Exercise-4.4, Class 12th, Maths, Part -2

Question 14 :

For the matrix $A=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]$, find the numbers $a$ and $b$ such that

$$A^2+aA+bI=0.$$

Solution:

Given:

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

We are to find $a$ and $b$ such that

$$A^2+aA+bI=0$$

Step 1: Compute $A^2$

$$A^2=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]$$Perform the multiplication:

$$A^2=\left[\begin{array}{cc}3\times 3+2\times 1& 3\times 2+2\times 1\\ 1\times 3+1\times 1& 1\times 2+1\times 1\end{array}\right]=\left[\begin{array}{cc}9+2& 6+2\\ 3+1& 2+1\end{array}\right]=\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]$$

Step 2: Substitute in the equation

$$A^2+aA+bI=0$$

That means:

$$\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]+a\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]+b\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$$

Step 3: Combine all matrices

$$\left[\begin{array}{cc}11+3a+b& 8+2a\\ 4+a& 3+a+b\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$$

Step 4: Equate corresponding elements to zero

From the first row, first column:

$$11+3a+b=0\text{(1)}$$

From the first row, second column:

$$8+2a=0\text{(2)}$$

From the second row, first column:

$$4+a=0\text{(3)}$$

From the second row, second column:

$$3+a+b=0\text{(4)}$$

Step 5: Solve for $a$ and $b$

From (3):

$$a=-4$$

Substitute $a=-4$:

$$3+(-4)+b=0⟹b=1$$

So $a=-4$ and $b=1$.

Step 6: Verification

Check equation (1):

$$11+3(-4)+1=11-12+1=0$$

True. Check (2):

$$8+2(-4)=8-8=0$$True.

Final Answer:

$$\boxed{{\displaystyle a=-4,b=1}}$$

 

Question 15 

For the matrix $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& -3\\ 2& -1& 3\end{array}\right]$, show that

$$A^3-6A^2+5A+11I=O\mathrm{.}$$

Hence, find $A^{-1}$.

Solution

Step 1 — compute $A^2$ and $A^3$.

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

Compute $A^2=A\cdot A$:

$$A^2=\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& 14\end{array}\right]$$

Compute $A^3=A^2\cdot A$:

$$A^3=\left[\begin{array}{ccc}8& 7& 1\\ -23& 27& -69\\ 32& -13& 58\end{array}\right]$$

Step 2 — form the combination $A^3-6A^2+5A+11I$.

Calculate termwise:

$$\begin{array}{rl}{\displaystyle A^3-6A^2+5A+11I}& {\displaystyle =\left[\begin{array}{ccc}8& 7& 1\\ -23& 27& -69\\ 32& -13& 58\end{array}\right]-6\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& 14\end{array}\right]+5\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& -3\\ 2& -1& 3\end{array}\right]+11\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}\\ {\displaystyle }& {\displaystyle =\left[\begin{array}{ccc}8-24+5+11& 7-12+5+0& 1-6+5+0\\ -23+18+5+0& 27-48+10+11& -69+84-15+0\\ 32-42+10+0& -13+18-5+0& 58-84+15+11\end{array}\right]}\\ {\displaystyle }& {\displaystyle =\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\mathrm{}}\end{array}$$

So the relation is verified:

$$\boxed{{\displaystyle A^3-6A^2+5A+11I=O}}$$

Step 3 — find $A^{-1}$.

Starting from

$$A^3-6A^2+5A+11I=O,$$

multiply on the right by $A^{-1}$ (which is valid because $A$ is invertible if we can find $A^{-1}$ from this equation):

$$A^2-6A+5I+11A^{-1}=O\mathrm{}$$

Rearrange to solve for $A^{-1}$:

$$11A^{-1}=-A^2+6A-5I,A^{-1}=\frac{1}{11}(6A-A^2-5I)\mathrm{}$$

Now substitute the matrices $A$ and $A^2$:

$$6A-A^2-5I=6\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& -3\\ 2& -1& 3\end{array}\right]-\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& 14\end{array}\right]-5\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$$
$$=\left[\begin{array}{ccc}6-4-5& 6-2-0& 6-1-0\\ 6-(-3)-0& 12-8-5& -18-(-14)-0\\ 12-7-0& -6-(-3)-0& 18-14-5\end{array}\right]=\left[\begin{array}{ccc}-3& 4& 5\\ 9& -1& -4\\ 5& -3& -1\end{array}\right]\mathrm{}$$

