Question 13 Marks
If $A=\left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{array}\right]$, then reduce it to $I_3$ by using row transformations.
Answer$|A|=\left|\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{array}\right|$
$= 2 (0 – 1) – 1(1 – 1) + 3 (1 – 0)$
$= -2 – 0 + 3 = 1 ≠ 0$
$∴ A$ is a non-singular matrix.
Hence, the required transformation is possible.Now, $A=\left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{array}\right]$
By $R_1 – R_2,$ we get,
$A \sim\left(\begin{array}{lll}1 & 1 & 2 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{array}\right)$
By $R_2-R_1$ and $R_3-R_1$, we get,
$A \sim\left(\begin{array}{rrr}1 & 1 & 2 \\0 & -1 & -1 \\0 & 0 & -1\end{array}\right)$
By $(-1) R_2$ and $(-1) R_3$, we get,
$A \sim\left(\begin{array}{lll}1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right)$
By $R_1-R_2$, we get,
$A \sim\left(\begin{array}{lll}1 & 0 & 1 \\0 & 1 & 1 \\0 & 0 & 1\end{array}\right)$
By $R_1-R_3$ and $B y R_2-R_3$, we get
$A \sim\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]=I_3\text {. }$
View full question & answer→Question 23 Marks
If $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1\end{array}\right]$ then reduce it to $I_3$ by using column transformations.
Answer$|A|=\left|\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1\end{array}\right| \\ =1(1-0)-0+0=1 \neq 0$
$∴ A$ is a non$-$singular matrix.
Hence, the required transformation is possible.
Now, $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1\end{array}\right]$
By $C_1-2 C_2$, we get, $A \sim\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ -3 & 3 & 1\end{array}\right]$
By $C_1 + 3C_3 $ and $C_2 – 3C_3, $ we get,
$A \sim\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I_3$.
View full question & answer→Question 33 Marks
Find the inverse of each of the following matrices (if they exist) : $\left[\begin{array}{lll}1 & 2 & -2 \\ 0 & -2 & 1 \\ -1 & 3 & 0\end{array}\right]$
AnswerLet $A=\left[\begin{array}{lll}1 & 2 & -2 \\ 0 & -2 & 1 \\ -1 & 3 & 0\end{array}\right]$
$\therefore A^{-1}=\left[\begin{array}{lll}1 & 2 & -2 \\ 0 & -2 & 1 \\ -1 & 3 & 0\end{array}\right]$
$=1\left|\begin{array}{ll}1 \\ 3 & 0\end{array}\right|-2\left|\begin{array}{ll}0 & 1 \\ -1 & 1\end{array}\right|-2\left|\begin{array}{ll}0 & -2 \\ -1 & 3\end{array}\right|$
$|A| = 1(0 – 3) – 2(0 + 1) – 2(0 – 2)$
$= -3 – 2 + 4$
$= -1 \neq 0$
$\therefore A^{-1}$ exists.
We have
$AA^{-1} = I$
$\left[\begin{array}{ccc}1 & 2 & -2 \\ 0 & -2 & 1 \\ -1 & 3 & 0\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$R_3 \rightarrow R_3 +R_1$
$\left[\begin{array}{ccc}1 & 2 & -2 \\ 0 & -2 & 1 \\ 0 & 5 & -2\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]$
$ R_3 \rightarrow R_3+2 R_2$
$\left[\begin{array}{ccc}1 & 2 & -2 \\ 0 & -2 & 1 \\ 0 & 1 & -0\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 2 & 1\end{array}\right]$
$R_2 \leftrightarrow R_3$
$\left[\begin{array}{ccc}1 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & -2 & 1\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 0\end{array}\right]$
$R_1 \rightarrow R_1-2 R_2 \quad R 3 \rightarrow R_3+2 R_2$
$\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ccc}-1 & -4 & -2 \\ 1 & 2 & 1 \\ 2 & 5 & 2\end{array}\right]$
$R_1 \rightarrow R_1+2 R_3$
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{lll}3 & 6 & 2 \\ 1 & 2 & 1 \\ 2 & 5 & 2\end{array}\right] \\ \therefore A^{-1}=\left[\begin{array}{lll}3 & 6 & 2 \\ 1 & 2 & 1 \\ 2 & 5 & 2\end{array}\right]$
View full question & answer→Question 43 Marks
Find the inverse of each of the following matrices $($if they exist$)$ : $\left[\begin{array}{lll}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$
AnswerLet $A=\left[\begin{array}{lll}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$
$ \therefore|A|=\left|\begin{array}{lll}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right|= 2(3 – 0) – 0 – 1(5 – 0)$
$= 6 – 0 – 5 = 1 \neq 0$
$\therefore A^{-1}$ exists.
Consider $AA^{-1} = I$
$\therefore\left[\begin{array}{rrr}2 & 0 & -1 \\5 & 1 & 0 \\0 & 1 & 3\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]$
By $R_1 \leftrightarrow R_2$, we get,
$\left[\begin{array}{rrr}5 & 1 & 0 \\2 & 0 & -1 \\0 & 1 & 3\end{array}\right] A^{-1}=\left[\begin{array}{lll}0 & 1 & 0 \\1 & 0 & 0 \\0 & 0 & 1\end{array}\right]$
By $R_1-2 R_2$, we get,
$\left[\begin{array}{rrr}1 & 1 & 2 \\2 & 0 & -1 \\0 & 1 & 3\end{array}\right] A^{-1}=\left[\begin{array}{rrr}-2 & 1 & 0 \\1 & 0 & 0 \\0 & 0 & 1\end{array}\right]$
By $R_2-2 R_1$, we get,
$\left[\begin{array}{rrr}1 & 1 & 2 \\0 & -2 & -5 \\0 & 1 & 3\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{rrr}-2 & 1 & 0 \\5 & -2 & 0 \\0 & 0 & 1\end{array}\right]$
By $R_2+3 R_3$, we get,
$\left[\begin{array}{lll}1 & 1 & 2 \\0 & 1 & 4 \\0 & 1 & 3\end{array}\right] A^{-1}=\left[\begin{array}{rrr}-2 & 1 & 0 \\5 & -2 & 3 \\0 & 0 & 1\end{array}\right]$
By $R_1-R_2$ and $R_3-R_2$, we get,
$\left[\begin{array}{rrr}1 & 0 & -2 \\0 & 1 & 4 \\0 & 0 & -1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}-7 & 3 & -3 \\5 & -2 & 3 \\-5 & 2 & -2\end{array}\right]$
By $(-1) R_3$, we get,
$\left[\begin{array}{rrr}1 & 0 & -2 \\0 & 1 & 4 \\0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}-7 & 3 & -3 \\5 & -2 & 3 \\5 & -2 & 2\end{array}\right]$
By $R_1+2 R_3$ and $R_2-4 R_3$, we get,
$\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}3 & -1 & 1 \\-15 & 6 & -5 \\5 & -2 & 2\end{array}\right]$
$\therefore A^{-1}=\left[\begin{array}{lll}3 & -1 & 1 \\-15 & 6 & -5 \\5 & -2 & 2\end{array}\right]$
View full question & answer→Question 53 Marks
Find the inverse of each of the following matrices (if they exist) : $\left[\begin{array}{lll}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right]$
Answer$\begin{array}{l}\text { Let } A=\left[\begin{array}{lll}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right] \\ \therefore|A|=\left|\begin{array}{lll}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right|\end{array}$$= 1(0 + 25) + 3(0 + 10) + 2(-15 – 0)$
$= 25 + 30 -30$
$= 25 \neq 0$
$\therefore A^{-1}$ exists.
