Question 15 Marks
Let $\text{A}=\begin{bmatrix}3 & 2&7 \\1 & 4&3\\-2&5&8 \end{bmatrix}.$ Find matrices X and Y such that X + Y = A, where X is a symmetric and Y is a skew-symmetric matrix.
AnswerGiven: $\text{A}=\begin{bmatrix}3 & 2&7 \\1 & 4&3\\-2&5&8 \end{bmatrix}$ $\Rightarrow\text{A}^{\text{T}}=\begin{bmatrix}3 & 1&-2 \\2 & 4&5\\7&3&8 \end{bmatrix}$Let $\text{x}=\frac{1}{2}(\text{A}+\text{A})^\text{T}$
$=\frac{1}{2}\begin{pmatrix}\begin{bmatrix}3&2&7 \\1&4&3\\-2&5&8 \end{bmatrix}+\begin{bmatrix}3&1 & -2 \\2 & 4&5\\7&3&8 \end{bmatrix}\ \end{pmatrix}=\begin{bmatrix}3&\frac{3}{2}&\frac{5}{2}\\\frac{3}{2}&4&4\\\frac{5}{2}&4&8 \end{bmatrix}$
Let $\text{y}=\frac{1}{2}(\text{A}+\text{A}^{\text{T}})$ $=\frac{1}{2}\begin{pmatrix}\begin{bmatrix}3&2&7 & \\1&4 & 3\\-2&5&8 \end{bmatrix}+\begin{bmatrix}3 & 1&-2 \\2 & 4&5\\7&3&8 \end{bmatrix}\end{pmatrix} =\begin{bmatrix}0&\frac{1}{2} & \frac{9}{2} \\\frac{-1}{2}& 0&-1\\\frac{-9}{2}&0&0 \end{bmatrix}$ $\text{x}^{\text{T}}=\begin{bmatrix}3 & \frac{3}{2}&\frac{5}{2} \\\frac{3}{2} & 4&4\\\frac{5}{2}&4&8 \end{bmatrix}^{\text{T}}=\begin{bmatrix}3&\frac{3}{2} & \frac{5}{2} \\\frac{3}{2} & 4&4\\\frac{5}{2}&4&8 \end{bmatrix}^{\text{T}}=\text{x}$ $\text{y}^{\text{T}}=\begin{bmatrix}0 & \frac{1}{2}&\frac{9}{2} \\\frac{-1}{2} & 0&-1\\\frac{-9}{2}&1&0 \end{bmatrix}=\begin{bmatrix}0 & \frac{-1}{2}&\frac{-9}{2} \\\frac{1}{2} & 0&1\\\frac{9}{2} &-1&0\end{bmatrix}=-\begin{bmatrix}0&\frac{1}{2}&\frac{9}{2} \\\frac{-1}{2} & 0&-1\\\frac{-9}{2}&1&0 \end{bmatrix}=\text{y}$ Thus, x is a symmetric matrix and y is skew- symmrteic matrix. Now, $\text{x}+{\text{y}}=\begin{bmatrix}3 & \frac{3}{2}&\frac{5}{2} \\\frac{-3}{2} & 4&4\\\frac{5}{2}&4&8 \end{bmatrix}=\begin{bmatrix}0 & \frac{-1}{2}&\frac{-9}{2} \\\frac{1}{2} & 0&-1\\\frac{9}{2} &1&0\end{bmatrix}$ $=\begin{bmatrix}0&\frac{1}{2}&\frac{9}{2} \\\frac{-1}{2} & 0&-1\\\frac{-9}{2}&1&0 \end{bmatrix}=\begin{bmatrix}3&2&7 \\1&4 & 0\\-2&5&8 \end{bmatrix}=\text{A}$ $\therefore\ \text{x}=\begin{bmatrix}3&\frac{3}{2}&\frac{5}{2} \\\frac{3}{2} & 4&4\\\frac{5}{2}&4&8\end{bmatrix}$ and $\text{y}=\begin{bmatrix}0&\frac{1}{2}&\frac{9}{2} \\\frac{-1}{2} &0&-1\\\frac{-9}{3}&1&0\end{bmatrix}$
View full question & answer→Question 25 Marks
If $\text{A}=\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix},\text{ B}=\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix},$ find matrix X such that 2A + 3X = 5B.
AnswerGiven, $2\text{A}+3\text{X}=5\text{B}$
$\Rightarrow2\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix}+3\text{X}=5\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix}$
$\Rightarrow\begin{bmatrix}4&-4\\8&4\\-10&2\end{bmatrix}+3\text{X}=\begin{bmatrix}40&0\\20&-10\\15&30\end{bmatrix}$
$\Rightarrow3\text{X}=\begin{bmatrix}40&0\\20&-10\\15&30\end{bmatrix}-\begin{bmatrix}4&-4\\8&4\\-10&2\end{bmatrix}$
$\Rightarrow3\text{X}=\begin{bmatrix}40-4&0+4\\20-8&-10-4\\15+10&30-2\end{bmatrix}$
$\Rightarrow3\text{X}=\begin{bmatrix}36&4\\12&-14\\25&28\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{3}\begin{bmatrix}36&4\\12&-14\\25&28\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}12&\frac{4}{3}\\4&\frac{-14}{3}\\\frac{25}{3}&\frac{28}{3}\end{bmatrix}$
View full question & answer→Question 35 Marks
If $\text{P}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix},$ prove that $\text{PQ}=\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}=\text{QP}$
AnswerGiven: $\text{P}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}$
Now,
$\text{P}\text{Q}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}$
$ =\begin{bmatrix}\text{xa}+0+0&0+0+0&0+0+0\\0+0+0&0+\text{y}\text{b}+0&0+0+0\\0+0+0&0+0+0&0+0+\text{zc}\end{bmatrix}$
$ =\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}\ \dots(4)$
Also,
$\text{Q}\text{P}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$
$=\begin{bmatrix}\text{ax}+0+0&0+0+0&0+0+0\\0+0+0&0+\text{by}+0&0+0+0\\0+0+0&0+0+0&0+0+\text{cz}\end{bmatrix}$
$=\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}\ \dots(5)$
From (4) and (5), we get
$\text{PQ}=\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}=\text{QP}$
View full question & answer→Question 45 Marks
If $\text{A}=\begin{bmatrix}2&3\\-1&0\end{bmatrix},$ show that $A^2 - 2A + 3I_2 = 0$.
AnswerGiven: $\text{A}=\begin{bmatrix}2&3\\-1&0\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}2&3\\-1&0\end{bmatrix}\begin{bmatrix}2&3\\-1&0\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4-3&6+0\\-2+0&-3+0\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&6\\-2&-3\end{bmatrix}$
$ \text{A}^2-2\text{A}+3\text{I}_2$
$\Rightarrow\text{A}^2-2\text{A}+3\text{I}_2=\begin{bmatrix}1&6\\-2&-3\end{bmatrix}-2\begin{bmatrix}2&3\\-1&0\end{bmatrix}+3\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$ \Rightarrow\text{A}^2-2\text{A}+3\text{I}_2=\begin{bmatrix}1&6\\-2&-3\end{bmatrix}-\begin{bmatrix}4&6\\-2&0\end{bmatrix}+\begin{bmatrix}3&0\\0&3\end{bmatrix}$
$ \Rightarrow\text{A}^2-2\text{A}+3\text{I}_2=\begin{bmatrix}1-4+3&6-6+0\\-2+2+0&-3+0+3\end{bmatrix}$
$\Rightarrow\text{A}^2-2\text{A}+3\text{I}_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0$
Hence proved.
View full question & answer→Question 55 Marks
If $\text{A}=\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix},$ show that $A^2 - 7A + 10I_3 = 0$.
AnswerGiven,
$\text{A}=\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}$
$ \text{A}^2-7\text{A}+10\text{I}_3$
$ =\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}-7\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}+10\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$ =\begin{bmatrix}9+2+0&6+8+0&0+0+0\\3+4+0&2+16+0&0+0+0\\0+0+0&0+0+0&0+0+25\end{bmatrix}-\begin{bmatrix}21&14&0\\7&28&0\\0&0&35\end{bmatrix}+\begin{bmatrix}10&0&0\\0&10&0\\0&0&10\end{bmatrix} $
$=\begin{bmatrix}11&14&0\\7&18&0\\0&0&25\end{bmatrix}-\begin{bmatrix}21&14&0\\7&28&0\\0&0&35\end{bmatrix}+\begin{bmatrix}10&0&0\\0&10&0\\0&0&10\end{bmatrix}$
$=\begin{bmatrix}11-21+10&14-14+0&0-0+0\\7-7+0&18-28+10&0-0+0\\0-0+0&0-0+0&25-35+10\end{bmatrix}$
$ =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0$
Hence,
$ \text{A}^2-7\text{A}+10\text{I}_3=0$
View full question & answer→Question 65 Marks
If $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then prove that $A^2 - A + 2I = O$.
AnswerGiven: $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}$
$\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}-\begin{bmatrix}3&-2\\4&-2\end{bmatrix}+2\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}1-3&-2+2\\4-4&-4+2\end{bmatrix}\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}-2&0\\0&-2\end{bmatrix}\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}-2+2&0+0\\0+0&-2+2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=0$
Hence proved.
