Exercise4.4, Class 12th, Maths, Chapter 4, NCERT

Exercise 4.4 

Question 1:
Find  $\mathrm{adj}\mathrm{A }$for

$$A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)\mathrm{}$$Solution:
For a $2\times \mathrm{2 }$matrix

$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ $$\mathrm{adj}A=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)\mathrm{}$$

So for

$A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$

$$\boxed{{\displaystyle \mathrm{adj}A=\left(\begin{array}{cc}4& -2\\ -3& 1\end{array}\right)\mathrm{}}}$$

(You can check  $A(\mathrm{adj}A)=(\det A)I$: 

$\det A=1\cdot 4-2\cdot 3=-2$, and indeed $A\mathrm{adj}A=(-2)\mathrm{I }$)

Question 2:
Find $\mathrm{adj} \mathrm{A }$

for$$A=\left(\begin{array}{ccc}1& -1& 2\\ 2& 3& 5\\ -2& 0& 1\end{array}\right)\mathrm{}$$

Below I show the cofactors, the adjoint, and a verification

$A\mathrm{adj}A=(\det A)I$

Minors and cofactors

Compute the  $2\times \mathrm{2\; }$minors and cofactors

$A_{ij}=(-1{)}^{i+j}{M}_{ij}$

$$\text{Cofactor matrix }[A_{ij}]=\left(\begin{array}{ccc}3& -12& 6\\ 1& 5& 2\\ -11& -1& 5\end{array}\right)\mathrm{}$$

(Each entry is the cofactor of the corresponding element of

$A$.)

$\mathrm{adj}\mathrm{A }$

(adjugate = transpose of cofactor matrix)

$$\boxed{{\displaystyle \mathrm{adj}A=\left(\begin{array}{ccc}3& 1& -11\\ -12& 5& -1\\ 6& 2& 5\end{array}\right)\mathrm{}}}$$

Determinant and verification

$$\mathrm{det }A=27$$

Now verify    

$A\mathrm{adj}A=(\det A)I=27I$

Multiplying gives

$$A\mathrm{adj}A=\left(\begin{array}{ccc}27& 0& 0\\ 0& 27& 0\\ 0& 0& 27\end{array}\right)=27I,$$

so the adjoint is correct.

___________________

Verify the identity for both matrices

$$\text{Show that }A(\mathrm{adj}A)=(\mathrm{adj}A)A=\mathrm{\mid }A\mathrm{\mid }I$$

Question 3:

$A=\left(\begin{array}{cc}2& 3\\ -4& -6\end{array}\right)$

Solution – Compute the determinant:

$$\mathrm{\mid }A\mathrm{\mid }=2(-6)-3(-4)=-12+12=0.$$

Compute the Adj (classical adjoint). For a $2\times \mathrm{2 }$matrix

$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$,

$\mathrm{adj}A=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$

Thus,

$$\mathrm{adj}A=\left(\begin{array}{cc}-6& -3\\ 4& 2\end{array}\right)\mathrm{}$$

Multiply:

$$A(\mathrm{adj}A)=\left(\begin{array}{cc}2& 3\\ -4& -6\end{array}\right)\left(\begin{array}{cc}-6& -3\\ 4& 2\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\mathrm{}$$

Similarly

$$(\mathrm{adj}A)A=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\mathrm{}$$

Since 

$\mathrm{\mid }A\mathrm{\mid }=0$, we have

$\mathrm{\mid }A\mathrm{\mid }I=0\cdot I=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$

Hence,

$$A(\mathrm{adj}A)=(\mathrm{adj}A)A=\mathrm{\mid }A\mathrm{\mid }I,$$

as required.

Question 4:

$A=\left(\begin{array}{ccc}1& -1& 2\\ 3& 0& -2\\ 1& 0& 3\end{array}\right)$

Solution – Compute the determinant –

$$\mathrm{\mid }A\mathrm{\mid }=11$$

Compute the adjugate (transpose of cofactor matrix). The cofactors lead to

$$\mathrm{adj}A=\left(\begin{array}{ccc}0& 3& 2\\ -11& 1& 8\\ 0& -1& 3\end{array}\right)\mathrm{.}$$