Consider $AA^{-1} = I$
$\therefore\left[\begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\begin{array}{l}\text { By } R_2 \rightarrow R_2+3 R_1 \\ {\left[\begin{array}{ccc}1 & 3 & -2 \\ 0 & 9 & -11 \\ 2 & 5 & 0\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]}\end{array}$
$\begin{array}{l}\text { By } R_3 \rightarrow R_3-2 R_1 \\ {\left[\begin{array}{ccc}1 & 3 & -2 \\ 0 & 9 & -11 \\ 0 & -1 & 4\end{array}\right] A^{-1}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1\end{array}\right]}\end{array}$
$\begin{array}{l}\text { By } R \rightarrow \frac{1}{9} R_2 \\ {\left[\begin{array}{ccc}1 & 3 & -2 \\ 0 & 1 & -\frac{11}{9} \\ 0 & -1 & 4\end{array}\right] A^{-1}=\left[\begin{array}{ccc}1 & 0 & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ -2 & 0 & 1\end{array}\right]}\end{array}$
$\begin{array}{l}\text { By } R_3 \rightarrow R_3+R_2 \\ {\left[\begin{array}{ccc}1 & 3 & -2 \\ 0 & 1 & -\frac{11}{9} \\ 0 & 0 & \frac{25}{9}\end{array}\right] A^{-1}=\left[\begin{array}{ccc}1 & 0 & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ -\frac{5}{3} & \frac{1}{9} & 1\end{array}\right]}\end{array}$
$\begin{array}{l}\text { By } R_1 \rightarrow R_1+3 R_2 \\ {\left[\begin{array}{ccc}1 & 0 & \frac{5}{3} \\ 0 & 1 & -\frac{11}{9} \\ 0 & 0 & \frac{25}{9}\end{array}\right] A^{-1}=\left[\begin{array}{ccc}0 & -\frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ -\frac{5}{3} & \frac{1}{9} & 1\end{array}\right]}\end{array}$
$\text { By } R_3 \rightarrow \frac{9}{25} R_3 \\ {\left[\begin{array}{ccc}1 & 0 & \frac{5}{3} \\ 0 & 1 & -\frac{11}{9} \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ccc}0 & -\frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25}\end{array}\right]} $
$ \text { By } R_2 \rightarrow R_2+\frac{11}{9} R_3 \\ {\left[\begin{array}{ccc}1 & 0 & \frac{5}{3} \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{ccc}0 & -\frac{1}{3} & 0 \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25}\end{array}\right]} $
$ \text { By } R_1 \rightarrow R_1-\frac{5}{3} R_3 \\ {\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]
A^{-1}=\left[\begin{array}{ccc}\frac{5}{5} & -\frac{6}{15} & -\frac{3}{5} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25}\end{array}\right]} $
$ A^{-1}=\left[\begin{array}{ccc}1 & -\frac{2}{5} & -\frac{3}{5} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25}\end{array}\right] $
$ A^{-1}=\frac{1}{25}\left[\begin{array}{ccc}25 & -10 & -15 \\ -10 & 4 & 11 \\ -15 & 1 & 9\end{array}\right] $
View full question & answer→Question 63 Marks
Find the inverse of each of the following matrices $($if they exist$) : \left[\begin{array}{lll}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$
Answer$\begin{array}{l}\text { Let } A=\left[\begin{array}{lll}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right] \\ \therefore|A|=\left|\begin{array}{lll}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right|\end{array}$
$= 2(4 + 6) +3(4 – 9) + 3(-4 – 6)$
$= 20 – 15 – 30 $
$= -25 ≠ 0$
$\therefore A^{-1} $ exists.