View full question & answer→Question 75 Marks
Show that $\text{AB}\neq\text{BA}$ in the following cases:
$\text{A}=\begin{bmatrix}10&-4&-1\\-11&5&0\\9&-5&1 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&1\\3&4&2\\1&3&2\end{bmatrix}$
Answer$\text{A}=\begin{bmatrix}10&-4&-1\\-11&5&0\\9&-5&1 \end{bmatrix},\ \text{B}=\begin{bmatrix}1&2&1\\3&4&2\\1&3&2\end{bmatrix}$$\text{AB}=\begin{bmatrix}10&-4&-1\\-11&5&0\\9&-5&1 \end{bmatrix}\begin{bmatrix}1&2&1\\3&4&2\\1&3&2\end{bmatrix}$
$=\begin{bmatrix}10-12-1&20-16-3&10-8-2\\-11+15+0&-22+20+0&-11+10+0\\9-15+1&18-20+3&9-10+2\end{bmatrix}$
$\text{AB}=\begin{bmatrix}-3&1&0\\4&-2&-1\\-5&1&1\end{bmatrix}\ \dots(\text{i})$
$\text{BA}=\begin{bmatrix}1&2&1\\3&4&2\\1&3&2\end{bmatrix}\begin{bmatrix}10&-4&-1\\-11&5&0\\9&-5&1 \end{bmatrix}$
$=\begin{bmatrix}10-22+9&-4+10-5&-9+0+1\\30-44+10&-12+20-10&-3+0+2\\10-33+18&-4+15-10&-1+0+2\end{bmatrix}$
$\text{BA}=\begin{bmatrix}-3&1&0\\4&-2&-1\\-5&1&1\end{bmatrix}\ \dots(\text{ii})$
From equation (i) and (ii),
$\text{AB}\neq\text{BA}$
View full question & answer→Question 85 Marks
If $\text{A}=\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix},$ show that $AB = BA = O_{3 \times 3}$
AnswerHere,
$\text{AB}=\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}0+\text{abc}-\text{abc}&0+\text{b}^2\text{c}-\text{b}^2\text{c}&0+\text{bc}^2-\text{bc}^2\\-\text{a}^2\text{c}+0+\text{a}^2\text{c}&-\text{abc}+0+\text{abc}&-\text{ac}^2+0+\text{ac}^2\\\text{a}^2\text{b}-\text{a}^2\text{b}+0&\text{ab}^2-\text{ab}^2+0&\text{abc}-\text{abc}+0\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\Rightarrow\text{AB}=\text{O}_{3\times3}\ \dots(1)$
$\text{BA}=\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix}\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0-\text{abc}+\text{abc}&\text{a}^2\text{c}+0-\text{a}^2\text{c}&-\text{a}^2\text{b}+\text{a}^2\text{b}+0\\0-\text{b}^2\text{c}+\text{b}^2\text{c}&\text{abc}+0-\text{abc}&-\text{ab}^2+\text{ab}^2+0\\0-\text{bc}^2+\text{bc}^2&\text{ac}^2+0-\text{ac}^2&-\text{abc}+\text{abc}+0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\Rightarrow\text{BA}=\text{O}_{3\times3}\ \dots(2)$
$ \Rightarrow\text{AB}=\text{BA}=0_{3\times3}$ [From eqs. (1) and (2)]
View full question & answer→Question 95 Marks
Find the matrix A such that
$\begin{bmatrix}1&1\\0&1\end{bmatrix}\text{A}=\begin{bmatrix}3&3&5\\1&0&1\end{bmatrix}$
AnswerLet $\text{A}=\begin{bmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&1\\0&1\end{bmatrix}\begin{bmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\end{bmatrix}=\begin{bmatrix}3&3&5\\1&0&1\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}\text{x}+\text{a}&\text{y}+\text{b}&\text{z}+\text{c}\\0+\text{a}&0+\text{b}&0+\text{c}\end{bmatrix}=\begin{bmatrix}3&3&5\\1&0&1\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}\text{x}+\text{a}&\text{y}+\text{b}&\text{z}+\text{c}\\\text{a}&\text{b}&\text{c}\end{bmatrix}=\begin{bmatrix}3&3&5\\1&0&1\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
⇒ x + a = 3 ...(1)
y + b = 3 ...(2)
z + c = 5 ...(3)
⇒ a = 1, b = 0 and c = 1
Putting the value of a in eq. (1), we get
x + 1 = 3
⇒ x = 3 - 1
$\therefore$ x = 2
Putting the value of b in eq. (2), we get
y + b = 3
⇒ y + 0 = 3
$\therefore$ y = 3
Putting the value of c in eq. (3), we get
z + 1 = 5
⇒ z = 5 - 1
$\therefore$ z = 4
$\therefore\ \text{A}=\begin{bmatrix}2&3&4\\1&0&1\end{bmatrix}$
View full question & answer→Question 105 Marks
If $\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix},$ prove that $\text{A}^\text{n}=\begin{bmatrix}1&\text{n}\\0&1\end{bmatrix}$ for all positive integers n.
AnswerGiven,
$\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}$
To prove $\text{A}^\text{n}=\begin{bmatrix}1&\text{n}\\0&1\end{bmatrix}$ we will use the principle of mathematical induction.
Step 1: Put n - 1
$\text{A}^1=\begin{bmatrix}1&1\\0&1\end{bmatrix}$
So,
$A^n$ is true for n = 1
Step 2: Let, $A^n$ be true for n = k, then
$\text{A}^\text{k}=\begin{bmatrix}1&\text{k}\\0&1\end{bmatrix}\ \dots(\text{i})$
Step 3: We have to show that $ \text{A}^\text{k+1}=\begin{bmatrix}1&\text{k}+1\\0&1\end{bmatrix}$
So,
$\text{A}^\text{k+1}=\text{A}^\text{k}\times\text{A}$
$=\begin{bmatrix}1&\text{k}\\0&1\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}$ {using equation (i) and given}
$ =\begin{bmatrix}1+0&1+\text{k}\\0+0&0+1\end{bmatrix}$
$\text{A}^{\text{k}+1}=\begin{bmatrix}1&1+\text{k}\\0&1\end{bmatrix}$
This shows that $A^n$ is true for n = k + 1 whenever it is true for n = k
Hence, by the principle of mathematical induction $A^n$ is true for all positive integer.
View full question & answer→Question 115 Marks
Find the value of x for which the matrix product
$\begin{bmatrix}2&0&7\\0&1&0\\1&-2&1\end{bmatrix}\begin{bmatrix}-\text{x}&14\text{x}&7\text{x}\\0&1&0\\\text{x}&-4\text{x}&-2\text{x}\end{bmatrix}$ equal an identity matrix.
AnswerHere,
$\begin{bmatrix}2&0&7\\0&1&0\\1&-2&1\end{bmatrix}\begin{bmatrix}-\text{x}&14\text{x}&7\text{x}\\0&1&0\\\text{x}&-4\text{x}&-2\text{x}\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}-2\text{x}+0+7\text{x}&28\text{x}+0-28\text{x}&14\text{x}+0-14\text{x}\\0+0+0&0+1-0&0+0-0\\-\text{x}-0+\text{x}&14\text{x}-2-4\text{x}&7\text{x}-0-2\text{x}\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}5\text{x}&0&0\\0&1&0\\0&10\text{x}-2&5\text{x}\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore\ 5\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{5}$
View full question & answer→Question 125 Marks
If $\text{A}=\begin{bmatrix}3&-5\\-4&2\end{bmatrix},$ find $A^2 - 5A - 14$.
AnswerGiven: $\text{A}=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}\begin{bmatrix}3&-5\\-4&2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9+20&-15-10\\-12-8&20+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}$
$\text{A}^2-5\text{A}-14\text{I}$
$\Rightarrow\text{A}^2-5\text{A}-14\text{I}=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}-5\begin{bmatrix}3&-5\\-4&2\end{bmatrix}-14\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}-14\text{I}=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}-\begin{bmatrix}15&-25\\-20&10\end{bmatrix}-\begin{bmatrix}14&0\\0&14\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}-14\text{I}=\begin{bmatrix}29-15-14&-25+25+0\\-20+20+0&24-10-14\end{bmatrix}$
$ \Rightarrow\text{A}^2-5\text{A}-14\text{I}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
View full question & answer→Question 135 Marks
Solve the matrix equations:
$\begin{bmatrix}2\text{x}&3\end{bmatrix}\begin{bmatrix}1&2\\-3&0\end{bmatrix}\begin{bmatrix}\text{x}\\8\end{bmatrix}=0$
Answer$\begin{bmatrix}2\text{x}&3\end{bmatrix}\begin{bmatrix}1&2\\-3&0\end{bmatrix}\begin{bmatrix}\text{x}\\8\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}2\text{x}-9&4\text{x}\end{bmatrix}\begin{bmatrix}\text{x}\\8\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}\text{x}(2\text{x}-9)+32\text{x}\end{bmatrix}=0$
$ \Rightarrow\begin{bmatrix}2\text{x}^2-9\text{x}+32\text{x}\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}2\text{x}^2+23\text{x}\end{bmatrix}=0$
$\Rightarrow2\text{x}^2+23\text{x}=0$
$\Rightarrow\text{x}(2\text{x}+23)=0$
$\Rightarrow\text{x}=0\ \text{or }\text{x}=-\frac{23}{2}$
$\therefore\ \text{x}=0\ \text{or }\text{x}=-\frac{23}{2}$
View full question & answer→Question 145 Marks
Express the following matrix as the sum of a symmetric and skew-symmetric matrix and verify your result:
$\text{A}=\begin{bmatrix}3 & -2 &-4\\3 & -2&-5\\-1&-1& 2\end{bmatrix}$
AnswerLet, $\text{A}=\begin{bmatrix}3 & -2 &-4\\3 & -2&-5\\-1&-1& 2\end{bmatrix}$
$\Rightarrow\text{A}^{\text{T}}=\begin{bmatrix}3 & 3 &-1\\-2&-2 & 1\\-4&-5&2 \end{bmatrix}$
Let, $\text{X}=\frac{1}{2}(\text{A}+\text{A}^{\text{T}})=\frac{1}{2}\begin{bmatrix}3&-2 & -4 \\3&-2 & -5\\-1&1&2 \end{bmatrix}+\begin{bmatrix}3&3&-1 \\-2&-2&1\\-4&-5&2 \end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}3+3 & -2+3&-4-1 \\3-2& -2-2&-5+1\\-1-4&1-5&2+2 \end{bmatrix}=\frac{1}{2}\begin{bmatrix}6 & 1&-5 \\1 & -4&-4\\-5&-4&4 \end{bmatrix}$
$=\begin{bmatrix}3 & \frac{1}{2}&\frac{-5}{2} \\\frac{1}{2} & -2&-2\\\frac{-5}{2}&-2&2 \end{bmatrix}$
Now, $\text{X}^{\text{T}}=\begin{bmatrix}3 & \frac{1}{2}&\frac{-5}{2} \\\frac{1}{2} & -2&2\\\frac{-5}{2}&-2&2 \end{bmatrix}^{\text{T}}=\begin{bmatrix}3& \frac{1}{2}&\frac{-5}{2} \\\frac{1}{2} & -2&-2\\\frac{-5}{2}&-2&2 \end{bmatrix}=\text{X}$
⇒ X is a symmatric matrix.
