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Maths Solve the Following Question.(4 Marks)

Matrics · Maharashtra Board STD 12 (MSBSHSE - Semi English Medium). 16 questions.

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Solve the Following Question.(4 Marks)

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16 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
$a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} = |A|$
Answer
$\begin{array}{l}\mathrm{A}_{11}=(-1)^{1+1} \mathrm{M}_{11}=\left|\begin{array}{ll}a_{22} & a_{23} \\ a_{32} & a_{33}\end{array}\right|\end{array}  $
$ \mathrm{A}_{12}=(-1)^{1+2} \mathrm{M}_{12}=-\left|\begin{array}{ll}a_{21} & a_{23} \\ a_{31} & a_{33}\end{array}\right|  $
$ \mathrm{A}_{13}=(-1)^{1+3} \mathrm{M}_{13}=\left|\begin{array}{ll}a_{21} & a_{22} \\ a_{31} & a_{32}\end{array}\right|$
$\begin{array}{l}\therefore a_{11} \mathrm{~A}_{11}+a_{12} \mathrm{~A}_{12}+a_{13} \mathrm{~A}_{13} \\ =a_{11}\left|\begin{array}{ll}a_{22} & a_{23} \\ a_{32} & a_{33}\end{array}\right|-a_{12}\left|\begin{array}{ll}a_{21} & a_{23} \\ a_{31} & a_{33}\end{array}\right|+a_{13}\left|\begin{array}{ll}a_{21} & a_{22} \\ a_{31} & a_{32}\end{array}\right|\end{array}$
$=\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|=|\mathrm{A}|$.
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Question 24 Marks
$a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0$
Answer
$\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{3 \times 3}=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$(i) $\mathrm{A}_{21}=(-1)^{2+1} \mathrm{M}_{21}=-\left|\begin{array}{ll}a_{12} & a_{13} \\ a_{32} & a_{33}\end{array}\right|$
$= -(a_{12}a_{33} – a_{13}a_{32})$
$= -a_{12}a_{33} + a_{13}a_{32}$
$\mathrm{A}_{22}=(-1)^{2+2} \mathrm{M}_{22}=\left|\begin{array}{ll}a_{11} & a_{13} \\ a_{31} & a_{33}\end{array}\right|$
= a11a33 – a13a31
$\mathrm{A}_{23}=(-1)^{2+3} \mathrm{M}_{23}=-\left|\begin{array}{ll}a_{11} & a_{12} \\ a_{31} & a_{32}\end{array}\right|$
$= -(a11a32 – a12a31)$
$= -a11a32+ a12a31$
$\therefore a11A21 + a12A22 + a13A23$
$= a11(-a1233 + a13a32) + a12(a11a33 – a13a31) + a13(-a11a32 + a12a31)$
$= -a11a12a33 + a11a13a32 + a11a12a33 – a12a13a31 – a11a13a32 + a12a13a31$
$= 0$
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Question 34 Marks
Solve the following equations by the method of reduction. $x + y + z = 1, 2x + 3y + 2z = 2$ and $x + y + 2z = 4.$
Answer
The above equation can be written in the form $AX = B$ as
${\left[\begin{array}{lll} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 1 & 1 & 2 \end{array}\right]\left[\begin{array}{l}
x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 4 \end{array}\right]} $
$\text { using } R _2 \rightarrow R _2- R _3 \quad \text { and } R _1 \rightarrow R _1-\frac{1}{2} R _3 $
$ {\left[\begin{array}{ccc} \frac{1}{2} & \frac{1}{2} & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} -1 \\ -2 \\
4\end{array}\right]} $
Now using $R _1 \rightarrow R _1-\frac{1}{4} R _2$ we get
$\left[\begin{array}{lll}\frac{1}{4} & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-\frac{1}{2} \\ -2 \\ 4\end{array}\right]$
Note that here we have reduced the original matrix $A$ to a lower triangular matrix.
Hence we can rewrite the equations in their original form as
$ \frac{x}{4}=-\frac{1}{2}$
$\text { (i) i.c. } x =-2x+2 y=-2 $
$\therefore 2 y=-2+2=0$
$\therefore y=0$ and  $x+y+2 z=4$
$\therefore 2 z=4+2+0$
$\therefore 2 z=6$
$\therefore z=3$
$\therefore x=-2, y=0, z=3 $ is the required solution. 
