3 Marks Question

Question 513 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: $(aI + bA)^{k + 1} = a^{k + 1}I + (k + 1)a^{k }bA$
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.

View full question & answer→

Question 523 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.
$\therefore\ \text{x}=-4$
$\therefore\ \text{A}=[-4]$

View full question & answer→

Question 533 Marks

Find the matrix A such that
$\text{A}=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\\11&10&9\end{bmatrix}$

Answer

Let $\text{A}=\begin{bmatrix}\text{x}&\text{a}\\\text{y}&\text{b}\\\text{z}&\text{c}\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}&\text{a}\\\text{y}&\text{b}\\\text{z}&\text{c}\end{bmatrix}\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\\11&10&9\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}+4\text{a}&2\text{x}+5\text{a}&3\text{x}+6\text{a}\\\text{y}+4\text{b}&2\text{y}+5\text{b}&3\text{y}+6\text{b}\\\text{z}+4\text{c}&2\text{z}+5\text{c}&3\text{z}+6\text{c}\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\\11&10&9\end{bmatrix}$
By comparing the corresponding elements, we get
x + 4a = -7 and 2x + 5a = -8
⇒ a = -2 and x = 1
Also,
y + 4b = 2 and 2y + 5b = 4
⇒ b = 0 and y = 2
And
z + 4c = 11 and 2z + 5b = 10
⇒ c = 4 and z = -5
$\therefore\ \text{A}=\begin{bmatrix}1&-2\\2&0\\-5&4\end{bmatrix}$

View full question & answer→

Question 543 Marks

Using elementary transformation, find the inverse of each of the matrices, $\begin{bmatrix}3 & 10 \\2 & 7 \end{bmatrix}$

Answer

Let $\text{A}=\begin{bmatrix}3&10\\2&7\end{bmatrix}$
since $\text{A = IA}\ \Rightarrow \begin{bmatrix}3&10\\2&7\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\text{A}$
$\Rightarrow\ \begin{bmatrix}1&3\\2&7\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\text{A}\ \ \ \ \ \ \ \ \left[ \text R_1\rightarrow \text R_1- \text R_2\right]$
$\Rightarrow\ \begin{bmatrix}1&3\\0&1\end{bmatrix}=\begin{bmatrix}1&-1\\-2&3\end{bmatrix}\text{A} \ \ \ \ \ \ \ \left[\text R_2\rightarrow \text R_2-2 \text R_1\right]$
$\Rightarrow\ \begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}7&-10\\-2&3\end{bmatrix}\text{A} \ \ \ \ \ \ \left[\text R_1\rightarrow \text R_1-3\text R_2\right]$
$\therefore\ \ \ \ \ \ \text{A}^{-1}=\begin{bmatrix}7&-10\\-2&3\end{bmatrix}$

View full question & answer→

Question 553 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} $
$= \begin{bmatrix}5 - 10 + 6& -1-0 + 0&2 - 5 + 0\\9 - 10 + 0&-2 - 5 + 6&5-15 + 0\\0-5 + 0&-1 + 5+0&-2-0 + 6\end{bmatrix}$
$=\begin{bmatrix}1&-1&-3\\-1&-1&-10\\-5&4&4\end{bmatrix}$

View full question & answer→

Question 563 Marks

If $\text A = \begin{bmatrix}3&-2\\4&-2\end{bmatrix} \text {and I} = \begin{bmatrix}1&0\\0&1\end{bmatrix},$ find $k$ so that $A^2 = kA - 2I.$

Answer

Given: $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix} \text{and}\ \text{I}\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\text{A}^2=k\text{A}-2\text{I}\Rightarrow\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}=k\begin{bmatrix}3&-2\\4&-2\end{bmatrix}-2\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}=\begin{bmatrix}3k&-2k\\4k&-2k\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&-2\\4&-4\end{bmatrix}=\begin{bmatrix}3k-2&-2k-0\\4k-0&-2k-2\end{bmatrix}$
Equating corresponding entries, we have
$3k - 2 = 1 \Rightarrow 3k = 3$
$k = 1$
And $4k = 4$
$ \Rightarrow k = 1$ and $-4 = -2k - 2$
$\Rightarrow 2k = 2 $
$\Rightarrow k = 1$
$\therefore k = 1$