Let, $\text{y}=\frac{1}{2}(\text{A}-\text{A}^{\text{T}})=\begin{bmatrix}3&-2&-4 \\3&-2&-5 \\-1&1&2 \end{bmatrix}-\begin{bmatrix}3&3&-1 \\-2&-2&1 \\-4&-5&2 \end{bmatrix}$ $=\frac{1}{2}\begin{bmatrix}3-3&-2-3&-4+1 \\3+2 & -2+2&-5-1\\-1+4&1+5&2-2 \end{bmatrix} $
$=\frac{1}{2}\begin{bmatrix}0&-5&-3 \\5&0&6\\3&6&0 \end{bmatrix}=\begin{bmatrix}0&\frac{-5}{2}&\frac{-3}{2} \\\frac{5}{2} &0&-3\\\frac{3}{2}&3&0 \end{bmatrix}$
$-\text{Y}^{\text{T}}=-\begin{bmatrix}0&\frac{-5}{2}&\frac{-3}{2} \\5&0&6\\3&6&0 \end{bmatrix}=\begin{bmatrix}0&\frac{-5}{2}&\frac{-3}{2} \\\frac{5}{2} &0&-3\\\frac{3}{2}&3&0 \end{bmatrix}=\text{Y}$
⇒ Y is a skew symmetric matrix
$\text{X}+\text{Y}=\begin{bmatrix}3 & \frac{1}{2}&\frac{-5}{2} \\\frac{1}{2} & -2&-2\\\frac{-5}{2}&-2&2 \end{bmatrix}+\begin{bmatrix}0&\frac{-5}{2}&\frac{-3}{2} \\\frac{5}{2} & 0&-3\\\frac{3}{2}&3&0 \end{bmatrix}=\begin{bmatrix}3+0&{\frac{1}{2}-\frac{5}{2}}&{\frac{-5}{-2}-\frac{3}{2}} \\{\frac{1}{2}+\frac{5}{2}} & -2+0&-2-3&\\{\frac{-5}{2}+\frac{3}{2}}&-2+3&2+0 \end{bmatrix}$
$=\begin{bmatrix}3 & -2&-4 \\3&-2&-5\\-1&1&2 \end{bmatrix}=\text{A}$
Hence, symmetric matrix $\text{X}=\begin{bmatrix}3 & \frac{1}{2}&\frac{-5}{2} \\\frac{1}{2}&-2&-2\\\frac{-5}{2}&-2&2 \end{bmatrix}$
View full question & answer→Question 155 Marks
If $\text{A}=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix},$ find $A^2 - 5A + 4I$ and hence find a matrix X such that $A^2 - 5A + 4I + X = 0$.
AnswerGiven: $\text{A}=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}$
$ \text{A}^2=\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}$
$=\begin{bmatrix}4+0+1&0+0-1&2+0+0\\4+2+3&0+1-3&2+3+0\\2-2+0&0-1-0&1-3+0\end{bmatrix}$
$ =\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}$
Now,
$ \text{A}^2-5\text{A}+4\text{I}=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-5\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}+4\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$ =\begin{bmatrix}5-10+4&-1-0+0&2-5+0\\9-10+0&-2-5+4&5-15+0\\0-5+0&-1+5+0&-2-0+4\end{bmatrix}$
$ =\begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix}$
Now, $ \text{A}^2-5\text{A}+4\text{I}+\text{X}=0$
$ =\text{X}=-(\text{A}^2-5\text{A}+4\text{I})$
$ \therefore\ \text{X}=-\begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix}=\begin{bmatrix}1&1&3\\1&3&10\\5&-4&-2\end{bmatrix}$
View full question & answer→Question 165 Marks
If $B, C$ are $n$ rowed square matrices and if $A = B + C, BC = CB, C^2 = O,$ then show that for every $n \in N, A^{n+1} = B^n(B + (n + 1)C).$
AnswerLet $P(n)$ be the statement given by $P(n) : A^{n+1} = B^n(B + (n + 1)C)$ For $n = 1,$ We have
$P(1) : A^2 = B(B + 2C)$
Here,
LHS $= A^2$
$= (B + C)(B + C)$
$= B(B + C)+ C(B + C)$
$= B^2+ BC + CB + C^2$
$= B^2 + 2BC [\because BC = CB$ and $C^2 = O]$
$= B(B + 2C) =$ RHS
RHS Hence, the statement is true for $n=1.$
If the statement is true for $n = k,$ then
$P(k) : A^{k+1} = B^k(B + (k + 2)C) ...(1)$
For $P(k + 1)$ to be true, we must have
$P(k + 1) : A^{k+2} = B^{k+1}(B + (k + 2)C)$
Now,
$A^{k+2} =A^{k+1}A$
$= [B^k(B + (k + 1)C)] (B + C) [$From eqn. $(1)]$
$= [B^{k+1} + (k + 1)B^kC] (B + C)$
$= B^{k+1}(B + C) + (k + 1)B^kC(B + C)$
$= B^{k+2} + B^{k+1}C + (k + 1)B^kCB + (k + 1)B^kC^2$
$= B^{k+2} + B^{k+1}C + (k + 1)B^kBC [\because BC = CB$ and $C^2 = 0]$
$= B^{k+2} + B^{k+1}C + (k + 1)B^{k+1}C$
$= B^{k+2} + (k + 2)B^{k+1}C$
$= B^{k+1}[B + (k + 2)C]$
So, the statement is true for $n = k + 1.$
Hence, by the principle of mathematical induction, $P(n)$ is true for all $n \in N.$
View full question & answer→Question 175 Marks
If $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix},$ show that $A^2 - 5A + 7I_2 = 0$.
AnswerGiven: $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
Now,
$ \text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$
$\text{A}^2-5\text{A}+7\text{I}_2$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}8-15+7&5-5+0\\-5+5+0&3-10+7\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=0$
Hence proved.
View full question & answer→Question 185 Marks
If $\text{A}=\begin{bmatrix} 3\\5\\2\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&0&4\end{bmatrix},$ verify that $(AB)^T = B^TA^T$.
AnswerGiven,
$\text{A}=\begin{bmatrix}3\\5\\2 \end{bmatrix},\text{B}=\begin{bmatrix}1&0&4\end{bmatrix}$
$\text{AB}^\text{T}=\text{B}^\text{T}\text{A}^\text{T }$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}3\\5\\2\end{bmatrix}\begin{bmatrix}1&0&4 \end{bmatrix}\end{pmatrix}^\text{T}=\begin{bmatrix}1&0&4\end{bmatrix}^\text{T}\begin{bmatrix}3\\5\\2\end{bmatrix}^\text{T}$
$\Rightarrow\begin{bmatrix}3&0&12\\5&0&20\\2&0&8\end{bmatrix}^\text{T}=\begin{bmatrix}1\\0\\4\end{bmatrix}\begin{bmatrix}3&5&2 \end{bmatrix}$
$\Rightarrow\begin{bmatrix} 3&5&2\\0&0&0\\12&20&8\end{bmatrix}=\begin{bmatrix} 3&5&2\\0&0&0\\12&20&8\end{bmatrix}$
$\Rightarrow\text{LHS}=\text{RHS}$
So,
$(\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$
View full question & answer→Question 195 Marks
For the following matrices verify the associativity of multiplication i.e., (AB) C = A(BC):
$\text{A}=\begin{bmatrix}1&2&0\\-1&0&1\end{bmatrix},\text{B}=\begin{bmatrix}1&0\\-1&2\\0&3\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1\\-1\end{bmatrix}$
Answer$(\text{AB})\text{C}=\text{A}(\text{BC})$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}1&2&0\\-1&0&1\end{bmatrix}\begin{bmatrix}1&0\\-1&2\\0&3\end{bmatrix}\end{pmatrix}\begin{bmatrix}1\\-1\end{bmatrix}$ $=\begin{bmatrix}1&2&0\\-1&0&1\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1&0\\-1&2\\0&3\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}\end{pmatrix}$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}1-2+0&0+4+0\\-1-0+0&0+0+3\end{bmatrix}\end{pmatrix}\begin{bmatrix}1\\-1\end{bmatrix}$ $=\begin{bmatrix}1&2&0\\-1&0&1\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1-0\\-1-2\\0-3\end{bmatrix}\end{pmatrix}$
$\Rightarrow\begin{bmatrix}-1&4\\-1&3\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}1&2&0\\-1&0&1\end{bmatrix}\begin{bmatrix}\begin{bmatrix}1\\-3\\-3\end{bmatrix}\end{bmatrix}$
$\Rightarrow\begin{bmatrix}-1-4\\-1-3\end{bmatrix}=\begin{bmatrix}1-6-0\\-1-0-3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}-5\\-4\end{bmatrix}=\begin{bmatrix}-5\\-4\end{bmatrix}$
$\therefore\ \text{LHS}=\text{RHS}$
Hence proved.
