5 Marks Questions

Question 15 Marks

Find matrix $X$ if : $X\left(\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right)=\left(\begin{array}{lll}-7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9\end{array}\right)$

Answer

Clearly order of $X$ is $3 \times 2$
$\text{Let}\quad X=\left(\begin{array}{ll}a & b \\c & d \\e & f\end{array}\right)$
$\begin{array}{l}\text { So } \quad\left(\begin{array}{ll}a & b \\ c & d \\ e & f\end{array}\right)\left(\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right)=\left(\begin{array}{lll}-7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9\end{array}\right) \\ \left.\begin{array}{lll}a+4 b=-7 & c+4 d=2 & e+4 f=11 \\ 2 a+5 b=-8 & 2 c+5 d=4 & 2 e+5 f=10\end{array}\right\}\end{array}$
$3 a+6 b=-9,3 c+6 d=6,3 e+6 f=9$
Solving we get
$a=1, b=-2, c=2, d=0, e=-5, f=4$
Thus, $X=\left(\begin{array}{cc}1 & -2 \\ 2 & 0 \\ -5 & 4\end{array}\right)$

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Question 25 Marks

If $A=\left(\begin{array}{ccc}1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3\end{array}\right)$, then show that $A^3-4 A^2-3 A+$ $11 I=0$.

Answer

$\begin{aligned} A & =\left[\begin{array}{ccc}1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3\end{array}\right] \\ A^2 & =\left[\begin{array}{ccc}1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3\end{array}\right] \\ & =\left[\begin{array}{lll}9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9\end{array}\right]\end{aligned}$
$\begin{aligned} A^3 & =\left[\begin{array}{lll}9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9\end{array}\right]\left[\begin{array}{ccc}1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3\end{array}\right] \\ & =\left[\begin{array}{ccc}28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34\end{array}\right]\end{aligned}$
$\therefore A^{3}-4A^{2}-3A+11A$
$=\left[\begin{array}{ccc}28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34\end{array}\right]-4\left[\begin{array}{ccc}9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9\end{array}\right]-3\left[\begin{array}{ccc}1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3\end{array}\right]$$+\left[\begin{array}{ccc}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right]$
$=\left[\begin{array}{ccc}28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34\end{array}\right]-\left[\begin{array}{ccc}36 & 28 & 20 \\ 4 & 16 & 4 \\ 32 & 36 & 36\end{array}\right]-\left[\begin{array}{ccc}3 & 9 & 6 \\ 6 & 0 & -3 \\ 3 & 6 & 9\end{array}\right]$ $+\left[\begin{array}{ccc}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right]$
$=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
$=0$

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Question 35 Marks

Express the following matrix as a sum of a symmetric and skew symmetric matrices and verify your result :
$
\left[\begin{array}{lll}
3 & -2 & -4 \\
3 & -2 & -5 \\
-1 & 1 & 2
\end{array}\right]
$

Answer

We know that
$A= \frac{1}{2} (A+A')+ \frac{1}{2}(A-A')$
Here, $\frac {1}{2}(A+A')$ is symmetric matrix and $\frac {1}{2}(A-A')$ is skew symmetric matrix.
$\text{Now,}\quad A=\left[\begin{array}{lll}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]$
$\Rightarrow A^{\prime}=\left[\begin{array}{lll}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]$
$\therefore \quad \frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{lll}3+3 & -2+3 & -4-1 \\ 3-2 & -2-2 & -5+1 \\ -1-4 & 1-5 & 2+2\end{array}\right]$
$=\frac{1}{2}\left[\begin{array}{lll}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]$
which is symmetric.
$\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{lll}3-3 & -2-3 & -4+1 \\ 3+2 & -2+2 & -5-1 \\ -1+4 & 1+5 & 2-2\end{array}\right]$
$=\frac{1}{2}\left[\begin{array}{lll}0 & -5 & -3 \\ 5 & 0 & -6 \\ 3 & 6 & 0\end{array}\right]$
which is skew-symmetric.
$\therefore A=\frac{1}{2}\left[\begin{array}{lll}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]+\frac{1}{2}\left[\begin{array}{lll}0 & -5 & -3 \\ 5 & 0 & -6 \\ 3 & 6 & 0\end{array}\right]$
$\Rightarrow A=\left[\begin{array}{lll}3 & 1 / 2 & -5 / 2 \\ 1 / 2 & -2 & -2 \\ -5 / 2 & -2 & 2\end{array}\right]$ $+\left[\begin{array}{lll}0 & -5 / 2 & -3 / 2 \\ 5 / 2 & 0 & -3 \\ 3 / 2 & 3 & 0\end{array}\right]$
$=\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]$

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Question 45 Marks

Express the matrix $A=\left[\begin{array}{ccc}2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4\end{array}\right]$ as the sum of a symmetric and skew symmetric matrix.

Answer

$\begin{aligned} A & =\left[\begin{array}{ccc}2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4\end{array}\right] \\ \text { then } A^{\prime} & =\left[\begin{array}{ccc}2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4\end{array}\right]\end{aligned}$
$\begin{array}{c}\text{Let}\quad P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}4 & 11 & -5 \\11 & 6 & 3 \\-5 & 3 & 8\end{array}\right] \\=\left[\begin{array}{ccc}2 & \frac{11}{2} & -\frac{5}{2} \\\frac{11}{2} & 3 & \frac{3}{2} \\-\frac{5}{2} & \frac{3}{2} & 4\end{array}\right]\end{array}$
Since $P^{\prime}=P$
$\therefore P$ is a symmetric matrix.
$\text{Let }\quad Q=\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}0 & -3 & -7 \\ 3 & 0 & 7 \\ 7 & -7 & 0\end{array}\right]$
$=\left[\begin{array}{ccc}0 & -\frac{3}{2} & -\frac{7}{2} \\ \frac{3}{2} & 0 & \frac{7}{2} \\ \frac{7}{2} & -\frac{7}{2} & 0\end{array}\right]$
$Q^{\prime}=-\left[\begin{array}{ccc}0 & -\frac{3}{2} & -\frac{7}{2} \\ \frac{3}{2} & 0 & \frac{7}{2} \\ \frac{7}{2} & -\frac{7}{2} & 0\end{array}\right]=-Q$
Since $Q^{\prime}=-Q$
$\therefore Q$ is a skew symmetric matrix.
$\text{Also}\quad P+Q=\left[\begin{array}{ccc}2 & \frac{11}{2} & -\frac{5}{2} \\\frac{11}{2} & 3 & \frac{3}{2} \\-\frac{5}{2} & \frac{3}{2} & 4\end{array}\right]+\left[\begin{array}{ccc}0 & -\frac{3}{2} & -\frac{7}{2} \\\frac{3}{2} & 0 & \frac{7}{2} \\\frac{7}{2} & -\frac{7}{2} & 0\end{array}\right]$
$=\left[\begin{array}{ccc}2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4\end{array}\right]=A$

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Applied Maths 5 Marks Questions

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