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Question 44 Marks
 Solve the following equations by the method of inversion $x – y + z = 4, 2x + y –3z = 0, x + y + z = 2.$
Answer
The required matrix equation is
$\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$
i.e. $AX = B$
Hence, by premultiplying the equation by $A ^{-1}$, we get,
i.e. $\left( A ^{-1} A \right) X = A ^{-1} B $
i.e. $ IX = A ^{-1} B $
i.e. $ X = A ^{-1} B$
Now as $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right],$ 
By definition $,\text{ adj } A=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$ and $|A|=10$
$A ^{-1}=\frac{1}{| A |}(\operatorname{adj} A )$
 i.e. $ =\frac{1}{10}\left[\begin{array}{ccc} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3
\end{array}\right] $
Hence using ${(i)}  =\frac{1}{10}\left[\begin{array}{ccc} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array}\right]\left[\begin{array}{l} 4 \\ 0 \\ 2 \end{array}\right]$
$X \quad=\frac{1}{10}\left[\begin{array}{c} 20 \\ -10 \\ 10 \end{array}\right]$
$ \therefore\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 2 \\ -1 \\ 1 \end{array}\right]$
Hence, by equality of matrices we get $x=2, y=-1$ and $z=1$
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Question 54 Marks
Solve the equations $2x + 5y = 1 $ and $3x + 2y = 7$ by the method of inversion
Answer
Using the given equations we get the corresponding matrix equation as
$ \left[\begin{array}{ll} 2 & 5 \\ 3 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 1 \\ 7 \end{array}\right]$
i.e. $  AX = B , $ 
where $ A =\left[\begin{array}{ll} 2 & 5 \\ 3 & 2 \end{array}\right], X =\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B =\left[\begin{array}{l} 1 \\ 7 \end{array}\right]$
Hence, premultiplying the above matrix equation by $A ^{-1}$, we get
$ \begin{array}{rll} \left( A ^{-1} A \right) X & = A ^{-1} B \\ \text { i.e. } & IX & = A ^{-1} B \\ \text { i.c. } & X & = A ^{-1} B \end{array}$
i.e. $X = A ^{-1} B$
Now as $A =\left[\begin{array}{ll}2 & 5 \\ 3 & 2\end{array}\right],| A |$
$=-11$ and $\operatorname{adj} A =\left[\begin{array}{cc}2 & -5 \\ -3 & 2\end{array}\right]$
$\therefore \quad A ^{-1}=\frac{1}{| A |}(\operatorname{adj} A )$
i.e. $A ^{-1}=\frac{1}{-11}\left[\begin{array}{cc}2 & -5 \\ -3 & 2\end{array}\right]$
Hence using $(i)$ we get
$ X =\frac{1}{-11}\left[\begin{array}{cc} 2 & -5 \\ -3 & 2 \end{array}\right]\left[\begin{array}{l}
1 \\ 7 \end{array}\right] $
$ X =\frac{1}{11}\left[\begin{array}{cc} -2 & 5 \\ 3 & -2 \end{array}\right]\left[\begin{array}{l} 1 \\ 7
\end{array}\right]$
i.c. $\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}33 \\ -11\end{array}\right]$
i.e. $\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}3 \\ -1\end{array}\right]$
Hence by equality of matrices we get $x=3$ and $y=-1$.
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Question 64 Marks
Solve the following equations by inversion method $:2x + 6y = 8, x + 3y = 5$
Answer
The given equations can be written in the matrix form as $:\left[\begin{array}{ll}2 & 6 \\ 1 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}8 \\ 5\end{array}\right]$
This is of the form $AX = B,$ where
$\mathrm{A}=\left[\begin{array}{ll}2 & 6 \\ 1 & 3\end{array}\right], \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}8 \\ 5\end{array}\right]$
Let us find $A^{-1}.$
$|A|=\left|\begin{array}{ll}2 & 6 \\ 1 & 3\end{array}\right|=6-6=0$
$\therefore A^{-1}$ does not exist.
Hence, $x$ and $y$ do not exist.