Multiply:

$$A(\mathrm{adj}A)=\left(\begin{array}{ccc}11& 0& 0\\ 0& 11& 0\\ 0& 0& 11\end{array}\right)=11I,$$

and likewise

$$(\mathrm{adj}A)A=11I\mathrm{}$$

Thus for this matrix also

$$A(\mathrm{adj}A)=(\mathrm{adj}A)A=\mathrm{\mid }A\mathrm{\mid }I\mathrm{}$$

Conclusion

Both verifications hold:

• For

  $A=(\begin{array}{cc}2& 3\\ -4& -6\end{array})\; <span\; style="font-family:\; var(--wp--preset--font-family--manrope);\; font-size:\; var(--wp--preset--font-size--large);\; letter-spacing:\; -0.1px;">\hspace{0.25em} </span>$ $A(\mathrm{adj}A)=(\mathrm{adj}A)A=0=\mathrm{\mid }A\mathrm{\mid }I\mathrm{}$

• For

  $A=\left(\begin{array}{ccc}1& -1& 2\\ 3& 0& -2\\ 1& 0& 3\end{array}\right)$

$A(\mathrm{adj}A)=(\mathrm{adj}A)A=11I\mathrm{.}$

Question 5:

$$\text{Find }{A}^{-1}\text{ if }A=\left(\begin{array}{cc}2& -2\\ 4& 3\end{array}\right)\mathrm{.}$$

Formula for inverse of a 2×2 matrix

$$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$$

its inverse (if 

$\mathrm{\mid }A\mathrm{\mid }\mathrm{\ne }\mathrm{0 }$

is

$$A^{-1}=\frac{1}{ad-bc}(\begin{array}{cc}d& -b\\ -c& a\end{array}$$

Compute determinant

$$\mathrm{\mid }A\mathrm{\mid }=(2)(3)-(-2)(4)=6+8=14$$

Since

$\mathrm{\mid }A\mathrm{\mid }\mathrm{\ne }\mathrm{0 }$

, the inverse exists.

Apply the formula

$$A^{-1}=\frac{1}{14}\left(\begin{array}{cc}3& 2\\ -4& 2\end{array}\right)\mathrm{}$$

Final Answer:

$$\boxed{{\displaystyle A^{-1}=\frac{1}{14}\left(\begin{array}{cc}3& 2\\ -4& 2\end{array}\right)\mathrm{}}}$$

Verification (optional):

$$A{A}^{-1}=\left(\begin{array}{cc}2& -2\\ 4& 3\end{array}\right)\frac{1}{14}\left(\begin{array}{cc}3& 2\\ -4& 2\end{array}\right)=\frac{1}{14}\left(\begin{array}{cc}14& 0\\ 0& 14\end{array}\right)=I\mathrm{}$$

✔ Verified.

 

Question 6:

$$\text{Find }{A}^{-1}\text{ if }A=\left(\begin{array}{cc}-1& 5\\ -3& 2\end{array}\right)\mathrm{}$$

Formula for inverse

For any 

$2\times \mathrm{2 }$

matrix

$$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right),$$

if 

$\mathrm{\mid }A\mathrm{\mid }=ad-bc\mathrm{\ne }\mathrm{0 }$

, then

$$A^{-1}=\frac{1}{\mathrm{\mid }A\mathrm{\mid }}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)\mathrm{}$$

Compute determinant

$$\mathrm{\mid }A\mathrm{\mid }=(-1)(2)-(5)(-3)=-2+15=13$$

$\mathrm{\mid }A\mathrm{\mid }=13\mathrm{\ne }\mathrm{0 }$

, so the inverse exists.

Apply the formula

$$A^{-1}=\frac{1}{13}\left(\begin{array}{cc}2& -5\\ 3& -1\end{array}\right)\mathrm{}$$

Final Answer:

$$\boxed{{\displaystyle A^{-1}=\frac{1}{13}\left(\begin{array}{cc}2& -5\\ 3& -1\end{array}\right)\mathrm{}}}$$

Verification (optional):

$$A{A}^{-1}=\left(\begin{array}{cc}-1& 5\\ -3& 2\end{array}\right)\frac{1}{13}\left(\begin{array}{cc}2& -5\\ 3& -1\end{array}\right)=\frac{1}{13}\left(\begin{array}{cc}13& 0\\ 0& 13\end{array}\right)=I\mathrm{}$$

✔ Verified.