Consider $AA^{-1} = I$
$\therefore\left(\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right) A^{-1}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$
By $R_1 \leftrightarrow R_3$, we get,
$\left[\begin{array}{rrr}3 & -2 & 2 \\2 & 2 & 3 \\2 & -3 & 3\end{array}\right] A^{-1}=\left[\begin{array}{lll}0 & 0 & 1 \\0 & 1 & 0 \\1 & 0 & 0\end{array}\right]$
By $R_1-R_2$, we get,
$\left[\begin{array}{rrr}1 & -4 & -1 \\2 & 2 & 3 \\2 & -3 & 3\end{array}\right] A^{-1}=\left[\begin{array}{rrr}0 & -1 & 1 \\0 & 1 & 0 \\1 & 0 & 0\end{array}\right]$
By $R_2-2 R_1$ and $R_3-2 R_1$, we get,
$\left[\begin{array}{rrr}1 & -4 & -1 \\0 & 10 & 5 \\0 & 5 & 5\end{array}\right] A^{-1}=\left[\begin{array}{rrr}0 & -1 & 1 \\0 & 3 & -2 \\1 & 2 & -2\end{array}\right]$
By $\left(\frac{1}{10}\right) R_2$, we get,
$\left[\begin{array}{rrr}1 & -4 & -1 \\0 & 1 & \frac{1}{2} \\0 & 5 & 5\end{array}\right] A^{-1}=\left(\begin{array}{rrr}0 & -1 & 1 \\0 & \frac{3}{10} & -\frac{1}{5} \\1 & 2 & -2\end{array}\right)$
By $R_1+4 R_2$ and $R_3-5 R_2$, we get,
$\left[\begin{array}{lll}1 & 0 & 1 \\0 & 1 & \frac{1}{2} \\0 & 0 & \frac{5}{2}\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{rrr}0 & \frac{1}{5} & \frac{1}{5} \\0 & \frac{3}{10} & -\frac{1}{5} \\1 & \frac{1}{2} & -1\end{array}\right]$
By $\left(\frac{2}{5}\right) R_3$, we get,
$\left[\begin{array}{lll}1 & 0 & 1 \\0 & 1 & \frac{1}{2} \\0 & 0 & 1\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{rrr}0 & \frac{1}{5} & \frac{1}{5} \\0 & \frac{3}{10} & -\frac{1}{5} \\\frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right]$
By $R_1-R_3$ and $R_2-\frac{1}{2} R_3$, we get,
$\begin{array}{l}{\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{rrr}-\frac{2}{5} & 0 & \frac{3}{5} \\-\frac{1}{5} & \frac{1}{5} & 0 \\\frac{2}{5} & \frac{1}{5} & -\frac{2}{5}\end{array}\right]} \end{array} $
$\therefore A^{-1}=\frac{1}{5}\left[\begin{array}{rrr}-2 & 0 & 3 \\-1 & 1 & 0 \\2 & 1 & -2\end{array}\right]$
View full question & answer→Question 73 Marks
Find the inverse of each of the following matrices $($if they exist$) : \left[\begin{array}{cc}3 & -10 \\ 2 & -7\end{array}\right]$
Answer$\begin{array}{l}\text { Let } A=\left[\begin{array}{ll}3 & -10 \\ 2 & -7\end{array}\right] \end{array} $
$ \therefore|A|=\left|\begin{array}{cc}3 & -10 \\ 2 & -7\end{array}\right|=-21+20=-1 \neq 0$
$\therefore A^{-1}$ exists.
Consider $AA^{-1} = I$
$\therefore\left[\begin{array}{rr}3 & -10 \\ 2 & -7\end{array}\right] A^{-1}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
By $R_1-R_2$, we get,
$\left[\begin{array}{ll}1 & -3 \\2 & -7\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{rr}1 & -1 \\0 & 1\end{array}\right]$
By $R_2-2 R_1$, we get,
$\left[\begin{array}{ll}1 & -3 \\0 & -1\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 & -1 \\-2 & 3\end{array}\right]$
View full question & answer→Question 83 Marks
Find the inverse of each of the following matrices (if they exist) : $\left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right]$
AnswerLet $A=\left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right]$$\therefore|\mathrm{A}|=\left|\begin{array}{ll}2 & 1 \\7 & 4\end{array}\right|=8-7=1 \neq 0$
$\therefore A^{-1} $ exists.
Consider $AA^{-1} = I$
$\therefore\left[\begin{array}{ll}2 & 1 \\7 & 4\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
By $R_1 \rightarrow R_1-\frac{1}{7} R_2$ we get,
$\left[\begin{array}{ll}1 & \frac{3}{7} \\7 & 4\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{cc}1 & \frac{-1}{7} \\0 & 1\end{array}\right]$
By $R_2 \rightarrow R_2-7 R_1$ we get,
$\left[\begin{array}{ll}1 & \frac{3}{7} \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{cc}1 & \frac{-1}{7} \\-7 & 2\end{array}\right]$
$\begin{array}{l}\text { By } R_1 \rightarrow R_1-\frac{3}{7} R_2 \\ {\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{cc}4 & -1 \\ -7 & 2\end{array}\right]} \\ \therefore A^{-1}=\left[\begin{array}{cc}4 & -1 \\ -7 & 2\end{array}\right]\end{array}$
View full question & answer→Question 93 Marks
Find the inverse of each of the following matrices (if they exist) : $\left[\begin{array}{ll}2 & -3 \\ 5 & 7\end{array}\right]$
Answer$\begin{array}{l}\text { Let } A=\left[\begin{array}{ll}2 & -3 \\ 5 & 7\end{array}\right] \\ \therefore|A|=\left|\begin{array}{ll}2 & -3 \\ 5 & 7\end{array}\right|=14+15=29 \neq 0\end{array}$
$\therefore A^{-1}$ exists.
Consider $AA^{-1} = I$
$\therefore\left[\begin{array}{rr}2 & -3 \\5 & 7\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
By $3 R_1$, we get,
$\left[\begin{array}{rr}6 & -9 \\5 & 7\end{array}\right] \mathrm{A}^{-1}=\left(\begin{array}{ll}3 & 0 \\0 & 1\end{array}\right)$
By $R_1-R_2$, we get,
$\left[\begin{array}{rr}1 & -16 \\5 & 7\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{rr}3 & -1 \\0 & 1\end{array}\right]$
By $R_2-5 R_1$, we get,
$\left(\begin{array}{rr}1 & -16 \\0 & 87\end{array}\right) A^{-1}=\left[\begin{array}{rr}3 & -1 \\-15 & 6\end{array}\right]$
By $\left(\frac{1}{87}\right) R_2$, we get,
$\left[\begin{array}{rr}1 & -16 \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}3 & -1 \\-5 / 29 & 2 / 29\end{array}\right]$
By $R_1+16 R_2$, we get,
$\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}7 / 29 & 3 / 29 \\-5 / 29 & 2 / 29\end{array}\right]$
$\therefore A^{-1}=\frac{1}{29}\left[\begin{array}{rr}7 & 3 \\ -5 & 2\end{array}\right]$
View full question & answer→Question 103 Marks
Find the inverse of each of the following matrices (if they exist) : $\left[\begin{array}{ll}1 & 3 \\ 2 & 7\end{array}\right]$
AnswerLet $A=\left[\begin{array}{ll}1 & 3 \\ 2 & 7\end{array}\right]$$\therefore|A|=\left|\begin{array}{cc}1 & 3 \\ 2 & 7\end{array}\right|=7-6=1 \neq 0$
$\therefore A^{-1}$ exists.