View full question & answer→Question 205 Marks
For the following matrices verify the distributivity of matrix, multiplication over matrix addtion i.e., A(B + C) = AB + AC.
$\text{A}=\begin{bmatrix}2&-1\\1&1\\-1&2\end{bmatrix},\text{B}=\begin{bmatrix}0&1\\1&1\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}$
Answer$\text{A}(\text{B}+\text{C})=\text{AB}+\text{AC}$
$\Rightarrow\begin{bmatrix}2&-1\\1&1\\-1&2\end{bmatrix}\begin{pmatrix}\begin{bmatrix}0&1\\1&1\end{bmatrix}+\begin{bmatrix}1&-1\\0&1\end{bmatrix}\end{pmatrix}$ $=\begin{bmatrix}2&-1\\1&1\\-1&2\end{bmatrix}\begin{bmatrix}0&1\\1&1\end{bmatrix}+\begin{bmatrix}2&-1\\1&1\\-1&2\end{bmatrix}\begin{bmatrix}1&-1\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2&-1\\1&1\\-1&2\end{bmatrix}\begin{bmatrix}0+1&1-1\\1+0&1+1\end{bmatrix}$ $=\begin{bmatrix}0-1&2-1\\0+1&1+1\\0+2&-1+2\end{bmatrix}+\begin{bmatrix}2-0&-2-1\\1+0&-1+1\\-1+0&1+2\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2&-1\\1&1\\-1&2\end{bmatrix}\begin{bmatrix}1&0\\1&2\end{bmatrix}=\begin{bmatrix}-1&1\\1&2\\2&1\end{bmatrix}+\begin{bmatrix}2&-3\\1&0\\-1&3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2-1&0-2\\1+1&0+2\\-1+2&0+4\end{bmatrix}=\begin{bmatrix}-1+2&1-3\\1+1&2+0\\2-1&1+3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&-2\\2&2\\1&4\end{bmatrix}=\begin{bmatrix}1&-2\\2&2\\1&4\end{bmatrix}$
$\therefore\ \text{LHS}=\text{RHS}$
Hence proved.
View full question & answer→Question 215 Marks
Evaluate the following:
$\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1&0&2\\2&0&1\end{bmatrix}-\begin{bmatrix}0&1&2\\1&0&2 \end{bmatrix}\end{pmatrix}$
Answer$\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1&0&2\\2&0&1\end{bmatrix}-\begin{bmatrix}0&1&2\\1&0&2 \end{bmatrix}\end{pmatrix}$
$=\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1-0&0-1&2-2\\2-1&0-0&1-2 \end{bmatrix}\end{pmatrix}$
$=\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\begin{bmatrix}1&-1&0\\1&0&1\end{bmatrix}$
$=\begin{bmatrix}1-1&-1+0&0+1\\0+2&0+0&0-2\\2+3&-2+0&0-3\end{bmatrix}$
$=\begin{bmatrix}0&-1&1\\2&0&-2\\5&-2&-3\end{bmatrix}$
Hence,
$\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1&0&2\\2&0&1\end{bmatrix}-\begin{bmatrix} 0&1&2\\1&0&2\end{bmatrix} \end{pmatrix}\begin{bmatrix}0&-1&1\\2&0&-2\\5&-2&-3\end{bmatrix}$
View full question & answer→Question 225 Marks
If $\begin{bmatrix}\text{x}&4&1\end{bmatrix}\begin{bmatrix}2&1&2\\1&0&2\\0&2&-4\end{bmatrix}\begin{bmatrix}\text{x}\\4\\-1\end{bmatrix}=0,$ find x.
AnswerGiven,
$\begin{bmatrix}\text{x}&4&1\end{bmatrix}\begin{bmatrix}2&1&2\\1&0&2\\0&2&-4\end{bmatrix}\begin{bmatrix}\text{x}\\4\\-1\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}2\text{x}+4+0&\text{x}+0+2&2\text{x}+8-4\end{bmatrix}\begin{bmatrix}\text{x}\\4\\-1\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}2\text{x}+4&\text{x}+2&2\text{x}+4\end{bmatrix}\begin{bmatrix}\text{x}\\4\\-1\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}(2\text{x}+4)\text{x}+4(\text{x}+2)-1(2\text{x}+4)\end{bmatrix}=0$
$\Rightarrow2\text{x}^2+4\text{x}+4\text{x}+8-2\text{x}-4=0$
$\Rightarrow2\text{x}+6\text{x}+4=0$
$\Rightarrow2\text{x}^2+2\text{x}+4\text{x}+4=0$
$\Rightarrow2\text{x}(\text{x}+1)+4(\text{x}+1)=0$
$\Rightarrow(\text{x}+1)+(2\text{x}+4)=0$
$\Rightarrow\text{x}+1=0\ \text{or }2\text{x}+4=0$
$\Rightarrow\text{x}=-1\ \text{or }\text{x}=-2$
Hence, x = -1 or -2
View full question & answer→Question 235 Marks
$=\begin{bmatrix}1&1&\text{x}\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&1&0\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}=0,$ find x.
AnswerGiven: $=\begin{bmatrix}1&1&\text{x}\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&1&0\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}1+0+2\text{x}&0+2+\text{x}&2+1+0\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}=0$
$\begin{bmatrix}1+2\text{x}&2+\text{x}&3\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}=0$
$\Rightarrow[1+2\text{x}+2+\text{x}+3]=0$
$\Rightarrow6+3\text{x}=0$
$\Rightarrow3\text{x}=-6$
$\Rightarrow\text{x}=\frac{-6}{3}$
$\therefore\ \text{x}=-2$
View full question & answer→Question 245 Marks
If $\text{A}=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix},\text{B}=\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix},$ verify that A(B - C) = AB - AC.
AnswerGiven, $\text{A}=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix},\text{B}=\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}$ $\text{C}=\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}$ $ \text{A}(\text{B}-\text{C})=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}-\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}\end{bmatrix}$ $=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}0-1&5-5&-4-2\\-2+1&1-1&3-0\\-1-0&0+1&2-1\end{bmatrix}$ $=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}-1&0&-6\\-1&0&3\\-1&1&1\end{bmatrix}$ $=\begin{bmatrix}-1+0+2&0+0-2&-6+0-2\\-3+1+0&0+0+0&-18-3+0\\2-1-1&0+0+1&12+3+1\end{bmatrix}$ $\text{A}(\text{B}-\text{C})=\begin{bmatrix}1&-2&-8\\-2&0&-21\\0&1&16\end{bmatrix}\ \dots(\text{i})$ $ \text{AB}-\text{AC}=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}-\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}$ $=\begin{bmatrix}0+0+2&5+0+0&-4+0-4\\0+2+0&15-1+0&-12-3+0\\0-2-1&-10+1+0&8+3+2\end{bmatrix}-\begin{bmatrix}1+0+0&5+0+2&2+0-2\\3+1+0&15-1+0&6+0+0\\0-2-1&-10+1+1&-4+0+1\end{bmatrix}$ $=\begin{bmatrix}2&5&-8\\2&14&-15\\-3&-9&13\end{bmatrix}-\begin{bmatrix}1&7&0\\4&14&6\\-3&-10&-3\end{bmatrix}$ $=\begin{bmatrix}2-1&5-7&-8-0\\2-4&14-14&-14-6\\-3+3&-9+10&3+3\end{bmatrix}$ $\text{AB}-\text{AC}=\begin{bmatrix}1&-2&-8\\-2&0&-21\\0&1&16\end{bmatrix}\ \dots(\text{ii})$ From equation (i) and (ii),$\text{A}(\text{B}-\text{C})=\text{AB}-\text{AC}$
View full question & answer→Question 255 Marks
If $\text{A}=\begin{bmatrix}0&0\\4&0\end{bmatrix},$ find $A^{16}$.