Question 7:

$$\text{Find }{A}^{-1}\text{ if }A=\left(\begin{array}{ccc}1& 2& 3\\ 0& 2& 4\\ 0& 0& 5\end{array}\right)\mathrm{}$$

Observation

$\mathrm{A }$

is an upper triangular matrix, and for triangular matrices,

$$\mathrm{\mid }A\mathrm{\mid }=\text{product of diagonal elements.}$$

Determinant

$$\mathrm{\mid }A\mathrm{\mid }=1\times 2\times 5=10$$

Since 

$\mathrm{\mid }A\mathrm{\mid }\mathrm{\ne }0$, $A^{-\mathrm{1 }}$exists.

Use properties of triangular matrices

For an upper triangular matrix, its inverse is also upper triangular.
We find 

$A^{-1}=[a_{ij}]$

 such that 

$A{A}^{-1}=I$

Let

$$A^{-1}=\left(\begin{array}{ccc}a& b& c\\ 0& d& e\\ 0& 0& f\end{array}\right)\mathrm{}$$

Then

$$A{A}^{-1}=\left(\begin{array}{ccc}1& 2& 3\\ 0& 2& 4\\ 0& 0& 5\end{array}\right)\left(\begin{array}{ccc}a& b& c\\ 0& d& e\\ 0& 0& f\end{array}\right)=\left(\begin{array}{ccc}a& b+2d& c+2e+3f\\ 0& 2d& 2e+4f\\ 0& 0& 5f\end{array}\right)\mathrm{}$$

We want this to equal the identity

$I=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$

Compare entries

From the bottom:

$f=1⟹f=\frac{1}{5}\mathrm{}$

$2d=1⟹d=\frac{1}{2}\mathrm{}$

$a=1$

Now use off-diagonal equations:

$e+4f=0⟹e=-2f=-\frac{2}{5}\mathrm{}$

$b+2d=0⟹b=-2d=-1$

$c+2e+3f=0⟹c=-2e-3f\mathrm{}$

Substitute 

$e=-\frac{2}{\mathrm{5 }}$and  $f=\frac{1}{5}$:

$$c=-2(-\frac{2}{5})-3(\frac{1}{5})=\frac{4}{5}-\frac{3}{5}=\frac{1}{5}\mathrm{}$$

Write the inverse

$$\boxed{{\displaystyle A^{-1}=\left(\begin{array}{ccc}1& -1& \frac{1}{5}\\ 0& \frac{1}{2}& -\frac{2}{5}\\ 0& 0& \frac{1}{5}\end{array}\right)\mathrm{}}}$$

Verification (optional):

$$A{A}^{-1}=\left(\begin{array}{ccc}1& 2& 3\\ 0& 2& 4\\ 0& 0& 5\end{array}\right)\left(\begin{array}{ccc}1& -1& \frac{1}{5}\\ 0& \frac{1}{2}& -\frac{2}{5}\\ 0& 0& \frac{1}{5}\end{array}\right)=I\mathrm{}$$

✔ Verified.

Question 8:

$$\text{Find }{A}^{-1}\text{ for }A=\left(\begin{array}{ccc}1& 0& 0\\ 3& 3& 0\\ 5& 2& -1\end{array}\right)\mathrm{}$$

Solution.

$\mathrm{A }$is lower triangular, so $A^{-\mathrm{1 }}$is also lower triangular.
Let

$$A^{-1}=\left(\begin{array}{ccc}a& 0& 0\\ b& c& 0\\ d& e& f\end{array}\right)$$

and solve

$A{A}^{-1}=I$

.Multiply  $\mathrm{A }$and $A^{-1}$: From the first row: $a=1$

.From the second row:

$$3a+3b=0,3c=1\Rightarrow b=-1,c=\frac{1}{3}\mathrm{}$$

From the third row:

$$5a+2b-d=0,2c-e=0,-f=1$$

Substituting

$a=1,b=-1,c=\frac{1}{\mathrm{3 }}$

, gives

$$5(1)+2(-1)-d=0\Rightarrow d=3,2\cdot \frac{1}{3}-e=0\Rightarrow e=\frac{2}{3},f=-1.$$

Thus

$$\boxed{{\displaystyle A^{-1}=\left(\begin{array}{ccc}1& 0& 0\\ -1& \frac{1}{3}& 0\\ 3& \frac{2}{3}& -1\end{array}\right)\mathrm{}}}$$

(You can verify 

$A{A}^{-1}=\mathrm{I }$

by multiplication.)