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Question 74 Marks
Solve the following equations by inversion method $:x + y = 4, 2x – y = 5$
Answer
$x + y = 4, 2x – y = 5$
The given equations can be written in the matrix form as $:\left[\begin{array}{cc}1 & 1 \\ 2 & -1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}4 \\ 5\end{array}\right]$
This is of the form $AX = B$
$\Rightarrow X$
$\Rightarrow A^{-1}B$
$\begin{array}{l}A=\left[\begin{array}{cc}1 & 1 \\ 2 & -1\end{array}\right] \end{array} $
$ |A|=-1-2=-3 \neq 0$
$\begin{array}{l}\operatorname{Adj} A=\left[\begin{array}{cc}-1 & -1 \\ -2 & 1\end{array}\right] \end{array} $
$ =\frac{1}{-3}\left[\begin{array}{cc}-1 & -1 \\ -2 & 1\end{array}\right]$
$=\left[\begin{array}{cc}1 & 1 \\ 2 & -1\end{array}\right]  $
$ X=A^{-1} B=\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y}\end{array}\right]=\left[\begin{array}{cc}1 & 1 \\ 2 & -1\end{array}\right]\left[\begin{array}{l}4 \\ 5\end{array}\right]  $
$ =\left[\begin{array}{cc}4 & 5 \\ 8 & -5\end{array}\right]$
$=\left[\begin{array}{l}9 \\ 3\end{array}\right]  $
$ =\left[\begin{array}{l}3 \\ 1\end{array}\right]$
By equality of matrices.
$x = 3, y = 1$
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Question 84 Marks
Solve the following equations by inversion method : $x + 2y = 2, 2x + 3y = 3$
Answer
The given equations can be written in the matrix form as :
$\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}2 \\ 3\end{array}\right]$
This is of the form $AX = B,$ where
$A=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right], \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}2 \\ 3\end{array}\right]$
Let us find $A^{-1}$.
$|A|=\left|\begin{array}{ll}1 & 2 \\2 & 3\end{array}\right|=3-4=-1 \neq 0$
$\therefore \mathrm{A}^{-1}$ exists.
Consider $\mathrm{AA}^{-1}=\mathrm{I}$
$\therefore\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
By $R_2-2 R_1$, we get,
$\left[\begin{array}{rr}1 & 2 \\0 & -1\end{array}\right] A^{-1}=\left(\begin{array}{rr}1 & 0 \\-2 & 1\end{array}\right)$
By $(-1) R_2$, we get,
$\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}1 & 0 \\ 2 & -1\end{array}\right]$
By $R_1-2 R_2$, we get,
$\begin{array}{l}\quad\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}-3 & 2 \\ 2 & -1\end{array}\right] \\ \therefore A^{-1}=\left[\begin{array}{rr}-3 & 2 \\ 2 & -1\end{array}\right]\end{array}$
Now, premultiply $AX = B$ by $A^{-1}$, we get,
$A^{-1}(AX) = A^{-1}B$
$\therefore (A^{-1}A)X = A^{-1}B$
$\therefore IX = A^{-1}B$
$\therefore \mathrm{X}=\left[\begin{array}{rr}-3 & 2 \\ 2 & -1\end{array}\right]\left[\begin{array}{l}2 \\ 3\end{array}\right] $
$ \therefore\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{r}-6+6 \\ 4-3\end{array}\right]=\left[\begin{array}{l}0 \\ 1\end{array}\right]$
By equality of matrices,
$x = 0, y = 1$ is the required solution.
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Question 94 Marks
Find the inverse of the following matrices by the adjoint method : $\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$
Answer
Let $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right] $
$ \therefore|A|=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$
$= 1(10 – 0) – 0 + 0 = 1(10) – 0 + 0$
$= 10 \neq 0$
$\therefore A^{-1}$ exists.