AnswerGiven,
$\text{A}=\begin{bmatrix}0&0\\4&0\end{bmatrix}$
$ \text{A}^2=\text{A}\times\text{A}$
$ =\begin{bmatrix}0&0\\4&0\end{bmatrix}\begin{bmatrix}0&0\\4&0\end{bmatrix}$
$ =\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$=0$
$ \text{A}^4=\text{A}^2\times\text{A}^2$
$=0\times0$
$=0$
$ \text{A}^{16}=\text{A}^4\times\text{A}^4$
$=0\times0$
$=0$
So,
$A^{16}$ is a null matrix
View full question & answer→Question 265 Marks
Evaluate the following:
$\begin{pmatrix}\begin{bmatrix}1&3\\-1&-4 \end{bmatrix}+\begin{bmatrix}3&-2\\-1&1 \end{bmatrix}\end{pmatrix}\begin{bmatrix}1&3&5\\2&4&6 \end{bmatrix}$
Answer$\begin{pmatrix}\begin{bmatrix}1&3\\-1&-4 \end{bmatrix}+\begin{bmatrix}3&-2\\-1&1 \end{bmatrix}\end{pmatrix}\begin{bmatrix}1&3&5\\2&4&6 \end{bmatrix}$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}1+3&3-2\\-1-1&-4+1\end{bmatrix}\end{pmatrix}\begin{bmatrix}1&3&5\\2&4&6 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}4&1\\-2&-3\end{bmatrix}\begin{bmatrix}1&3&5\\2&4&6\end{bmatrix}$
$\Rightarrow\begin{bmatrix}4+2&12+4&20+6\\-2-6&-6-12&-10-18\end{bmatrix}$
$\Rightarrow\begin{bmatrix}6&16&26\\-8&-18&-28\end{bmatrix}$
View full question & answer→Question 275 Marks
If $\text{A}=\begin{bmatrix}-2\\4\\5\end{bmatrix},\text{B}=\begin{bmatrix}1&3&-6\end{bmatrix},$ verify that $(AB)^T = B^TA^T$
AnswerGiven: $\text{A}=\begin{bmatrix}-2\\4\\5\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}-2&4&5\end{bmatrix}$
$\text{B}=\begin{bmatrix}1&3&-6\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}1\\3\\-6\end{bmatrix}$
Now,
$\text{AB}=\begin{bmatrix}-2\\4\\5\end{bmatrix}\begin{bmatrix}1&3&-6\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}-2&-6&12\\4&12&-24\\5&15&-30\end{bmatrix}$
$\Rightarrow(\text{AB})^\text{T}=\begin{bmatrix}-2&4&5\\-6&12&15\\12&-24&-30\end{bmatrix}\ \dots(1)$
$\text{B}^\text{T}\text{A}^\text{A}=\begin{bmatrix}1\\3\\-6\end{bmatrix}\begin{bmatrix}-2&4&5\end{bmatrix}$
$\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}-2&4&5\\-6&12&15\\12&-24&-30\end{bmatrix}\ \dots(2)$
$\therefore\ (\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$ [From eqs. (1) and (2)]
View full question & answer→Question 285 Marks
If $l_i, m_i, n_i; i = 1, 2, 3$ denotes the direction cosines of three mutually perpendicular vectors in space, prove that $AA^T = I$, where
$\text{A}=\begin{bmatrix}\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\\\text{l}_3&\text{m}_3&\text{n}_3 \end{bmatrix}.$
AnswerGiven,
$l_i, m_i, n_i$ are direction cosines of three mutually perpendicular vectors
$\begin{matrix}\text{l}_1\text{l}_2+\text{m}_1\text{m}_2+\text{n}_1\text{n}_2=0 \\\text{l}_2\text{l}_3+\text{m}_2\text{m}_3+\text{n}_2\text{n}_3=0\\\text{l}_1\text{l}_3+\text{m}_1\text{m}_3+\text{n}_1\text{n}_3=0\end{matrix}\Bigg\}\ \dots(\text{A})$
And,
$\begin{matrix}\text{l}_1{^2}+\text{m}_1{^2}+\text{n}_1{^2}=1\\\text{l}_2{^2}+\text{m}_2{^2}+\text{n}_2{^2}=1\\\text{l}_3{^2}+\text{m}_3{^2}+\text{n}_3{^2}=1\end{matrix}\Bigg\}\ \dots(\text{B})$
Given
$\text{A}=\begin{bmatrix}\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\\\text{l}_3&\text{m}_3&\text{n}_3 \end{bmatrix}$
$\text{AA}^{T}=\begin{bmatrix}\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\\\text{l}_3&\text{m}_3&\text{n}_3 \end{bmatrix}\begin{bmatrix}\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\\\text{l}_3&\text{m}_3&\text{n}_3 \end{bmatrix}^\text{T}$
$=\begin{bmatrix}\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\\\text{l}_3&\text{m}_3&\text{n}_3 \end{bmatrix}\begin{bmatrix}\text{l}_1&\text{l}_1&\text{l}_3\\\text{m}_1&\text{m}_2&\text{m}_3\\\text{n}_1&\text{n}_2&\text{n}_3 \end{bmatrix}$
$=\begin{bmatrix}\text{l}_1{^2}+\text{m}_1{^2}+\text{n}_1{^2}&\text{l}_1\text{l}_2+\text{m}_1\text{m}_2+\text{n}_1\text{n}_2&\text{l}_1\text{l}_3+\text{m}_1\text{m}_3+\text{n}_1\text{n}_3\\\text{l}_1\text{l}_2+\text{m}_1\text{m}_2+\text{n}_1\text{n}_2&\text{l}_2{^2}+\text{m}_2{^2}+\text{n}_2{^2}&\text{l}_2\text{l}_3+\text{m}_2\text{m}_3+\text{n}_2\text{n}_3\\\text{l}_1\text{l}_3+\text{m}_1\text{m}_3+\text{n}_1\text{n}_3&\text{l}_3\text{l}_2+\text{m}_3\text{m}_2+\text{n}_3\text{n}_2&\text{l}_3{^2}+\text{m}_3{^2}+\text{n}_3{^2}\end{bmatrix}$
$=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \\ \end{bmatrix}$ {using (A) and (B)}
$=\text{l}$
Hence,
$\text{AA}^{\text{T}}=\text{l}$
View full question & answer→Question 295 Marks
If $\text{A}=\begin{bmatrix}4&2\\-1&-1 \end{bmatrix},$ prove that (A - 2I)(A - 3I) = 0
AnswerGiven, $\text{A}=\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}$
$(\text{A}-2\text{I})(\text{A}-3\text{I})$
$=\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-2\begin{bmatrix}1&0\\0&1 \end{bmatrix}\end{pmatrix}\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-3\begin{bmatrix}1&0\\0&1 \end{bmatrix} \end{pmatrix}$
$=\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-\begin{bmatrix}2&0\\0&2 \end{bmatrix} \end{pmatrix}\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-\begin{bmatrix}3&0\\0&3 \end{bmatrix} \end{pmatrix}$
$=\begin{pmatrix}\begin{bmatrix}4-2&2-0\\-1-0&1-2 \end{bmatrix} \end{pmatrix}\begin{pmatrix}\begin{bmatrix}4-3&2-0\\-1-0&1-3 \end{bmatrix} \end{pmatrix}$
$=\begin{bmatrix}2&2\\-1&-1 \end{bmatrix}\begin{bmatrix}1&2\\-1&-2 \end{bmatrix}$
$=\begin{bmatrix}2-2&4-4\\-1+1&-2+2 \end{bmatrix}$
$=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}$
$=0$
Hence,
$(\text{A}-2\text{I})(\text{A}-3\text{I})=0$
View full question & answer→Question 305 Marks
If $\text{A}=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix},$ show that AB = A and BA = B.
AnswerGiven, $\text{A}=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix},\text{B}=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}$
$\text{AB}=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}$
$=\begin{bmatrix}4+3-5&-4-9+10&-8-12+15\\-2-4+5&2+12-10&4+16-15\\2+3-4&-2-9+18&-4-12+12\end{bmatrix}$
$=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$
$\text{AB}=\text{A}$
$\text{BA}=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$
$=\begin{bmatrix}4+2-4&-6-8+12&-10-10+16\\-2-3+4&3+12-12&5+15-16\\2+2-3&-3-8+9&-5-10+12\end{bmatrix}$
$=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}$
$\text{BA}=\text{B}$
View full question & answer→Question 315 Marks
If $\text{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{B}\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ and $\text{C}=\begin{bmatrix}0&1\\1&0\end{bmatrix},$ then show that $A^2 = B^2 = C^2 = l_2$.