Consider $AA^{-1} = I$
$\therefore\left[\begin{array}{ll}1 & 3 \\ 2 & 7\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
By $R_2-2 R_1$, we get,
$\left[\begin{array}{ll}1 & 3 \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{cc}1 & 0 \\-2 & 1\end{array}\right]$
By $R_1-3 R_2$ we get,
$\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{cc}7 & -3 \\-2 & 1\end{array}\right]$
$A^{-1}=\left[\begin{array}{cc}7 & -3 \\ -2 & 1\end{array}\right]$
View full question & answer→Question 113 Marks
Find the inverse of each of the following matrices $($if they exist$) : \left[\begin{array}{ll}2 & 1 \\ 1 & -1\end{array}\right]$
Answer$\begin{array}{l}\text { Let } A=\left[\begin{array}{ll}2 & 1 \\ 1 & -1\end{array}\right] \\ \therefore|A|=\left|\begin{array}{ll}2 & 1 \\ 1 & -1\end{array}\right|=-2-1=-3 \neq 0\end{array}$
$\therefore A^{-1}$ exists.
Consider $AA^{-1} = I$
$\therefore\left[\begin{array}{rr}2 & 1 \\1 & -1\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
By $R_1 \leftrightarrow R_2$, we get,
$\left[\begin{array}{rr}1 & -1 \\2 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ll}0 & 1 \\1 & 0\end{array}\right]$
By $R_2-2 R_1$, we get,
$\left[\begin{array}{rr}1 & -1 \\0 & 3\end{array}\right] A^{-1}=\left[\begin{array}{rr}0 & 1 \\1 & -2\end{array}\right]$
By $\left(\frac{1}{3}\right) \mathrm{R}_2$, we get,
$\left[\begin{array}{rr}1 & -1 \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}0 & 1 \\1 / 3 & -2 / 3\end{array}\right]$
By $R_1+R_2$, we get,
$\begin{aligned}& {\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{rr}1 / 3 & 1 / 3 \\1 / 3 & -2 / 3\end{array}\right] }\end{aligned} $
$\therefore A^{-1}=\frac{1}{3}\left[\begin{array}{rr}1 & 1 \\1 & -2\end{array}\right]$
View full question & answer→Question 123 Marks
Find the inverse of each of the following matrices $($if they exist$) : \left[\begin{array}{ll}1 & -1 \\ 2 & 3\end{array}\right]$
Answer$\begin{array}{l}\text { Let } A=\left[\begin{array}{ll}1 & -1 \\ 2 & 3\end{array}\right] \\ \therefore|A|=\left|\begin{array}{ll}1 & -1 \\ 2 & 3\end{array}\right|=3+2=5 \neq 0\end{array}$
$\therefore \mathrm{A}^{-1}$ exists.
Consider $\mathrm{AA}^{-1}=1$
$\therefore\left[\begin{array}{rr}1 & -1 \\2 & 3\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
By $R_2-2 R_1$, we get,
$\left[\begin{array}{rr}1 & -1 \\0 & 5\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 & 0 \\-2 & 1\end{array}\right]$
By $\left(\frac{1}{5}\right) R_2$, we get,
$\left(\begin{array}{rr}1 & -1 \\0 & 1\end{array}\right) A^{-1}=\left(\begin{array}{rr}1 & 0 \\-\frac{2}{5} & \frac{1}{5}\end{array}\right)$
By $R_1+R_2$, we get,
$\begin{array}{l}{\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}\frac{3}{5} & \frac{1}{5} \\2 & \frac{1}{5}\end{array}\right]} \end{array}$
$\therefore A^{-1}=\frac{1}{5}\left[\begin{array}{rr}3 & 1 \\-2 & 1\end{array}\right] $
View full question & answer→Question 133 Marks
The cost of $2$ books and $6$ note books is $Rs. 34$ and the cost of $3$ books and $4$ notebooks is
$Rs. 31.$
Using matrices, find the cost of one book and one note$-$book.
AnswerHence, using the above information we get the following equations
and $2 x+6 y=34$
$3 x+4 y=31$
The above equations can be expressed in the form
$\left[\begin{array}{ll}2 & 6 \\ 3 & 4\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}34 \\ 31\end{array}\right]$
i.e. $A X=B$
Now using $R_1 \rightarrow R_2-\frac{3}{2} R_1 \quad$ we get
$\left[\begin{array}{cc} 2 & 6 \\ 0 & -5 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 34 \\ -20\end{array}\right]$
As the above matrix $'A\ '$ is reduced to an upper triangular matrix, we can write the equations in their original form as $2 x+6 y=34$
and $-5 y=-20$
$\therefore y=4$
and $2 x=34-6 y=34-24$
$\therefore 2 x=10$
$\therefore x=5$
$\therefore$ the cost of a book is $Rs. 5$ and that of a note book is $Rs. 4 .$
View full question & answer→Question 143 Marks
Solve the equation $2x + 3y = 9$ and $y – x = –2$ using the method of reduction.
View full question & answer→Question 153 Marks
Find the adjoint of matrix $A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 3 & 1 & 2 \\ -1 & 1 & 2\end{array}\right]$
AnswerHere
$a _{11}=2 \therefore M _{11}=0$
$ \therefore A _{11}=(-1)^{1+1} M _{11}=0$
$a _{12}=0 \therefore M _{12}=8$
$ \therefore A _{12}=(-1)^{1+2} M _{12}=-8$
$a _{13}=-1 \therefore M _{13}=4$
$ \therefore A _{13}=(-1)^{1+3} M _{13}=4$
$a _{21}=3 \therefore M _{21}=1$
$ \therefore A _{21}=(-1)^{2+1} M _{21}=-1$
$a _{22}=1 \therefore M _{22}=3$
$ \therefore A _{21}=(-1)^{2+2} M _{22}=3$
$a _{23}=2 \therefore M _{23}=2$
$ \therefore A_{23}=(-1)^{2+3} M_{23}=-2$
$a _{31}=-1 \therefore M_{31}=1$
$ \therefore A_{31}=(-1)^{3+1} M_{31}=1$
$a _{32}=1 \therefore M_{32}=7$
$ \therefore A_{32}=(-1)^{3+2} M_{32}=-7$
$a_{33}=2 \therefore M_{33}=2$
$ \therefore A_{33}=(-1)^{3+3} M_{33}=2$
$\begin{array}{l}_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{52} & A_{33} \end{array}]$
$=\left[\begin{array}{ccc} 0 & -8 & 4 \\ -1 & 3 & -2 \\ 1 & -7 & 2 \end{array}\right]$
$\therefore \quad \text { adj } A=\left[A_{11}\right]_{303}^T=\left[\begin{array}{ccc} 0 & -1 & 1 \\ -8 & 3 & -7 \\ 4 & -2 & 2
\end{array}\right]$
We know that a determinant can be expanded with the help of any row.