Question 9:

find $A^{-\mathrm{1 }}$

for$$A=\left(\begin{array}{ccc}2& 1& 3\\ 4& -1& 0\\ -7& 2& 1\end{array}\right)$$

Determinant

Compute $\mathrm{\mid }A\mathrm{\mid }$

(expand along the first row):

$$\mathrm{\mid }A\mathrm{\mid }=2\mid \begin{array}{cc}-1& 0\\ 2& 1\end{array}\mid -1\mid \begin{array}{cc}4& 0\\ -7& 1\end{array}\mid +3\mid \begin{array}{cc}4& -1\\ -7& 2\end{array}\mid$$

$$=2(-1)-1(4)+3(1)=-2-4+3=-3$$

So

$\mathrm{\mid }A\mathrm{\mid }=-3$

 (non zero) — inverse exists.

Cofactor / Adjugate

Compute the matrix of cofactors (I show the cofactors 

$C_{ij}$):

$$[C_{ij}]=\left(\begin{array}{ccc}-1& -4& 1\\ 5& 23& -11\\ 3& 12& -6\end{array}\right)$$

The adjugate (adj $A$) is the transpose of this:

$$\mathrm{adj}A=\left(\begin{array}{ccc}-1& 5& 3\\ -4& 23& 12\\ 1& -11& -6\end{array}\right)\mathrm{}$$

Inverse

$$A^{-1}=\frac{1}{\mathrm{\mid }A\mathrm{\mid }}\mathrm{adj}A=-\frac{1}{3}\left(\begin{array}{ccc}-1& 5& 3\\ -4& 23& 12\\ 1& -11& -6\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{3}& -\frac{5}{3}& -1\\ \frac{4}{3}& -\frac{23}{3}& -4\\ -\frac{1}{3}& \frac{11}{3}& 2\end{array}\right)\mathrm{}$$

Equivalently-

$A^{-1}={\displaystyle \frac{1}{3}}\left(\begin{array}{ccc}1& -5& -3\\ 4& -23& -12\\ -1& 11& 6\end{array}\right)$

Quick check (first-row × first-column)

$$2\cdot \frac{1}{3}+1\cdot \frac{4}{3}+3\cdot (-\frac{1}{3})=\frac{2+4-3}{3}=1,$$

so the product gives the identity as expected.

Question 10.

$$\text{Find }{A}^{-1}\text{ for }A=\left(\begin{array}{ccc}1& -1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}\right)\mathrm{}$$

Solution:

Compute 

$\mathrm{\mid }A\mathrm{\mid }$

. Expanding along the first row,

$$\begin{array}{rl}{\displaystyle \mathrm{\mid }A\mathrm{\mid }}& {\displaystyle =1\cdot \mid \begin{array}{cc}2& -3\\ -2& 4\end{array}\mid -(-1)\cdot \mid \begin{array}{cc}0& -3\\ 3& 4\end{array}\mid +2\cdot \mid \begin{array}{cc}0& 2\\ 3& -2\end{array}\mid }\\ {\displaystyle }& {\displaystyle =1(2\cdot 4-(-3)(-2))+1(0\cdot 4-(-3)\cdot 3)+2(0\cdot (-2)-2\cdot 3)}\\ {\displaystyle }& {\displaystyle =1(8-6)+1(9)+2(-6)=2+9-12=-1}\end{array}$$

So 

$\mathrm{\mid }A\mathrm{\mid }=-1$

 (nonzero) and the inverse exists.

Compute cofactors (minors 

$M_{ij}$

 and cofactors

$C_{ij}=(-1{)}^{i+j}{M}_{ij}$

$$\begin{array}{ll}{M}_{11}=\mid \begin{array}{cc}2& -3\\ -2& 4\end{array}\mid =2,& C_{11}=+2,\\ M_{12}=\mid \begin{array}{cc}0& -3\\ 3& 4\end{array}\mid =9,& C_{12}=-9,\\ M_{13}=\mid \begin{array}{cc}0& 2\\ 3& -2\end{array}\mid =-6,& C_{13}=-6,\\ M_{21}=\mid \begin{array}{cc}-1& 2\\ -2& 4\end{array}\mid =0,& C_{21}=0,\\ M_{22}=\mid \begin{array}{cc}1& 2\\ 3& 4\end{array}\mid =-2,& C_{22}=-2,\\ M_{23}=\mid \begin{array}{cc}1& -1\\ 3& -2\end{array}\mid =1,& C_{23}=-1,\\ M_{31}=\mid \begin{array}{cc}-1& 2\\ 2& -3\end{array}\mid =-1,& C_{31}=-1,\\ M_{32}=\mid \begin{array}{cc}1& 2\\ 0& -3\end{array}\mid =-3,& C_{32}=+3,\\ M_{33}=\mid \begin{array}{cc}1& -1\\ 0& 2\end{array}\mid =2,& C_{33}=+\mathrm{2,}\end{array}$$