First we have to find the co $-$ factor matrix
$=\left[\mathrm{A}_{\mathrm{ij}}\right]_{3 \times 3^{\prime}} \text { where } \mathrm{A}_{\mathrm{ij}}=(-1)^{\mathrm{i}+\mathrm{j}} \mathrm{M}_{\mathrm{ij}} $
Now, $ \mathrm{A}_{11}=(-1)^{1+1} \mathrm{M}_{11}=\left|\begin{array}{ll}2 & 4 \\ 0 & 5\end{array}\right|=10-0=10 $
$ \mathrm{~A}_{12}=(-1)^{1+2} \mathrm{M}_{12}=-\left|\begin{array}{ll}0 & 4 \\ 0 & 5\end{array}\right|=-0-0=0 $
$\mathrm{~A}_{13}=(-1)^{1+3} \mathrm{M}_{13}=\left|\begin{array}{ll}0 & 2 \\ 0 & 0\end{array}\right|=0-0=0 $
$ \mathrm{~A}_{21}=(-1)^{2+1} \mathrm{M}_{21}=-\left|\begin{array}{ll}2 & 3 \\ 0 & 5\end{array}\right|=-10-0=-10 $
$ \mathrm{~A}_{22}=(-1)^{2+2} \mathrm{M}_{22}=\left|\begin{array}{ll}1 & 3 \\ 0 & 5\end{array}\right|=5-0=5 $
$ \mathrm{~A}_{23}=(-1)^{2+3} \mathrm{M}_{23}=-\left|\begin{array}{ll}1 & 2 \\ 0 & 0\end{array}\right|=-0-0=0 $
$ \mathrm{~A}_{31}=(-1)^{3+1} \mathrm{M}_{31}=\left|\begin{array}{ll}2 & 3 \\ 2 & 4\end{array}\right|=8-6=2 $
$ \mathrm{~A}_{32}=(-1)^{3+2} \mathrm{M}_{32}=-\left|\begin{array}{ll}1 & 3 \\ 0 & 4\end{array}\right|=-4-0=-4 $
$ \mathrm{~A}_{33}=(-1)^{3+3} \mathrm{M}_{33}=\left|\begin{array}{ll}1 & 2 \\ 0 & 2\end{array}\right|=2-0=2$
$\therefore$ the co $-$ factor matrix
$=\left[\begin{array}{lll}A_{11} & A_{12} & A_{13} \\A_{21} & A_{22} & A_{23} \\A_{31} & A_{32} & A_{33}\end{array}\right]=\left[\begin{array}{ccc}10 & 0 & 0 \\-10 & 5 & 0 \\2 & -4 & 2\end{array}\right]$
$\therefore \operatorname{adj} A=\left[\begin{array}{ccc}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right] $
$ \therefore A^{-1}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A}) \\ =\frac{1}{10}\left(\begin{array}{rrr}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right) $
$ \therefore A^{-1}=\frac{1}{10}\left(\begin{array}{rrr}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right)$
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Question 104 Marks
Find the inverse of the following matrices by the adjoint method : $\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$
Answer
Let $ A  =\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right] $
$ \therefore|A|  =\left|\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right| $
$ =1(-3-0)-0+0$
$ =-3 \neq 0$
$\therefore \mathrm{A}^{-1}$ exists.
First we have to find the co $-$ factor matrix
$=\left[\mathrm{A}_{i j}\right]_{3 \times 3}$, where $\mathrm{A}_{i j}=(-1)^{i+i} \mathrm{M}_{i j}$
Now $ \mathrm{A}_{11}=(-1)^{1+1} \mathrm{M}_{11}=\left|\begin{array}{lr}3 & 0 \\ 2 & -1\end{array}\right|=-3-0=-3$
$ \mathrm{~A}_{12}=(-1)^{1+2} \mathrm{M}_{12}=-\left|\begin{array}{rr}3 & 0 \\ 5 & -1\end{array}\right|=-(-3-0)=3 $
$ \mathrm{~A}_{13}=(-1)^{1+3} \mathrm{M}_{13}=\left|\begin{array}{ll}3 & 3 \\ 5 & 2\end{array}\right|=6-15=-9 $
$ \mathrm{~A}_{21}=(-1)^{2+1} \mathrm{M}_{21}=-\left|\begin{array}{lr}0 & 0 \\ 2 & -1\end{array}\right|=-(0-0)=0 $
$ \mathrm{~A}_{22}=(-1)^{2+2} \mathrm{M}_{22}=\left|\begin{array}{ll}1 & 0 \\ 5 & -1\end{array}\right|=-1-0=-1 $
$ \mathrm{~A}_{23}=(-1)^{2+3} \mathrm{M}_{23}=-\left|\begin{array}{ll}1 & 0 \\ 5 & 2\end{array}\right|=-(2-0)=-2 $
$ \mathrm{~A}_{31}=(-1)^{3+1} \mathrm{M}_{31}=\left|\begin{array}{ll}0 & 0 \\ 3 & 0\end{array}\right|=0-0=0 $