AnswerHere,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+0&0+0\\0+0&0+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\ \dots(1)$
$\text{B}^2=\text{BB}$
$\Rightarrow\text{B}^2=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}1&0\\0&-1\end{bmatrix}$
$\Rightarrow\text{B}^2=\begin{bmatrix}1+0&0-0\\0-0&0+1\end{bmatrix}$
$\Rightarrow\text{B}^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\ \dots(2)$
$\text{C}^2=\text{CC}$
$\Rightarrow\text{C}^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\Rightarrow\text{C}^2=\begin{bmatrix}0+1&0+0\\0+0&1+0\end{bmatrix}$
$\Rightarrow\text{C}^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\ \dots(3)$
We know,
$\text{I}_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\ \dots(4)$
$\Rightarrow\text{A}^2=\text{B}^2=\text{C}^2=\text{I}^2$ [From eqs. (1), (2), (3) and (4)]
View full question & answer→Question 325 Marks
For the following matrices verify the associativity of multiplication i.e., (AB) C = A(BC):
$\text{A}=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix},\text{B}=\begin{bmatrix}1&-1&1\\0&1&2\\2&-1&1\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}$
AnswerGiven,
$\text{A}=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix},\text{B}=\begin{bmatrix}1&-1&1\\0&1&2\\2&-1&1\end{bmatrix},\text{C}=\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}$
$(\text{AB})\text{C}=\begin{pmatrix}\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix}\begin{bmatrix}1&-1&1\\0&1&2\\2&-1&1\end{bmatrix}\end{pmatrix}\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}4+0+6&-4+2-3&4+4+3\\1+0+4&-1+1-2&1+2+2\\3+0+2&-3+0-1&3+0+1\end{bmatrix}\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}10&-5&11\\5&-2&5\\5&-4&4\end{bmatrix}\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}10-15+0&20+0+0&-10+5+11\\5-6+0&10+0+0&-5-2+5\\5-12+0&10+0+0&-5-4+4\end{bmatrix}$
$(\text{AB})\text{C}=\begin{bmatrix}-5&20&-4\\-1&10&-2\\-7&10&-5\end{bmatrix}\ \dots(\text{i})$
$\text{A}(\text{BC})=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1&-1&1\\0&1&2\\2&-1&1\end{bmatrix}\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}\end{pmatrix}$
$=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix}\begin{bmatrix}1-3+0&2+0+0&-1-1+1\\0+3+0&0+0+0&0+1+2\\2-3+0&4+0+0&-2-1+1\end{bmatrix}$
$=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix}\begin{bmatrix}-2&2&-1\\3&0&3\\-1&4&-2\end{bmatrix}$
$=\begin{bmatrix}-8+6-3&8+0+12&-4+6-6\\-2+3-2&2+0+8&-1+3-4\\-6+0-1&6+0+4&-3+0-2\end{bmatrix}$
$\text{A}(\text{BC})=\begin{bmatrix}-5&20&-4\\-1&10&-2\\-7&10&-5\end{bmatrix}\ \dots\text{(ii)}$
From equation (i) and (ii),
$(\text{AB})\text{C}=\text{A}(\text{BC})$
View full question & answer→Question 335 Marks
Solve the matrix equations:
$\begin{bmatrix}\text{x}&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}\text{x}\\4\\1\end{bmatrix}=0$
Answer$\begin{bmatrix}\text{x}&-5&-1\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}\text{x}\\4\\1\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}\text{x}-0-2&0-10-0&2\text{x}-5-3\end{bmatrix}\begin{bmatrix}\text{x}\\4\\1\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}\text{x}-2&-10&2\text{x}-8\end{bmatrix}\begin{bmatrix}\text{x}\\4\\1\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}\text{x}^2-2\text{x}-40+2\text{x}-8\end{bmatrix}=0$
$ \Rightarrow\text{x}^2-48=0$
$\Rightarrow\text{x}^2=48$
$ \Rightarrow\text{x}=\pm\sqrt{48}$
View full question & answer→Question 345 Marks
If $\text{A}=\begin{bmatrix}2&-1\\3&2 \end{bmatrix} $ and $\text{B}=\begin{bmatrix}0&4\\-1&7\end{bmatrix} ,$ find $3A^2 - 2B + l$
AnswerGiven: $\text{A}=\begin{bmatrix}2&-1\\3&2 \end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}2&-1\\3&2 \end{bmatrix}\begin{bmatrix}2&-1\\3&2 \end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4-3&-2-2\\6+6&-3+4 \end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&-4\\12&1 \end{bmatrix}$
$3\text{A}^2-2\text{B}+\text{I}$
$\Rightarrow3\text{A}^2-2\text{B}+\text{I}=3\begin{bmatrix}1&-4\\12&1 \end{bmatrix}-2\begin{bmatrix}0&4\\-1&7 \end{bmatrix}+\begin{bmatrix}1&0\\0&1 \end{bmatrix}$
$\Rightarrow3\text{A}^2-2\text{B}+\text{I}=\begin{bmatrix}3&-12\\36&3 \end{bmatrix}-\begin{bmatrix}0&8\\-2&14 \end{bmatrix}+\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow3\text{A}^2-2\text{B}+\text{I}=\begin{bmatrix}3-0+1&-12-8+0\\36+2+0&3-14+1\end{bmatrix}$
$\Rightarrow3\text{A}^2-2\text{B}+\text{I}=\begin{bmatrix}4&-20\\38&-10 \end{bmatrix}$
View full question & answer→Question 355 Marks
If $\text{A}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix},\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ and $x^2 = -1$ then show that $(A + B)^2 = A^2 + B^2$.
AnswerGiven: $\text{A}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix},\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ and $x^2 = -1$ To show: $(A + B)^2 = A^2 + B^2$
L.H.S
$\text{A}+\text{B}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix}+\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$=\begin{bmatrix}0+0&-\text{x}+1\\\text{x}+1&0+0\end{bmatrix}$
$=\begin{bmatrix}0&-\text{x}+1\\\text{x}+1&0\end{bmatrix}$
$(\text{A}+\text{B})^2=\begin{bmatrix}0&-\text{x}+1\\\text{x}+1&0\end{bmatrix}\begin{bmatrix}0&-\text{x}+1\\\text{x}+1&0\end{bmatrix}$
$=\begin{bmatrix}0+(1-\text{x})(1+\text{x})&0+0\\0+0&(\text{x}+1)(1-\text{x})\end{bmatrix}$
$=\begin{bmatrix}1-\text{x}^2&0\\0&1-\text{x}^2\end{bmatrix}\ \dots(1)$
R.H.S
$ \text{A}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix}$
$ \text{A}^2=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix}\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix}$
$=\begin{bmatrix}0-\text{x}^2&0+0\\0+0&-\text{x}^2+0\end{bmatrix}$
$=\begin{bmatrix}-\text{x}^2&0\\0&-\text{x}^2\end{bmatrix}\ \dots(2)$
$\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\text{B}^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$=\begin{bmatrix}0+1&0+0\\0+0&1+0\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}\ \dots(3)$
Adding (2) and (3), we get
$\text{A}^2+\text{B}^2=\begin{bmatrix}-\text{x}^2&0\\0&-\text{x}^2\end{bmatrix}+\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$ =\begin{bmatrix}1-\text{x}^2&0\\0&1-\text{x}^2\end{bmatrix}\ \dots(4)$
Comparing (1) and (4), we get
$(A + B)^2 = A^2 + B^2$
View full question & answer→Question 365 Marks
Show that the matrix $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$ satisfies the equation $A^3 - 4A^2 + A = 0$.
AnswerGiven, $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}4+3&6+6\\2+2&3+4\end{bmatrix}$
$ =\begin{bmatrix}7&12\\4&7\end{bmatrix}$
$ \text{A}^3=\text{A}^2.\text{A}$
$ =\begin{bmatrix}7&12\\4&7\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}14+12&21+24\\8+7&12+14\end{bmatrix}$
$=\begin{bmatrix}26&45\\15&26\end{bmatrix}$
Hence, $\text{A}^3-4\text{A}^2+\text{A}$
$=\begin{bmatrix}26&45\\15&26\end{bmatrix}-4\begin{bmatrix}7&12\\4&7\end{bmatrix}+\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}26-28+2&45-48+3\\15-16+1&26-28+2\end{bmatrix}$
$ =\begin{bmatrix}0&0\\0&0\end{bmatrix}$
So, $ \text{A}^3-4\text{A}+\text{A}=0$
View full question & answer→Question 375 Marks
Compute the elements $a_{43}$ and $a_{22}$ of the matrix:
$\text{A}=\begin{bmatrix}0&1&0\\2&0&2\\0&3&2\\4&0&4\end{bmatrix}\begin{bmatrix}2&-1\\-3&2\\4&3\end{bmatrix}\begin{bmatrix}0&1&-1&2&-2\\3&-3&4&-4&0\end{bmatrix}$
AnswerWe have,
Given: $\text{A}=\begin{bmatrix}0&1&0\\2&0&2\\0&3&2\\4&0&4\end{bmatrix}\begin{bmatrix}2&-1\\-3&2\\4&3\end{bmatrix}\begin{bmatrix}0&1&-1&2&-2\\3&-3&4&-4&0\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}0&1&0\\2&0&2\\0&3&2\\4&0&4\end{bmatrix}\begin{bmatrix}0-3&2+3&-2-4&4+4&-4-0\\0+6&-3-6&3+8&-6-8&6+0\\0+9&4-9&-4+12&8-12&-8+0\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}0&1&0\\2&0&2\\0 &3&2\\4&0&4\end{bmatrix}\begin{bmatrix}-3&5&-6&8&-4\\6&-9&11&-14&6\\9&-5&8&-4&-8\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}0+6+0&0-9-0&0+11+0&0-14-0&0+6-0\\-6+0+18&10-0-10&-12+0+16&16-0-8&-8+0-16\\0+18+18&0-27-10&0+33+16&0-42-8&0+18-16\\-12+0+36&20-0-20&-24+0+32&32-0-16&-16+0-32\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}6&-9&11&-14&6\\12&0&4&8&-24\\36&-37&49&-50&2\\24&0&8&16&-48\end{bmatrix}$
$\therefore\ \text{a}_{43}=8\text{ and a}_{22}=0$
View full question & answer→Question 385 Marks
If $\text{A}=\begin{bmatrix}\cos\theta&\text{i}\sin\theta\\\text{i}\sin\theta&\cos\theta\end{bmatrix},$ then prove by principle of mathematical induction that $\text{A}^\text{n}=\begin{bmatrix}\cos\text{n}\theta&\text{i}\sin\text{n}\theta\\\text{i}\sin\text{n}\theta&\cos\text{n}\theta\end{bmatrix}$ for all $\text{n}\in\text{N}.$
AnswerWe shall prove the result by the principle of mathematical induction on n.
Step 1: If n = 1, by definition of integral power of a matrix, we have
$\text{A}^1=\begin{bmatrix}\cos1\theta&\text{i}\sin1\theta\\\text{i}\sin1\theta&\cos1\theta\end{bmatrix}=\begin{bmatrix}\cos\theta&\text{i}\sin\theta\\\text{i}\sin\theta&\cos\theta\end{bmatrix}=\text{A}$
Thus, the result is true for n = 1.