Cofactor matrix

$C=[C_{ij}]$

is

$$C=\left(\begin{array}{ccc}2& -9& -6\\ 0& -2& -1\\ -1& 3& 2\end{array}\right)\mathrm{}$$

Adjugate is transpose of cofactor matrix:

$$\mathrm{adj}A=C^T=\left(\begin{array}{ccc}2& 0& -1\\ -9& -2& 3\\ -6& -1& 2\end{array}\right)\mathrm{}$$

Inverse:

$$A^{-1}=\frac{1}{\mathrm{\mid }A\mathrm{\mid }}\mathrm{adj}A=-1\cdot \mathrm{adj}A=\left(\begin{array}{ccc}-2& 0& 1\\ 9& 2& -3\\ 6& 1& -2\end{array}\right)\mathrm{}$$

Question 11.

$$\text{Find }{A}^{-1}\text{ for }A=\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \alpha & \sin \alpha \\ 0& \sin \alpha & -\cos \alpha \end{array}\right)\mathrm{}$$

Solution:

Write 

$A=\mathrm{diag}(1,M)$

where

$$M=\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{array}\right)\mathrm{}$$

Compute 

$\det M$

:

$$\det M=(\cos \alpha )(-\cos \alpha )-(\sin \alpha )(\sin \alpha )=-({\cos }^2\alpha +{\sin }^2\alpha )=-1$$

So 

$\mathrm{\mid }A\mathrm{\mid }=1\cdot \det M=-1$, hence 

$\mathrm{A }$is invertible.

Now compute $M^{-1}$using the $2\times \mathrm{2 }$

formula:

$$M^{-1}=\frac{1}{\det M}\left(\begin{array}{cc}-\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right)$$

$$=-1\left(\begin{array}{cc}-\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right)=\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{array}\right)=M\mathrm{}$$

Thus 

$M^{-1}=M$

 Equivalently 

$M^2=I_2$

Therefore

$$A^{-1}=\mathrm{diag}(1,M^{-1})=\mathrm{diag}(1,M)=A\mathrm{}$$

Final answer

$$\boxed{{\displaystyle A^{-1}=\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \alpha & \sin \alpha \\ 0& \sin \alpha & -\cos \alpha \end{array}\right)=A}}$$

(Verification:

$A^2=\mathrm{diag}(1,M^2)=\mathrm{diag}(1,I_2)=I_3$, so indeed 

$A^{-1}=A$

Question 12:

$$\text{Let }A=\left(\begin{array}{cc}3& 7\\ 2& 5\end{array}\right)\text{ and }B=\left(\begin{array}{cc}6& 8\\ 7& 9\end{array}\right)\mathrm{.}$$

$$\text{Verify that }(AB{)}^{-1}=B^{-1}{A}^{-1}\mathrm{.}$$

Let’s verify 

$$(AB{)}^{-1}=B^{-1}{A}^{-1}$$

Compute

$A^{-\mathrm{1 }}$

and 

$B^{-1}$

(a)  $A^{-1}$ $$\mathrm{\mid }A\mathrm{\mid }=3(5)-7(2)=15-14=1$$

$$A^{-1}=\left(\begin{array}{cc}5& -7\\ -2& 3\end{array}\right)\mathrm{}$$

(b)  $B^{-1}$ $$\mathrm{\mid }B\mathrm{\mid }=6(9)-8(7)=54-56=-2$$

$$B^{-1}=\frac{1}{-2}\left(\begin{array}{cc}9& -8\\ -7& 6\end{array}\right)=\left(\begin{array}{cc}-\frac{9}{2}& 4\\ \frac{7}{2}& -3\end{array}\right)\mathrm{}$$