View full question & answer→

Question 573 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 $\text{B}\neq\text{C}.$

View full question & answer→

Question 583 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 $P.M.I.$

View full question & answer→

Question 593 Marks

If $\text{A}=\begin{bmatrix}4&-1&-4\\3&0&-4\\3&-1&-3\end{bmatrix},$ show that $A^2 = I_3.$

Answer

Given, $\text{A}=\begin{bmatrix}4&-1&-4\\3&0&-4\\3&-1&-3\end{bmatrix}$
$\text{A}^2=\text{A.A}$ $=\begin{bmatrix}4&-1&-4\\3&0&-4\\3&-1&-3\end{bmatrix}\begin{bmatrix}4&-1&-4\\3&0&-4\\3&-1&-3\end{bmatrix}$
$=\begin{bmatrix}16-3-12&-4+0+4&-16+4+12\\12+0-12&-3+0+4&-12+0+12\\12-3-9&-3+0+3&-12+4+9\end{bmatrix}$
$=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$=\text{I}_3$
Hence$,\text{A}^2=\text{I}_3$

View full question & answer→

Question 603 Marks

If $\text{A} = \begin{bmatrix}3 & 1 \\-1 & 2\end{bmatrix},$ show that $A^2 - 5A + 7I = 0.$

Answer

Given: $\text{A} = \begin{bmatrix}3 & 1 \\-1 & 2\end{bmatrix},$
$\therefore \ \text{A}^{2}-5\text{A}+7\text{I}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$= \begin{bmatrix}9-1&3 + 2\\-3-2&-1 + 4\end{bmatrix}-\begin{bmatrix}15 & 5 \\-5 & 10\end{bmatrix} + \begin{bmatrix}7 & 0 \\0 & 7 \end{bmatrix} $
$= \begin{bmatrix}8 & 5 \\-5 & 3 \end{bmatrix} - \begin{bmatrix}15 & 5\\-5& 10 \end{bmatrix}+ \begin{bmatrix}7 & 0 \\0 & 7 \end{bmatrix}$
$=\begin{bmatrix}8-15&5-5\\-5+5&3-10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$=\begin{bmatrix}-7&0\\0&-7\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$=\begin{bmatrix}-7+7&0+7\\0+7&-7+7\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$=0 $
$= \text{R.H.S.}$

View full question & answer→

Question 613 Marks

Find the values of a, b, c and d if:$3\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}=\begin{bmatrix}\text{a}&6\\-1&2\text{d}\end{bmatrix}+\begin{bmatrix}4&\text{a}+\text{b}\\\text{c}+\text{d}&3\end{bmatrix}.$

Answer

Consider, $3\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}=\begin{bmatrix}\text{a}&6\\-1&2\text{d}\end{bmatrix}+\begin{bmatrix}4&\text{a}+\text{b}\\\text{c}+\text{d}&3\end{bmatrix}$$\Rightarrow\ \begin{bmatrix}3\text{a}&3\text{b}\\3\text{c}&3\text{d}\end{bmatrix}=\begin{bmatrix}\text{a}+4&6+\text{a}+\text{b}\\\text{c}+\text{d}-1&3+2\text{d}\end{bmatrix}$
By equality of matrices, we get
3a = a + 4, 3b = 6 + a + b and 3d = 3 + 2d
⇒ a = 2, b = 4 and d = 3.

View full question & answer→

Question 623 Marks

If $\text{A}=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix},$ then verify that $A^TA = I_2.$

Answer

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

View full question & answer→

MATHS 3 Marks Question