For example, expansion by $2{ }^{\text {nd }}$ row $a_{21} A_{21}+a_{22} A_{22}+\ldots+a_{2 n} A_{2 n}=|A|$.
But if we multiply the row by a different row of cofactors, then the sum is zero.
For example, $a_{21} A_{31}+a_{22} A_{12}+\ldots . a_{2 n} A_{3 n}=0$
This helps us to prove that $A ^{-1}=\frac{\operatorname{adj} A }{| A |}$
$\therefore \text { A.adj } A =\left|\begin{array}{ccccc}
A \mid & 0 & 0 & \ldots & 0 \\ 0 & |A| & 0 & \ldots & 0 \\ \vdots & & & & \\ 0 & 0 & 0 & \ldots & \mid A
\end{array}\right|=|A| \cdot I$
$\therefore \quad A ^{-1}=\frac{\operatorname{adj} A }{| A |}$
Thus, if $A =\left[a_e\right]_{\operatorname{man}}$ is a non$-$singular square matrix then its inverse exists and it is given by $A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)$
Think why $A ^{-1}$ does not exist if $A$ is singular.
View full question & answer→Question 163 Marks
Find the inverse of $A=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$ by elementary column transformation.
AnswerAs $A ^{-1}$ is required by column transformations therefore we have to consider $A ^{-1} A =1$ and have to perform column transformations on $A$.
Consider
$\therefore \quad A ^{-1}\left[\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
Using $C _2 \rightarrow C _2-3 C _1$ and $C _3 \rightarrow C _3-3 C _1$
$\therefore \quad A^{-1}\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}
1 & -3 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] $
Use $c _1 \rightarrow C _1- C _2$
$\therefore \quad A ^{-1}\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc} 4 & -3 & -3 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
Use $C _1 \rightarrow C _1- C _3$
$\therefore \quad A ^{-1}\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]$
$\therefore \quad A ^{-1} I =\left[\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]$
$\therefore \quad A^{-1}=\left[\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]$
View full question & answer→Question 173 Marks
Find the inverse of $A=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]$ by using elementary row transformations.
Answerbave ba perfor: now tramfiemations de A.
Censider
i.e.
$
\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right] A^{-3}=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
$
Use $R _1 \leftrightarrow R _1$
$
\left[\begin{array}{lll}
1 & 1 & 2 \\
3 & 2 & 6 \\
2 & 2 & 5
\end{array}\right] A^{\prime}=\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right]
$
Une $R_2 \rightarrow R_2-3 R_1$ and $R_n \rightarrow R_1-2 R_1$
$
\therefore \quad\left[\begin{array}{ccc}
1 & 1 & 2 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{array}\right] A^{-}=\left[\begin{array}{ccc}
0 & 1 & 0 \\
1 & -3 & 0 \\
0 & -2 & 1
\end{array}\right]
$
Now we $R_2 \rightarrow-R_2$
$
\therefore \quad\left[\begin{array}{lll}
1 & 1 & 2 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \quad A^{\prime}=\left[\begin{array}{ccc}
0 & 1 & 0 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]
$
$Usc R _1 \rightarrow R _2- R _2$
$
\therefore \quad\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \quad A^{\prime}=\left[\begin{array}{ccc}
1 & -2 & 0 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]
$
$Uxt _{ x } R _1 \rightarrow R _1-2 R _1$
$
\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 1
\end{array}\right] \quad A ^{-1}=\left[\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]
$
$
\therefore A ^{\prime}=\left[\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right] \quad \therefore A^{+}=\left[\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]
$
CYou eas verify thet ${A A^{-1}}^2=1$,
View full question & answer→Question 183 Marks
Find the inverse of $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
AnswerAs $|A|=\left|\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right|=-2$
$\therefore \quad| A | \neq 0 \quad \therefore A ^{-1}$ exists.
Let $\quad AA ^{-1}= I ($Here we can use only row transformation$)$
Using $R_2 \rightarrow R_2-3 R_1$
$\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \quad A ^{-1}=\left[\begin{array}{ll}
1 & 0 \\ 0 & 1\end{array}\right]$ becomes
$\left[\begin{array}{cc} 1 & 2 \\ 0 & -2 \end{array}\right] A ^{-1}=\left[\begin{array}{cc} 1 & 0 \\ -3 & 1 \end{array}\right]$
Using $-\frac{1}{2} R_2$ we get
$\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right] A ^{-1}=\left[\begin{array}{cc} 1 & 0 \\ \frac{3}{2} & -\frac{1}{2}
\end{array}\right]$
Using $R_1 \rightarrow R_1-2 R_2$
We get $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \quad A ^{-1}=\left[\begin{array}{cc}-2 & 1 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right]$
$\therefore \quad A ^{-1}=\left[\begin{array}{cc}-2 & 1 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right] \quad ($Verify the answer.$)$
View full question & answer→Question 193 Marks
Solve the following equations by reduction method : 5x + 2y = 4, 7x + 3y = 5
Answer5x + 2y = 4 ………..(1)
7x + 3y = 5 …………(2)
Multiplying Eq. (1) with 7 and Eq. (2) with 5\begin{array}{c}
35 x+14 y=28 \\
35 x+15 y=25 \\
-\quad- \\
\hline-1 y=3 \\
y=-3
\end{array}
Put y = -3 into Eq. (1)
5x + 2y = 4
5x + 2(-3) = 4
5x – 6 = 4
5x = 4 + 6
5x = 10
$x=\frac{10}{5}$
x = 2
Hence, x = 2, y = -3 is the required solution.
