 603 Marks

If A, B are square matrices of same order and B is a skew-symmetric matrix, show that A′BA is skew symmetric.

Answer

Since, A and B are square matrices of same order and B is a skew symmetric matrix i.e. B’ = -B. Now, we have to prove that A’BA is a skew-symmetric matrix. $\therefore$ A'BA' = A'BA' = BA'A' $[\because$ AB' = B'A'$]$= A'BA' = A'(-B)A = -A'BA
Hence, A’BA is a skew-symmetric matrix.

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Question 613 Marks

Let $\text{A}=\begin{bmatrix}2&-3\\-7&5\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&0\\2&-4\end{bmatrix},$ verify that
$(\text{A}+\text{B})^\text{T}=\text{A}^\text{T}+\text{B}^\text{T}$

Answer

Given: $\text{A}=\begin{bmatrix}2&-3\\-7&5\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}2&-7\\-3&5\end{bmatrix}$
$\text{B}=\begin{bmatrix}1&0\\2&-4\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}1&2\\0&-4\end{bmatrix}$
Given,
$(\text{A}+\text{B})^\text{T}=\text{A}^\text{T}+\text{B}^\text{T}$
$\begin{pmatrix} \begin{bmatrix} 2&-3\\-7&5\end{bmatrix}+\begin{bmatrix} 1&0\\2&-4\end{bmatrix}\end{pmatrix}^\text{T}$ $=\begin{bmatrix}2&-3\\-7&5\end{bmatrix}^\text{T}+\begin{bmatrix}1&0\\2&-4\end{bmatrix}^\text{T}$
$\Rightarrow\begin{bmatrix}2+1&-3+0\\-7+2&5-4\end{bmatrix}^\text{T}$ $=\begin{bmatrix}2&-7\\-3&5\end{bmatrix}+\begin{bmatrix}1&2\\0&-4\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3&-3\\-5&1\end{bmatrix}^\text{T}=\begin{bmatrix}2+1&-7+2\\-3+0&5-4\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3&-5\\-3&1\end{bmatrix}=\begin{bmatrix}3&-5\\-3&1\end{bmatrix}$
$\Rightarrow\text{LHS}=\text{RHS}$
So,
$(\text{A}+\text{B})^\text{T}=\text{A}^\text{T}+\text{B}^\text{T}$

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Question 623 Marks

If $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then find $\lambda$ so that $\text{A}^2 = 5\text{A} + \lambda\text{I}.$

Answer

Given, $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix},\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
And
$\text{A}^2=5\text{A}+\lambda\text{I}$
$\Rightarrow\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}=5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+\lambda\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}9-1&3+2\\-3+2&-1+4\end{bmatrix}=\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}\lambda&0\\0&\lambda\end{bmatrix}$
$\Rightarrow\begin{bmatrix}8&5\\-5&3\end{bmatrix}=\begin{bmatrix}15+\lambda&5\\-5&10+\lambda\end{bmatrix}$
Since, Corresponding entries of equal matrices are equal, So
$8=15+\lambda$
$\lambda=8-15$
$\lambda=-7$

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MATHS 3 Marks Question