3 Marks Question

Question 13 Marks

A trust fund has ₹ 35,000 is to be invested in two different types of bonds. The first bond pays $8 \%$ interest per annum which will be given to orphanage and second bond pays $10 \%$ interest per annum which will be given to an N.G.O. (Cancer Aid Society). Use matrix multiplication, determine how to divide ₹ 35,000 among two types of bonds if the trust fund obtains an annual total interest of ₹ 3,200.

Answer

Let investment in first type of bond be ₹ $x.$
$\therefore$ The investment in second type of bond = ₹ $(35,000 - x)$
$\therefore \quad\left[\begin{array}{ll}x & 35,000-x\end{array}\right]\left[\begin{array}{l}\frac{8}{100} \\ \frac{10}{100}\end{array}\right]=[3,200]$
or $\frac{8}{100} x+(35,000-x) \frac{10}{100}=3,200$
or $x=$ ₹ $15,000$
$\therefore$ Investment in first bond = ₹ $15,000$
and Investment in second bond
= ₹ $(35,000-15,000)$
= ₹ $20,000$

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Question 23 Marks

Show that the matrix $B^T A B$ is symmetric or skew-symmetric accordingly when $A$ is symmetric or skew-symmetric.

Answer

$\text{Case I}$: Let $A$ be a symmetric matrix. Then $A^T=A$.
$\text{Now,}\quad$ $\left(B^T A B\right)^T=B^T A^T\left(B^T\right)^T\quad\text{[By reversal law]}$
$=B^T A^T B \quad\left[\because(B^T\right)^T=B]$
$\text{or}\quad$ $\left(B^T A B\right)^T=B^T A B\quad$ $\left[\because A^T=A\right]$
$\therefore B^T A B$ is a symmetric matrix.
$\text{Case II}$: Let $A$ be a skew-symmetric matrix. Then, $A^T=-A.$
$\text{Now, }\quad\left(B^T A B\right)^T=B^T A^T\left(B^T\right)^T\quad\text{[By reversal law]}$
$\text{or}\quad\left(B^T A B\right)^T=B^T A^T B\quad$ $\left[\because(B^T\right)^T=Bt]$
$\text{or}\quad\left(B^T A B\right)^T=B^T(-A) B \quad\left[\because A^T=-A\right]$
$\text{or}\quad\left(B^T A B\right)^T=-B^T A B$
$\therefore B^{T}AB$ is a skew-symmetric matrix.

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Question 33 Marks

Find matrix $X$ so that
$X\left(\begin{array}{lll}1 & 2 & 3 \\4 & 5 & 6\end{array}\right)=\left(\begin{array}{lll}-7 & -8 & -9 \\2 & 4 & 6\end{array}\right)$

Answer

Let $X=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$
then, $\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\left(\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right)=\left(\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right)$
$\Rightarrow\left(\begin{array}{lll}a+4 b & 2 a+5 b & 3 a+6 b \\ c+4 d & 2 c+5 d & 3 c+6 d\end{array}\right)=\left(\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right)$
$\begin{array}{lll}a+4 b=-7 & 3 a+6 b=-9 & c+4 d=2 \\ 2 a+5 b=-8 & 3 c+6 d=6 & 2 c+5 d=4\end{array}$
Equating and solving to get $a=1,b=-2,c=2,d=0$
$X=\left(\begin{array}{cc}1 & -2 \\ 2 & 0\end{array}\right)$

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Question 43 Marks

If $A=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]$, find $A^2-5 A+4 I$ and hence find a matrix $X$ such that $A^2-5 A+4 I+X=0$.

Answer

Getting $A^2=A \cdot A$
$=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]$
$=\left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]$
$\therefore A^2-5 A+4 I=\left[\begin{array}{ccl}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]-5\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]$
$+4\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]+\left[\begin{array}{ccc}-10 & 0 & -5 \\ -10 & -5 & -15 \\ -5 & 5 & 0\end{array}\right]$$+\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$
$=\left[\begin{array}{ccc}-1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2\end{array}\right]$
$\therefore X=-\left(A^2-5 A+4 I\right)$
$\text{or}\quad X=\left[\begin{array}{ccc}1 & 1 & 3 \\ 1 & 3 & 10 \\ 5 & -4 & -2\end{array}\right]$

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Applied Maths 3 Marks Question

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