Step 2: Let the result be true for n = m. Then,
$\text{A}^\text{m}=\begin{bmatrix}\cos\text{m}\theta&\text{i}\sin\text{m}\theta\\\text{i}\sin\text{m}\theta&\cos\text{m}\theta\end{bmatrix}$
Now we shall show that the result is true for n = m + 1.
Here,
$ \text{A}^\text{m+1}=\begin{bmatrix}\cos\text{m}+1\theta&\text{i}\sin\text{m}+1\theta\\\text{i}\sin\text{m}+1\theta&\cos\text{m}+1\theta\end{bmatrix}$
By definition of integral power of matrix, we have
$\text{A}^\text{m+1}=\text{A}^\text{m}\text{A}$
$\Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}\cos\text{m}\theta&\text{i}\sin\text{m}\theta\\\text{i}\sin\text{m}\theta&\cos\text{m}\theta\end{bmatrix}\begin{bmatrix}\cos\theta&\text{i}\sin\theta\\\text{i}\sin\theta&\cos\theta\end{bmatrix}$ [From eq. (1)]
$\Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}\cos\text{m}\theta.\cos\theta+\text{i}\sin\text{m}\theta.\text{i}\sin\theta&\cos\text{m}\theta.\text{i}\sin\theta+\text{i}\sin\text{m}\theta.\cos\theta\\\text{i}\sin\text{m}\theta.\cos\theta+\cos\text{m}\theta.\text{i}\sin\theta&\text{i}\sin\text{m}\theta.\text{i}\sin\theta+\cos\text{m}\theta.\cos\theta\end{bmatrix}$
$ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\cos\text{m}\theta.\cos\theta-\sin\text{m}\theta.\sin\theta&\text{i}(\cos\text{m}\theta.\sin\theta+\sin\text{m}\theta.\cos\theta)\\\text{i}(\sin\text{m}\theta.\cos\theta+\cos\text{m}\theta.\sin\theta)&-\sin\text{m}\theta.\sin\theta+\cos\text{m}\theta.\cos\theta\end{bmatrix}$
$\Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}\cos\text{m}\theta.\cos\theta-\sin\text{m}\theta.\sin\theta&\text{i}(\cos\text{m}\theta.\sin\theta+\sin\text{m}\theta.\cos\theta)\\\text{i}(\sin\text{m}\theta.\cos\theta+\cos\text{m}\theta.\sin\theta)&\cos\text{m}\theta.\cos\theta-\sin\text{m}\theta.\sin\theta\end{bmatrix}$
$\Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}\cos(\text{m}\theta+\theta)&\text{i}\sin(\text{m}\theta+\theta)\\\text{i}\sin(\text{m}\theta+\theta)&\cos(\text{m}\theta+\theta)\end{bmatrix}$
$ \Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}\cos(\text{m}+1)\theta&\text{i}\sin(\text{m}+1)\theta\\\text{i}\sin(\text{m}+1)\theta&\cos(\text{m}+1)\theta\end{bmatrix}$
This shows that when the result is true for n = m, it is true for n = m + 1.
Hence, by the principle of mathematical induction, the result is valid for all $\text{n}\in\text{N}.$
View full question & answer→Question 395 Marks
If $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix},$ show that $A^2 - 5A + 7I = 0$ use this to find $A^4$.
AnswerGiven: $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$
$\text{A}^2-5\text{A}+7\text{I}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}=\begin{bmatrix}8-15+7&5-5+0\\-5+5+0&3-10+7\end{bmatrix}$
$ \Rightarrow\text{A}^2-5\text{A}+7\text{I}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0$
Hence proved.
Given: $\text{A}^2-5\text{A}+7\text{I}=0$
$\Rightarrow\text{A}^2=5\text{A}-7\text{I}\ \dots(1)$
$\Rightarrow\text{A}^3=\text{A}(5\text{A}-7\text{I})$ (Multipying by A on both sides)
$\Rightarrow\text{A}^3=5\text{A}^2-7\text{AI}$
$\Rightarrow\text{A}^3=5(5\text{A}-7\text{I})-7\text{A}$ [From eq. (1)]
$\Rightarrow\text{A}^3=25\text{A}-35\text{I}-7\text{A}$
$\Rightarrow\text{A}^3=18\text{A}-35\text{I}$
$\Rightarrow\text{A}^4=(18\text{A}-35\text{I})\text{A}$ (Multipying by A on both sides)
View full question & answer→Question 405 Marks
If $\text{A}=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix},$ then show that A is a root of the polynomial $f(x) = x^3 - 6x^2 + 7x + 2$.
AnswerGiven: $f(x) = x^3 - 6x^2 + 7x + 2$
$f(A) = A^3 - 6A^2 + 7A + 2I_3$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}$
$\text{A}^3=\text{A}^2\text{A}$
$\Rightarrow\text{A}^3=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$ \Rightarrow\text{A}^3=\begin{bmatrix}5+0+16&0+0+0&10+0+24\\2+0+10&0+8+0&4+4+15\\8+0+26&0+0+0&16+0+39\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}$
$\Rightarrow\text{A}^3-6\text{A}^2+7\text{A}+2\text{I}_3=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}+7\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}+2\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{A}^3-6\text{A}^2+7\text{A}+2\text{I}_3=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}+\begin{bmatrix}7&0&14\\0&14&7\\14&0&21\end{bmatrix}+\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$
$ \Rightarrow\text{A}^3-6\text{A}^2+7\text{A}+2\text{I}_3=\begin{bmatrix}21-30+7+2&0-0+0+0&34-48+14+0\\12-12+0+0&8-24+14+2&23-30+7+0\\34-48+14+0&0-0+0+0&55-78+21+2\end{bmatrix}$
$ \Rightarrow\text{A}^3-6\text{A}^2+7\text{A}+2\text{I}_3=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0$
Since f(A) = 0, A is the root of $f(x) = x^3 - 6x^2 + 7x + 2$.
View full question & answer→Question 415 Marks
Express the matrix $\text{A}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.
AnswerGiven: $\text{A}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}$
$\text{A}^{\text{T}}=\begin{bmatrix}3 & 1 \\-4 & -1 \end{bmatrix}$
Let $\text{X}=\frac{1}{2}(\text{A}+\text{A}^{\text{T}})=\frac{1}{2}\begin{pmatrix}\begin{bmatrix}3& -4 \\1 &-1 \end{bmatrix}+\begin{bmatrix}3 & 1 \\-4 & -1 \end{bmatrix} \end{pmatrix}=\begin{bmatrix}3 & \frac{-3}{2} \\\frac{-3}{2} & -1 \end{bmatrix}$
$\text{X}^{\text{T}}=\begin{bmatrix}3 & \frac{-3}{2} \\\frac{-3}{2} & -1 \end{bmatrix}^{\text{T}}=\begin{bmatrix}3 & \frac{-3}{2} \\\frac{-3}{2} & -1 \end{bmatrix}=\text{X}$
Let $\text{Y}=\frac{1}{2}(\text{A}-\text{A}^{\text{T}})=\frac{1}{2}\begin{pmatrix}\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}-\begin{bmatrix}3 & 1 \\-4 & -1 \end{bmatrix} \end{pmatrix}=\begin{bmatrix}0 & \frac{-5}{2} \\\frac{5}{2} & 0 \end{bmatrix}$
$\text{Y}^{\text{T}}=\begin{bmatrix}0 & \frac{-5}{2} \\\frac{5}{2} & 0 \end{bmatrix}^{\text{T}}=\begin{bmatrix}0 & \frac{5}{2} \\\frac{-5}{2} & 0 \end{bmatrix}=\begin{bmatrix}0 & \frac{-5}{2} \\\frac{5}{2} & 0 \end{bmatrix}=\text{Y}$
$\therefore$ x is a symmetric matric x and y is as skew-symmrtric matrix.
$\text{X}+\text{Y}=\begin{bmatrix}3 & \frac{-3}{2} \\\frac{-3}{2} & -1 \end{bmatrix}+\begin{bmatrix}0 & \frac{-5}{2} \\\frac{5}{2} & 0 \end{bmatrix}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}=\text{A}$
View full question & answer→Question 425 Marks
If $\text{A}=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix},$ then prove that $A^2 - 4A - 5I = 0$.
AnswerGiven: $\text{A}=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}$
$ \Rightarrow\text{A}^2=\begin{bmatrix}1+4+4&2+2+4&2+4+2\\2+2+4&4+1+4&4+2+2\\2+4+2&4+2+2&4+4+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}$
$\text{A}^2-4\text{A}-5\text{I}$
$\Rightarrow\text{A}^2-4\text{A}-5\text{I}=\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}-4\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}-5\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-4\text{A}-5\text{I}=\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}-\begin{bmatrix}4&8&8\\8&4&8\\8&8&4\end{bmatrix}-\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}$
$ \Rightarrow\text{A}^2-4\text{A}-5\text{I}=\begin{bmatrix}9-4-5&8-8-0&8-8-0\\8-8-0&9-4-5&8-8-0\\8-8-0&8-8-0&9-4-5\end{bmatrix}$
$ \Rightarrow\text{A}^2-4\text{A}-5\text{I}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0$
Hence proved.
View full question & answer→Question 435 Marks
If $\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix},$ show that $\text{A}^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $\text{A}^3=\begin{bmatrix}1&3\\0&1\end{bmatrix}.$
AnswerGiven, $\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&1\\0&1\end{bmatrix}\begin{bmatrix}1&1\\0&1 \end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+0&1+1\\0+0&0+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$
$ \text{A}^3=\text{A}^2\text{A}$
$\Rightarrow\text{A}^3=\begin{bmatrix}1&2\\0&1 \end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}1+0&1+2\\0+0&0+1 \end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}1&3\\0&1 \end{bmatrix}$
Hence proved.