View full question & answer→Question 203 Marks
Solve the following equations by reduction method $:3x – y = 1, 4x + y = 6$
AnswerThe given equations can be written in the matrix form as $:\left[\begin{array}{rr}3 & -1 \\ 4 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}1 \\ 6\end{array}\right]$
By $4 R_1$ and $3 R_2$, we get,
$\left[\begin{array}{rr}12 & -4 \\ 12 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{r}4 \\ 18\end{array}\right]$
By $R_2-R_1$, we get
$\left[\begin{array}{rr}12 & -4 \\ 0 & 7\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{r}4 \\ 14\end{array}\right] $
$\therefore \left[\begin{array}{r}12 x-4 y \\ 0+7 y\end{array}\right]=\left[\begin{array}{r}4 \\ 14\end{array}\right]$
By equality of matrices,
$12x – 4y = 4 … (1)$
$7y = 14 … (2)$
From $(2), y = 2$
Substituting $y = 2$ in $(1),$ we get,
$12x – 8 = 4$
$\therefore 12x = 12 $
$\therefore x = 1$
Hence, $x = 1, y = 2$ is the required solution.
View full question & answer→Question 213 Marks
Solve the following equations by reduction method $:x + 3y = 2, 3x + 5y = 4.$
AnswerThe given equations can be written in the matrix form as $:\left[\begin{array}{ll}1 & 3 \\ 3 & 5\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}2 \\ 4\end{array}\right]$
By $R_2 – 3R_1$, we get
$\left[\begin{array}{lr}1 & 3 \\ 0 & -4\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left(\begin{array}{r}2 \\ -2\end{array}\right)$
$\therefore\left[\begin{array}{l}x+3 \\ 0-4 y\end{array}\right]=\left[\begin{array}{r}2 \\ -2\end{array}\right]$
By equality of matrices,
$\begin{array}{l}x+3 y=2 \ldots(1) \\ -4 y=-2\end{array}$
From $(2),y=\frac{1}{2}$
Substituting $y=\frac{1}{2}$ in $(1),$ we get,
$x+\frac{3}{2}=2$
$\therefore x=2-\frac{3}{2}=\frac{1}{2}$
Hence, $x=\frac{1}{2}, y=\frac{1}{2}$ is the required solution.
View full question & answer→Question 223 Marks
Solve the following equations by reduction method $: 2x + y = 5, 3x + 5y = -3$
AnswerThe given equations can be written in the matrix form as $:\left[\begin{array}{ll}2 & 1 \\ 3 & 5\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{r}5 \\ -3\end{array}\right]$
By $2 \mathrm{R}_2$, we get,
$\left[\begin{array}{rr}2 & 1 \\ 6 & 10\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{r}5 \\ -6\end{array}\right]$
By $R_2-3 R_1$, we get,
$\left[\begin{array}{ll}2 & 1 \\ 0 & 7\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{r}5 \\ -21\end{array}\right]$
$\therefore\left[\begin{array}{l}2 x+y \\ 0+7 y\end{array}\right]=\left[\begin{array}{r}5 \\ -21\end{array}\right]$
By equality of matrices,
$2x + y = 5 …(1)$
$7y = -21 …(2)$
From $(2), y = -3$
Substituting $y = -3$ in $(1),$ we get,
$2x – 3 = 5$
$\therefore 2x = 8 $
$\therefore x = 4$
Hence, $x = 4, y = -3$ is the required solution.
View full question & answer→Question 233 Marks
Find the matrix of co$-$factors for the following matrices : $\left[\begin{array}{rrr}1 & 0 & 2 \\ -2 & 1 & 3 \\ 0 & 3 & -5\end{array}\right]$
AnswerLet $A=\left[\begin{array}{rrr}1 & 0 & 2 \\ -2 & 1 & 3 \\ 0 & 3 & -5\end{array}\right]$
Here, $a_{11}=1$,
$\begin{array}{l}A_{11}=\left|\begin{array}{cc}1 & 3 \\ 3 & -5\end{array}\right|=-14 \\ a_{12}=0 \text {, } \end{array}$
$ A_{12}=\left|\begin{array}{cc}-2 & 0 \\ 3 & -5\end{array}\right|=-10 \\ a_{13}=2 \text {, } $
$ A_{13}=\left|\begin{array}{cc}-2 & 1 \\ 0 & 3\end{array}\right|=-6 \\ a_{21}=-2 $
$ A_{21}=\left|\begin{array}{cc}0 & 2 \\ 3 & -5\end{array}\right|=6 \\ a_{22}=1 $
$ A_{22}=\left|\begin{array}{cc}1 & 0 \\ 2 & -5\end{array}\right|=-5 \\ a_{23}=3 $
$ A_{23}=\left|\begin{array}{ll}1 & 0 \\ 0 & 3\end{array}\right|=-3 \\ a_{31}=0 $
$ A_{31}=\left|\begin{array}{ll}0 & 2 \\ 1 & 3\end{array}\right|=-2 \\ a_{32}=3 $
$ A_{32}=\left|\begin{array}{cc}1 & 2 \\ -2 & 3\end{array}\right|=-7 \\$
$\begin{array}{l}a_{33}=-5 \\ A_{33}=\left|\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right|\end{array}$
$A_{11} = -14, A_{12} = -10, A_{13} = -6,$
$A_{21} = 6, A_{22} = -5, A_{23} = -3,$
$A_{31} = -2, A_{32} = -7, A_{33} = 1.$
$\therefore $ the co$-$factor matrix
$=\left[\begin{array}{lll}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right]=\left[\begin{array}{rrr}-14 & -10 & -6 \\ 6 & -5 & -3 \\ -2 & -7 & 1\end{array}\right]$
View full question & answer→Question 243 Marks
Find the matrix of co$-$factors for the following matrices $:\left[\begin{array}{rr}1 & 3 \\ 4 & -1\end{array}\right]$
AnswerLet $A=\left[\begin{array}{rr}1 & 3 \\ 4 & -1\end{array}\right]$
Here $,a_{11} = 1, M_{11} = -1$
$\therefore A_{11} = (-1)^{1+1}(-1) = -1$
$a_{12} = 3, M_{12} = 4$
$\therefore A_{12} = (-1)^{1+2}(4) = -4$
$a_{21} = 4, M_{21} = 3$
$\therefore A_{21} = (-1)^{2+1}(3) = -3$
$a_{22} = -1, M_{22} = 1$
$\therefore A_{22} = (-1)^{2+1}(1) = 1$
$\therefore$ the co$-$factor matrix $=\left[\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right]$
$=\left(\begin{array}{rr}-1 & -4 \\ -3 & 1\end{array}\right)$
View full question & answer→Question 253 Marks
Find the inverse of the following matrices$\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$
AnswerLet $ A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right] \\ \therefore|A|=\left|\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right|$