Compute

$AB$ $$AB=\left(\begin{array}{cc}3& 7\\ 2& 5\end{array}\right)\left(\begin{array}{cc}6& 8\\ 7& 9\end{array}\right)=\left(\begin{array}{cc}3\cdot 6+7\cdot 7& 3\cdot 8+7\cdot 9\\ 2\cdot 6+5\cdot 7& 2\cdot 8+5\cdot 9\end{array}\right)=\left(\begin{array}{cc}67& 87\\ 47& 61\end{array}\right)\mathrm{}$$

Compute

$(AB{)}^{-1}$ $$\mathrm{\mid }AB\mathrm{\mid }=67(61)-87(47)=4087-4089=-2$$

$$(AB{)}^{-1}=\frac{1}{-2}\left(\begin{array}{cc}61& -87\\ -47& 67\end{array}\right)=\left(\begin{array}{cc}-\frac{61}{2}& \frac{87}{2}\\ \frac{47}{2}& -\frac{67}{2}\end{array}\right)\mathrm{}$$

Compute

$B^{-1}{A}^{-1}$ $$B^{-1}{A}^{-1}=\left(\begin{array}{cc}-\frac{9}{2}& 4\\ \frac{7}{2}& -3\end{array}\right)\left(\begin{array}{cc}5& -7\\ -2& 3\end{array}\right)\mathrm{}$$

Multiply:

$$\begin{array}{rl}{\displaystyle B^{-1}{A}^{-1}}& {\displaystyle =\left(\begin{array}{cc}(-\frac{9}{2})(5)+4(-2)& (-\frac{9}{2})(-7)+4(3)\\ (\frac{7}{2})(5)+(-3)(-2)& (\frac{7}{2})(-7)+(-3)(3)\end{array}\right)}\\ {\displaystyle }& {\displaystyle =\left(\begin{array}{cc}-\frac{61}{2}& \frac{87}{2}\\ \frac{47}{2}& -\frac{67}{2}\end{array}\right)\mathrm{}}\end{array}$$

Compare

$$(AB{)}^{-1}=\left(\begin{array}{cc}-\frac{61}{2}& \frac{87}{2}\\ \frac{47}{2}& -\frac{67}{2}\end{array}\right)=B^{-1}{A}^{-1}\mathrm{}$$

Hence verified:

$$\boxed{{\displaystyle (AB{)}^{-1}=B^{-1}{A}^{-1}\mathrm{}}}$$

Question 13 :

If $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$, show that $A^2-5A+7I=O$. Hence, find $A^{-1}$.

Solution:

Let’s solve step-by-step carefully and clearly.

Given:

$$A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$$

We have to show that

$$A^2-5A+7I=O$$

and then use this result to find $A^{-1}$

Step 1: Compute $A^2$

$$A^2=A\cdot A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$$

Multiply carefully:

$$A^2=\left[\begin{array}{cc}(3)(3)+(1)(-1)& (3)(1)+(1)(2)\\ (-1)(3)+(2)(-1)& (-1)(1)+(2)(2)\end{array}\right]=\left[\begin{array}{cc}9-1& 3+2\\ -3-2& -1+4\end{array}\right]=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]$$

So,

$$A^2=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]$$

Step 2: Compute $5A$ and $7I$

$$5A=5\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]$$
$$7I=7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$$

Step 3: Compute $A^2-5A+7I$

$$A^2-5A+7I=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$$

First subtract $A^2-5A$:

$$\left[\begin{array}{cc}8-15& 5-5\\ -5-(-5)& 3-10\end{array}\right]=\left[\begin{array}{cc}-7& 0\\ 0& -7\end{array}\right]$$

Now add $7I$:

$$\left[\begin{array}{cc}-7& 0\\ 0& -7\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$$

Hence proved:

$$A^2-5A+7I=O$$

Step 4: Finding $A^{-1}$

From the equation:

$$A^2-5A+7I=0$$

Rearrange it as:

$$A^2-5A=-7I$$Multiply both sides by $A^{-1}$(on the right):

$$A-5I=-7A^{-1}$$

So,$$A^{-1}=\frac{1}{7}(5I-A)$$

Step 5: Substitute values

$$A^{-1}=\frac{1}{7}(5\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right])$$
$$A^{-1}=\frac{1}{7}\left[\begin{array}{cc}5-3& 0-1\\ 0-(-1)& 5-2\end{array}\right]=\frac{1}{7}\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]$$

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