View full question & answer→Question 445 Marks
If $\text{A}^\text{T}=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix},$ find $A^T - B^T$.
AnswerGiven: $\text{A}^\text{T}=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}$
Now,
$\text{A}^\text{T}-\text{B}^\text{T}=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}-\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}$
$=\begin{bmatrix}3+1&4-1\\-1-2&2-2\\0-1&1-3\end{bmatrix}$
$=\begin{bmatrix}4&3\\-3&0\\-1&-2\end{bmatrix}$
Therefore,
$\text{A}^\text{T}-\text{B}^\text{T}=\begin{bmatrix}4&3\\-3&0\\-1&-2\end{bmatrix}$
View full question & answer→Question 455 Marks
If $f(x) = x^2 - 2x$, find f(A), where $\text{A}=\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
AnswerGiven: $f(x) = x^2 - 2x$
$f(A) = A^2 - 2A$
Now,
$\text{A}^2=\text{AA}$
$ \Rightarrow\text{A}^2=\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}0+4+0&0+5+4&0+0+6\\0+20+0&4+25+0&8+0+0\\0+8+0&0+10+6&0+0+9\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4&9&6\\20&29&8\\8&16&9\end{bmatrix}$
$\text{f(A)}=\text{A}^2-2\text{A}$
$\Rightarrow\text{f(A)}=\begin{bmatrix}4&9&6\\20&29&8\\8&16&9\end{bmatrix}-2\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
$ \Rightarrow\text{f(A)}=\begin{bmatrix}4&9&6\\20&29&8\\8&16&9\end{bmatrix}-\begin{bmatrix}0&2&4\\8&10&0\\0&4&6\end{bmatrix}$
$ \Rightarrow\text{f(A)}=\begin{bmatrix}4-0&9-2&6-4\\20-8&29-10&8-0\\8-0&16-4&9-6\end{bmatrix}$
$\Rightarrow\text{f(A)}=\begin{bmatrix}4&7&2\\12&19&8\\8&12&3\end{bmatrix}$
View full question & answer→Question 465 Marks
For the matrices A and B, verify that $(AB)^T = B^TA^T$, where $\text{A}=\begin{bmatrix}1&3\\2&4\end{bmatrix},\text{B}=\begin{bmatrix}1&4\\2&5\end{bmatrix}$
AnswerGiven: $\text{A}=\begin{bmatrix}1&3\\2&4\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}1&2\\3&4\end{bmatrix}$
$\text{B}=\begin{bmatrix}1&4\\2&5\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}1&2\\4&5\end{bmatrix}$
Now,
$\text{AB}=\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}1&4\\2&5\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}1+6&4+15\\2+8&8+20\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}7&19\\10&28\end{bmatrix}$
$\Rightarrow(\text{AB})^\text{T}=\begin{bmatrix}7&10\\19&28\end{bmatrix}\ \dots(1)$
Also,
$\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}1&2\\4&5\end{bmatrix}\begin{bmatrix}1&2\\3&4\end{bmatrix}$
$\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}1+6&2+8\\4+15&8+20\end{bmatrix}$
$\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}7&10\\19&28\end{bmatrix}\ \dots(2)$
$\therefore\ (\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$ [From eqs. (1) and (2)]
View full question & answer→Question 475 Marks
Solve the matrix equations:
$\begin{bmatrix}\text{x}&1\end{bmatrix}\begin{bmatrix}1&0\\-2&-3\end{bmatrix}\begin{bmatrix}\text{x}\\5\end{bmatrix}=0$
AnswerHere,
$\begin{bmatrix}\text{x}&1\end{bmatrix}\begin{bmatrix}1&0\\-2&-3\end{bmatrix}\begin{bmatrix}\text{x}\\5\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}\text{x}-2&0-3\end{bmatrix}\begin{bmatrix}\text{x}\\5\end{bmatrix}=0$
$ \Rightarrow\begin{bmatrix}(\text{x}-2)\text{x}-15\end{bmatrix} =0$
$\Rightarrow\text{x}^2-2\text{x}-15=0$
$ \Rightarrow\text{x}^2-5\text{x}+3\text{x}-15=0$
$ \Rightarrow\text{x}(\text{x}-5)+3(\text{x}-5)=0$
$\Rightarrow(\text{x}-5)(\text{x}+3)=0$
$ \Rightarrow\text{x}-5=0\ \text{or}\ \text{x}+3=0$
$ \Rightarrow\text{x}=5\ \text{or}\ \text{x}=-3$
So,
$\text{x}=5\text{ or }-3$
View full question & answer→Question 485 Marks
Let $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix},$ Find $A^T, B^T$ and verify that.$(\text{A}\text{B})^\text{T}=\text{B}^\text{T}+\text{A}^\text{T}$
AnswerGiven: $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}$ $\text{A}^\text{T}=\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}$ and $\text{B}^\text{T}=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}$ $\text{AB}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}$
$\Rightarrow\ \text{AB}=\begin{bmatrix}1-2+0&2-1+0&3-3+0\\2+2+0&4+1+3&6+3+3\\1+4+0&2+2+1&3+6+1\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}-1&1&0\\4&8&12\\5&5&10\end{bmatrix}$
$\Rightarrow(\text{AB})^\text{T}=\begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}\ \dots(1)$
Now, $\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}$
$\Rightarrow\text{B}^\text{A}\text{A}^\text{T}=\begin{bmatrix}1-2+0&2+2+0&1+4+0\\2-1+0&4+1+3&2+2+1\\3-3+0&6+3+3&3+6+1\end{bmatrix}$ $\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}\ \dots(2)$
$\Rightarrow(\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$ [from eqs. (1) and (2)]
View full question & answer→Question 495 Marks
Without using the concept of inverse of a matrix, find the matrix $\begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{u}\end{bmatrix}$ such that $\begin{bmatrix}5&-7\\-2&3\end{bmatrix}\begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{u}\end{bmatrix}=\begin{bmatrix}-16&-6\\7&2\end{bmatrix}$
AnswerGiven: $\begin{bmatrix}5&-7\\-2&3\end{bmatrix}\begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{u}\end{bmatrix}=\begin{bmatrix}-16&-6\\7&2\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}5\text{x}-7\text{z}&5\text{y}-7\text{u}\\-2\text{x}+3\text{z}&-2\text{y}+3\text{u}\end{bmatrix}=\begin{bmatrix}-16&-6\\7&2\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$ \therefore\ 5\text{x}-7\text{z}=-16\ \dots(1)$
$5\text{y}-7\text{u}=-6\ \dots(2)$
$-2\text{y}+3\text{u}=2$
$ \Rightarrow3\text{u}=2+2\text{y}$
$\Rightarrow\text{u}=\frac{2+2\text{y}}{3}\ \dots(3)$
$-2\text{x}+3\text{z}=7$
$\Rightarrow3\text{z}=7+2\text{x}$
$ \Rightarrow\text{z}=\frac{7+2\text{x}}{3}\ \dots(4)$
Putting the value of z in eq. (1), we get
$5\text{x}-7\Big(\frac{7+2\text{x}}{3}\Big)=-16$
$\Rightarrow5\text{x}-\frac{49+14\text{x}}{3}=-16$
$ \Rightarrow\frac{15\text{x}-49+14\text{x}}{3}=-16$
$\Rightarrow\text{x}-49=-48$
$\Rightarrow\text{x}=-48+49$
$\therefore\ \text{x}=1$
Putting the value of x in eq. (4), we get
$\text{z}=\frac{7+2(1)}{3}$
$\text{z}=\frac{9}{3}=3$
Putting the value of u in eq. (2), we get
$5\text{y}-7\Big(\frac{2+2\text{y}}{3}\Big)=-6$
$\Rightarrow5\text{y}-\frac{14+14\text{y}}{3}=-6$
$\Rightarrow\frac{15\text{y}-14+14\text{y}}{3}=-6$
$ \Rightarrow\text{y}-14=-18$
$\Rightarrow\text{y}=-18+14$
$ \Rightarrow\text{y}=-4$
Putting the value of y in eq. (3), we get
$ \text{u}=\frac{2+2(-4)}{3}$
$ \Rightarrow\text{u}=-2$
$\therefore\ \begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{u}\end{bmatrix}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix}$
View full question & answer→Question 505 Marks
If $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then find $\lambda,\mu$ so that $\text{A}^2=\lambda\text{A}+\mu\text{I}$
AnswerGiven: $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4+3&6+6\\2+2&3+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}7&12\\4&7\end{bmatrix}$
$ \text{A}^2=\lambda\text{A}=\mu\text{I}$
$\Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\lambda\begin{bmatrix}2&3\\1&2\end{bmatrix}+\mu\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\begin{bmatrix}2\lambda&3\lambda\\1\lambda&2\lambda\end{bmatrix}+\begin{bmatrix}\mu&0\\0&\mu\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\begin{bmatrix}2\lambda+\mu&3\lambda+0\\\lambda+0&2\lambda+\mu\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\begin{bmatrix}2\lambda+\mu&3\lambda\\\lambda&2\lambda+\mu\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$ \therefore\ 7=2\lambda+\mu\ \dots(1)$
$ 12=3\lambda$
$\Rightarrow\lambda=\frac{12}{3}=4$
Putting the value of $\lambda$ in eq. (1), we get
$7=2(4)+\mu$
$\Rightarrow7-8=\mu$
$\therefore\ \mu=-1$
View full question & answer→