$ \mathrm{~A}_{32}=(-1)^{3+2} \mathrm{M}_{32}=-\left|\begin{array}{ll}1 & 0 \\ 3 & 0\end{array}\right|=-(0-0)=0 $
$ \mathrm{~A}_{33}=(-1)^{3+3} \mathrm{M}_{33}=\left|\begin{array}{ll}1 & 0 \\ 3 & 3\end{array}\right|=3-0=3$
$\therefore$ 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}-3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3\end{array}\right] $
$ \therefore \operatorname{adj} A=\left[\begin{array}{rrr}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right] $
$ \therefore A^{-1}=\frac{1}{|A|}(\text { adj } A) \\ =\frac{1}{-3}\left[\begin{array}{rrr}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right] $
$ \therefore A^{-1}=\frac{1}{3}\left[\begin{array}{rrr}3 & 0 & 0 \\ -3 & 1 & 0 \\ 9 & 2 & -3\end{array}\right]$
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Question 114 Marks
Find the inverse of the following matrices by the adjoint method : $\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$
Answer
Let $ A=\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right] \\ |A|==6+8=14 \neq 0 $
$ \therefore A^{-1} $ exist  First we have to find the co $-$ factor matrix
$= [A_{ij}] _{2\times 2}$ where $A_{ij} = (-1)^{i+j}M_{ij}$
Now $, A_{11} = (-1)^{1+1}M_{11} = 3$
$A_{12} = (-1)^{1+2}M = -4$
$A_{21} = (-2)^{2+1}M_{21} = (-2) = 2$
$A_{22} = (-1)^{2+2}M_{22} = 2$
Hence the co $-$ factor matrix
$=\left[\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right]=\left[\begin{array}{cc}3 & -4 \\ 2 & 2\end{array}\right] $
$ \therefore \operatorname {adj} A=\left[\begin{array}{cc}3 & 2 \\ -4 & 2\end{array}\right]$
$\therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{14}\left(\begin{array}{cc}3 & 2 \\ -4 & 2\end{array}\right)$
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Question 124 Marks
Find the inverse of the following matrices by the adjoint method : $\left[\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right]$
Answer
Let $A=\left[\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right]$$\therefore|A|=\left|\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right|=-2+15=13 \neq 0$
$\therefore A^{-1}$​​​​​​​ exists.
First we have to find the co-factor matrix
$= [A_{ij}]_{2\times 2}​​​​​​​$​​​​​​​, where $A_{ij} = (-1)^{i+j}M_{ij}$
$Now, A_{11} = (-1)^{1+1}M_{11} = 2$
$A_{12} = (-1)^{1+2}M_{12} = -(-3) = 3$
$A_{21} = (-1)^{2+1}M_{21} = -5$
$A_{22} = (-1)^{2+2}M_{22} = -1$
Hence, the co-factor matrix
$=\left[\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right]=\left[\begin{array}{cc}2 & 3 \\ -5 & -1\end{array}\right]$
$\therefore$ adj $A=\left[\begin{array}{ll}2 & -5 \\ 3 & -1\end{array}\right]$
$\therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{13}\left[\begin{array}{ll}2 & -5 \\ 3 & -1\end{array}\right]$
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Question 134 Marks
Find the inverse of the following matrices by the adjoint method : $\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$
Answer
 
Let $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right] $
$ \therefore|A|=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$
$= 1(10 – 0) – 0 + 0 = 1(10) – 0 + 0$
$= 10 \neq 0$
$\therefore A^{-1}$ exists.