$=2(3-0)-0(15-0)-1(5-0)$
$=6-0-5=1 \neq 0 \text {. }$
$\therefore \mathrm{A}^{-1} \text { exists. }$
$\text { Consider } \mathrm{AA}^{-1}=\mathrm{I}$
$\therefore\left(\begin{array}{rrr}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right) A^{-1}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$
By $3 R_1$, we get,
$\left[\begin{array}{rrr}6 & 0 & -3 \\5 & 1 & 0 \\0 & 1 & 3\end{array}\right] \quad A^{-1}=\left(\begin{array}{lll}3 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right)$
By $R_1-R_2$, we get,
$\left[\begin{array}{rrr}1 & -1 & -3 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right] A^{-1}=\left[\begin{array}{rrr}3 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
By $R_2-5 R_1$, we get,
$\left[\begin{array}{rrr}1 & -1 & -3 \\ 0 & 6 & 15 \\ 0 & 1 & 3\end{array}\right] A^{-1}=\left[\begin{array}{rrr}3 & -1 & 0 \\ -15 & 6 & 0 \\ 0 & 0 & 1\end{array}\right]$
By $R_2-5 R_3$, we get,
$\left[\begin{array}{rrr}1 & -1 & -3 \\ 0 & 1 & 0 \\ 0 & 1 & 3\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{rrr}3 & -1 & 0 \\ -15 & 6 & -5 \\ 0 & 0 & 1\end{array}\right]$
By $R_1+R_2$ and $R_3-R_2$, we get,
$\left[\begin{array}{rrr}1 & 0 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 3\end{array}\right] A^{-1}=\left[\begin{array}{rrr}-12 & 5 & -5 \\ -15 & 6 & -5 \\ 15 & -6 & 6\end{array}\right]$
By $\left(\frac{1}{3}\right) R_3$, we get,
$\left[\begin{array}{rrr}1 & 0 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}-12 & 5 & -5 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$
By $R_1+3 R_3$, we get
${\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]} $
$\therefore A^{-1}=\left[\begin{array}{rrr}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right].$
View full question & answer→Question 263 Marks
Find the inverse of the following matrices$\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$
AnswerLet $A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right] \\ \therefore|\mathrm{A}|=\left|\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right| $
$ =0(2-3)-1(1-9)+2(1-6)$
$=0+8-10$
$=-2 \neq 0 \text {. } $
$\therefore \mathrm{A}^{-1}$ exists.
Consider $\mathrm{AA}^{-1}=\mathrm{I}$
$\therefore\left(\begin{array}{lll}0 & 1 & 2 \\1 & 2 & 3 \\3 & 1 & 1\end{array}\right) A^{-1}=\left(\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right)$
By $R_1 \leftrightarrow R_2$, we get,
$\left(\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1\end{array}\right) A^{-1}=\left(\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right)$
By $R_3-3 R_1$, we get,
$\left[\begin{array}{rrr}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8\end{array}\right] A^{-1}=\left[\begin{array}{rrr}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1\end{array}\right]$
By $R_1-2 R_2$ and $R_3+5 R_2$, we get,
$\left[\begin{array}{rrr}1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 2\end{array}\right] A^{-1}=\left[\begin{array}{rrr}-2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1\end{array}\right]$
By $\left(\frac{1}{2}\right) R_3$, we get,
$\left[\begin{array}{rrr}1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}-2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 / 2 & -3 / 2 & 1 / 2\end{array}\right]$
By $R_1+R_3$ and $R_2-2 R_3$, we get,
${\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}1 / 2 & -1 / 2 & 1 / 2 \\ -4 & 3 & -1 \\ 5 / 2 & -3 / 2 & 1 / 2\end{array}\right]} $
$ \therefore A^{-1}=\frac{1}{2}\left[\begin{array}{rrr}1 & -1 & 1 \\ -8 & 6 & -2 \\ 5 & -3 & 1\end{array}\right]$
View full question & answer→Question 273 Marks
Find the inverse of the following matrices$\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$
Answer$\begin{array}{l}\text { Let } A=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right] \\ \therefore|A|=\left|\begin{array}{rr}2 & -3 \\ -1 & 2\end{array}\right|=4-3=1 \neq 0\end{array}$
$\therefore \mathrm{A}^{-1} \text { exists. }$
Consider $\mathrm{AA}^{-1}=\mathrm{I}$
$\therefore\left(\begin{array}{rr}2 & -3 \\-1 & 2\end{array}\right) \mathrm{A}^{-1}=\left(\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right)$
By $R_1+R_2$, we get,
$\left[\begin{array}{rr}1 & -1 \\ -1 & 2\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$
By $R_2+R_1$, we get,
$\left[\begin{array}{rr}1 & -1 \\0 & 1\end{array}\right) \mathrm{A}^{-1}=\left(\begin{array}{ll}1 & 1 \\1 & 2\end{array}\right)$
By $R_1+R_2$, we get,
$\begin{array}{l}{\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \mathrm{A}^{-1}=\left(\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right)} \end{array} $
$ \therefore A^{-1}=\left(\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right)$
View full question & answer→Question 283 Marks
Find the inverse of the following matrices $\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]$
Answer$\begin{array}{l}\text { Let } A=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right] \\ \therefore|A|=\left|\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right|=-1-4=-5 \neq 0\end{array}$
$\therefore \mathrm{A}^{-1}$ exists.