First we have to find the co $-$ factor matrix
$=\left[\mathrm{A}_{\mathrm{ij}}\right]_{3 \times 3^{\prime}} \text { where } \mathrm{A}_{\mathrm{ij}}=(-1)^{\mathrm{i}+\mathrm{j}} \mathrm{M}_{\mathrm{ij}} $
Now, $ \mathrm{A}_{11}=(-1)^{1+1} \mathrm{M}_{11}=\left|\begin{array}{ll}2 & 4 \\ 0 & 5\end{array}\right|=10-0=10 $
$ \mathrm{~A}_{12}=(-1)^{1+2} \mathrm{M}_{12}=-\left|\begin{array}{ll}0 & 4 \\ 0 & 5\end{array}\right|=-0-0=0 $
$\mathrm{~A}_{13}=(-1)^{1+3} \mathrm{M}_{13}=\left|\begin{array}{ll}0 & 2 \\ 0 & 0\end{array}\right|=0-0=0 $
$ \mathrm{~A}_{21}=(-1)^{2+1} \mathrm{M}_{21}=-\left|\begin{array}{ll}2 & 3 \\ 0 & 5\end{array}\right|=-10-0=-10 $
$ \mathrm{~A}_{22}=(-1)^{2+2} \mathrm{M}_{22}=\left|\begin{array}{ll}1 & 3 \\ 0 & 5\end{array}\right|=5-0=5 $
$ \mathrm{~A}_{23}=(-1)^{2+3} \mathrm{M}_{23}=-\left|\begin{array}{ll}1 & 2 \\ 0 & 0\end{array}\right|=-0-0=0 $
$ \mathrm{~A}_{31}=(-1)^{3+1} \mathrm{M}_{31}=\left|\begin{array}{ll}2 & 3 \\ 2 & 4\end{array}\right|=8-6=2 $
$ \mathrm{~A}_{32}=(-1)^{3+2} \mathrm{M}_{32}=-\left|\begin{array}{ll}1 & 3 \\ 0 & 4\end{array}\right|=-4-0=-4 $
$ \mathrm{~A}_{33}=(-1)^{3+3} \mathrm{M}_{33}=\left|\begin{array}{ll}1 & 2 \\ 0 & 2\end{array}\right|=2-0=2$
$\therefore$ the co $-$ factor matrix
$=\left[\begin{array}{lll}A_{11} & A_{12} & A_{13} \\A_{21} & A_{22} & A_{23} \\A_{31} & A_{32} & A_{33}\end{array}\right]=\left[\begin{array}{ccc}10 & 0 & 0 \\-10 & 5 & 0 \\2 & -4 & 2\end{array}\right]$
$\therefore \operatorname{adj} A=\left[\begin{array}{ccc}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right] $
$ \therefore A^{-1}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A}) \\ =\frac{1}{10}\left(\begin{array}{rrr}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right) $
$ \therefore A^{-1}=\frac{1}{10}\left(\begin{array}{rrr}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right)$
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Question 144 Marks
Find the inverse of the following matrices by the adjoint method : $\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]$
Answer
 Let $ A  =\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right] $
$ \therefore|A|  =\left|\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right| $
$ =1(-3-0)-0+0$
$ =-3 \neq 0$$\therefore \mathrm{A}^{-1}$ exists.
First we have to find the co $-$ factor matrix
$=\left[\mathrm{A}_{i j}\right]_{3 \times 3}$, where $\mathrm{A}_{i j}=(-1)^{i+i} \mathrm{M}_{i j}$
Now $ \mathrm{A}_{11}=(-1)^{1+1} \mathrm{M}_{11}=\left|\begin{array}{lr}3 & 0 \\ 2 & -1\end{array}\right|=-3-0=-3$
$ \mathrm{~A}_{12}=(-1)^{1+2} \mathrm{M}_{12}=-\left|\begin{array}{rr}3 & 0 \\ 5 & -1\end{array}\right|=-(-3-0)=3 $
$ \mathrm{~A}_{13}=(-1)^{1+3} \mathrm{M}_{13}=\left|\begin{array}{ll}3 & 3 \\ 5 & 2\end{array}\right|=6-15=-9 $
$ \mathrm{~A}_{21}=(-1)^{2+1} \mathrm{M}_{21}=-\left|\begin{array}{lr}0 & 0 \\ 2 & -1\end{array}\right|=-(0-0)=0 $
$ \mathrm{~A}_{22}=(-1)^{2+2} \mathrm{M}_{22}=\left|\begin{array}{ll}1 & 0 \\ 5 & -1\end{array}\right|=-1-0=-1 $
$ \mathrm{~A}_{23}=(-1)^{2+3} \mathrm{M}_{23}=-\left|\begin{array}{ll}1 & 0 \\ 5 & 2\end{array}\right|=-(2-0)=-2 $
$ \mathrm{~A}_{31}=(-1)^{3+1} \mathrm{M}_{31}=\left|\begin{array}{ll}0 & 0 \\ 3 & 0\end{array}\right|=0-0=0 $
$ \mathrm{~A}_{32}=(-1)^{3+2} \mathrm{M}_{32}=-\left|\begin{array}{ll}1 & 0 \\ 3 & 0\end{array}\right|=-(0-0)=0 $
$ \mathrm{~A}_{33}=(-1)^{3+3} \mathrm{M}_{33}=\left|\begin{array}{ll}1 & 0 \\ 3 & 3\end{array}\right|=3-0=3$
$\therefore$ 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}-3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3\end{array}\right] $
$ \therefore \operatorname{adj} A=\left[\begin{array}{rrr}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right] $
$ \therefore A^{-1}=\frac{1}{|A|}(\text { adj } A) \\ =\frac{1}{-3}\left[\begin{array}{rrr}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right] $
$ \therefore A^{-1}=\frac{1}{3}\left[\begin{array}{rrr}3 & 0 & 0 \\ -3 & 1 & 0 \\ 9 & 2 & -3\end{array}\right]$
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Question 154 Marks
Find the inverse of the following matrices by the adjoint method : $\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$
Answer
Let $\mathrm{A}=\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$
$\begin{array}{l}|A|=6+8=14 \neq 0 \\ \therefore A^{-1} \text { exist }\end{array}$
First we have to find the co $-$ factor matrix
$= [A_{ij}] _{2\times 2}$ where $A_{ij} = (-1)^{i+j}M_{ij}$
Now $A_{11} = (-1)^{1+1}M_{11} = 3$
$A_{12} = (-1)^{1+2}M = -4$
$A_{21} = (-2)^{2+1}M_{21} = (-2) = 2$
$A_{22} = (-1)^{2+2}M_{22} = 2$
Hence the co $-$ factor matrix
$=\left[\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right]=\left[\begin{array}{cc}3 & -4 \\ 2 & 2\end{array}\right] $
$ \therefore \operatorname{adj} \mathrm{A}=\left[\begin{array}{cc}3 & 2 \\ -4 & 2\end{array}\right] $
$ \therefore \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A})=\frac{1}{14}\left(\begin{array}{cc}3 & 2 \\ -4 & 2\end{array}\right)$
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Question 164 Marks
Find the inverse of the following matrices by the adjoint method : $\left[\begin{array}{rr}-1 & 5 \\ -3 & 2\end{array}\right]$
Answer
Let $A=\left[\begin{array}{rr}-1 & 5 \\ -3 & 2\end{array}\right] $
$\therefore|A|=\left|\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right|=-2+15=13 \neq 0$
$\therefore A^{-1} $ exists.
First we have to find the co $-$ factor matrix
$= [A_{ij}]_{2\times 2}$, where $A_{ij} = (-1)^{i+j}M_{ij}$​​​​​​​
Now,$ A_{11} = (-1)^{1+1}M_{11} = 2$
$A_{12} = (-1)^{1+2}M_{12} = -(-3) = 3$
$A_{21} = (-1)^{2+1}M_{21} = -5$
$A_{22} = (-1)^{2+2}M_{22} = -1$
Hence, the co $-$ factor matrix
$=\left[\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right]=\left[\begin{array}{cc}2 & 3 \\ -5 & -1\end{array}\right] $
$ \therefore \operatorname{adj} A=\left[\begin{array}{ll}2 & -5 \\ 3 & -1\end{array}\right]$
$ \therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{13}\left[\begin{array}{cc}2 & -5 \\ 3 & -1\end{array}\right]$
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Solve the Following Question.(4 Marks) - Maths STD 12 Questions - Vidyadip