Consider $\mathrm{AA}^{-1}=\mathrm{I}$
$\therefore\left[\begin{array}{rr}1 & 2 \\2 & -1\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
By $R_2-2 R_1$, we get,
$\left[\begin{array}{rr}1 & 2 \\0 & -5\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 & 0 \\-2 & 1\end{array}\right]$
By $\left(-\frac{1}{5}\right) R_2$, we get,
$\left[\begin{array}{ll}1 & 2 \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 & 0 \\2 / 5 & -1 / 5\end{array}\right]$
By $R_1-2 R_2$, we get,
$\begin{array}{l}{\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 / 5 & 2 / 5 \\ 2 / 5 & -1 / 5\end{array}\right]} \\ \therefore A^{-1}=\frac{1}{5}\left[\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right] .\end{array}$
The answer can be checked by finding the product $\mathrm{AA}^{-1}$
$\mathrm{AA}^{-1}=\left(\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right]\left[\begin{array}{rr}1 / 5 & 2 / 5 \\ 2 / 5 & -1 / 5\end{array}\right] $
$=\left[\begin{array}{ll}1\left(\frac{1}{5}\right)+2\left(\frac{2}{5}\right) & 1\left(\frac{2}{5}\right)+2\left(-\frac{1}{5}\right) \\ 2\left(\frac{1}{5}\right)-1\left(\frac{2}{5}\right) & 2\left(\frac{2}{5}\right)-1\left(-\frac{1}{5}\right)\end{array}\right] $
$ =\left[\begin{array}{lll}\frac{1}{5}+\frac{4}{5} & \frac{2}{5}-\frac{2}{5} \\ \frac{2}{5}-\frac{2}{5} & \frac{4}{5}+\frac{1}{5}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\mathrm{I} $
Hence, $\mathrm{A}^{-1}$ is the required answer.
View full question & answer→Question 293 Marks
Find the inverse of the following matrices $\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]$
Answer$\begin{array}{l}\text { Let } A=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right] \\ \therefore|A|=\left|\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right|=-1-4=-5 \neq 0\end{array}$
$\therefore \mathrm{A}^{-1} \text { exists. }$
Consider $\mathrm{AA}^{-1}=\mathrm{I}$
$\therefore\left[\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right] A^{-1}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
By $R_2-2 R_1$, we get,
$\left[\begin{array}{rr}1 & 2 \\0 & -5\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 & 0 \\-2 & 1\end{array}\right]$
By $\left(-\frac{1}{5}\right) R_2$, we get,
$\left[\begin{array}{ll}1 & 2 \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 & 0 \\2 / 5 & -1 / 5\end{array}\right]$
By $R_1-2 R_2$, we get,
$\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{rr}1 / 5 & 2 / 5 \\2 / 5 & -1 / 5\end{array}\right]$
The answer can be checked by finding the product
$\mathrm{AA}^{-1}=\left[\begin{array}{rr}1 & 2 \\ 2 & -1\end{array}\right]\left[\begin{array}{rr}1 / 5 & 2 / 5 \\ 2 / 5 & -1 / 5\end{array}\right] $
$ =\left[\begin{array}{rr}1\left(\frac{1}{5}\right)+2\left(\frac{2}{5}\right) & 1\left(\frac{2}{5}\right)+2\left(-\frac{1}{5}\right) \\ 2\left(\frac{1}{5}\right)-1\left(\frac{2}{5}\right) & 2\left(\frac{2}{5}\right)-1\left(-\frac{1}{5}\right)\end{array}\right] $
$ =\left[\begin{array}{ll}\frac{1}{5}+\frac{4}{5} & \frac{2}{5}-\frac{2}{5} \\ \frac{2}{5}-\frac{2}{5} & \frac{4}{5}+\frac{1}{5}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\mathrm{I} $
Hence, $\mathrm{A}^{-1}$ is the required answer.
View full question & answer→Question 303 Marks
Find the adjoint of the following matrices.: $\left[\begin{array}{ccc}1 & -1 & 2 \\ -2 & 3 & 5 \\ -2 & 0 & -1\end{array}\right]$
AnswerNow $, M_{11}=\left|\begin{array}{rr}3 & 5 \\ 0 & -1\end{array}\right|=-3-0=-3 $
$ \therefore A_{11}=(-1)^{1+1}(-3)=-3 \\ M_{12}=\left|\begin{array}{rr}-2 & 5 \\ -2 & -1\end{array}\right|=2+10=12 $
$\therefore A_{12}=(-1)^{1+2}(12)=-12 \\ M_{13}=\left|\begin{array}{ll}-2 & 3 \\ -2 & 0\end{array}\right|=0+6=6 $
$ \therefore \begin{array}{l} A_{13}=(-1)^{1+3}(6)=6\end{array}$
$ {M}_{21}=\left|\begin{array}{rr}-1 & 2 \\ 0 & -1\end{array}\right|=1-0=1 $
$ \therefore A_{21}=(-1)^{2+1}(1)=-1 \\ M_{22}=\left|\begin{array}{rr}1 & 2 \\ -2 & -1\end{array}\right|=-1+4=3 $
$ \therefore A_{22}=(-1)^{2+2}(3)=3 \\ M_{23}=\left|\begin{array}{rr}1 & -1 \\ -2 & 0\end{array}\right|=0-2=-2 $
$ \therefore A_{23}=(-1)^{2+3}(-2)=2 \\ M_{31}=\left|\begin{array}{rr}-1 & 2 \\ 3 & 5\end{array}\right|=-5-6=-11 $
$ \therefore A_{31}=(-1)^{3+1}(-11)=-11 \\ M_{32}=\left|\begin{array}{rr}1 & 2 \\ -2 & 5\end{array}\right|=5+4=9 $
$ \therefore A_{32}=(-1)^{3+2}(9)=-9 \\ M_{33}=\left|\begin{array}{rr}1 & -1 \\ -2 & 3\end{array}\right|=3-2=1 $
$ \therefore A_{33}=(-1)^{3+3}(1)=1 \text {. } \\$
$\mathrm{A}_{11}=-3, \mathrm{~A}_{12}=-12, \mathrm{~A}_{13}=6, $
$ \mathrm{~A}_{21}=-1, \mathrm{~A}_{22}=3, \mathrm{~A}_{23}=2, $
$ \mathrm{~A}_{31}=-11, \mathrm{~A}_{32}=-9, \mathrm{~A}_{33}=1 $
$\therefore \text { the co-factor matrix }=\left[\begin{array}{lll}\mathrm{A}_{11} & \mathrm{~A}_{12} & \mathrm{~A}_{15} \\ \mathrm{~A}_{21} & \mathrm{~A}_{22} & \mathrm{~A}_{23} \\ \mathrm{~A}_{31} & \mathrm{~A}_{32} & \mathrm{~A}_{33}\end{array}\right]$
$ =\left[\begin{array}{rrr}-3 & -12 & 6 \\ -1 & 3 & 2 \\ -11 & -9 & 1\end{array}\right] $
$ \therefore \operatorname{adj} A=\left[\begin{array}{rrr}-3 & -1 & -11 \\ -12 & 3 & -9 \\ 6 & 2 & 1\end{array}\right]$
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