Thus

$$A^{-1}=\frac{1}{11}\left[\begin{array}{ccc}-3& 4& 5\\ 9& -1& -4\\ 5& -3& -1\end{array}\right]\mathrm{}$$

You can write it explicitly as

$$\boxed{{\displaystyle A^{-1}=\left[\begin{array}{ccc}-\frac{3}{11}& \frac{4}{11}& \frac{5}{11}\\ \frac{9}{11}& -\frac{1}{11}& -\frac{4}{11}\\ \frac{5}{11}& -\frac{3}{11}& -\frac{1}{11}\end{array}\right]}}$$

(One may verify $A{A}^{-1}=I$)

Question 16 : If

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

verify that

$$A^3-6A^2+9A-4I=O,$$

and hence find $A^{-1}$.

Solution

Step 1 — compute $A^2$.

$$A^2=A\cdot A=\left[\begin{array}{ccc}2& -1& 1\\ -1& 2& -1\\ 1& -1& 2\end{array}\right]\left[\begin{array}{ccc}2& -1& 1\\ -1& 2& -1\\ 1& -1& 2\end{array}\right]=\left[\begin{array}{ccc}6& -5& 5\\ -5& 6& -5\\ 5& -5& 6\end{array}\right]\mathrm{}$$(Each entry calculated: e.g. top-left $=2\cdot 2+(-1)(-1)+1\cdot 1=4+1+1=6$, top-middle $=2(-1)+(-1)2+1(-1)=-2-2-1=-5$

Step 2 — compute $A^3$.

$$A^3=A^2\cdot A=\left[\begin{array}{ccc}22& -21& 21\\ -21& 22& -21\\ 21& -21& 22\end{array}\right]\mathrm{.}$$

(For example, top-left $=6\cdot 2+(-5)(-1)+5\cdot 1=12+5+5=22$, top-middle $=6(-1)+(-5)2+5(-1)=-6-10-5=-21$

Step 3 — form the combination $A^3-6A^2+9A-4I$

Compute termwise:

$$\begin{array}{rl}{\displaystyle A^3-6A^2+9A-4I}& {\displaystyle =\left[\begin{array}{ccc}22& -21& 21\\ -21& 22& -21\\ 21& -21& 22\end{array}\right]-6\left[\begin{array}{ccc}6& -5& 5\\ -5& 6& -5\\ 5& -5& 6\end{array}\right]+9\left[\begin{array}{ccc}2& -1& 1\\ -1& 2& -1\\ 1& -1& 2\end{array}\right]-4\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}\\ {\displaystyle }& {\displaystyle =\left[\begin{array}{ccc}22-36+18-4& -21+30-9+0& 21-30+9+0\\ -21+30-9+0& 22-36+18-4& -21+30-9+0\\ 21-30+9+0& -21+30-9+0& 22-36+18-4\end{array}\right]}\\ {\displaystyle }& {\displaystyle =\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\mathrm{}}\end{array}$$

Thus the identity is verified:

$$\boxed{{\displaystyle A^3-6A^2+9A-4I=O}}\mathrm{}$$

Step 4 — find $A^{-1}$.

Start from the matrix polynomial

$$A^3-6A^2+9A-4I=O\mathrm{}$$

Right-multiply by $A^{-1}$ (valid since the relation will imply invertibility):

$$A^2-6A+9I-4A^{-1}=O\mathrm{}$$

Rearrange to solve for $A^{-1}$:

$$4A^{-1}=A^2-6A+9I\Rightarrow A^{-1}=\frac{1}{4}(A^2-6A+9I)\mathrm{}$$

Now substitute the matrices $A$ and $A^2$:

$$A^2-6A+9I=\left[\begin{array}{ccc}6& -5& 5\\ -5& 6& -5\\ 5& -5& 6\end{array}\right]-6\left[\begin{array}{ccc}2& -1& 1\\ -1& 2& -1\\ 1& -1& 2\end{array}\right]+9\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\mathrm{}$$Compute it:

$$A^2-6A+9I=\left[\begin{array}{ccc}6-12+9& -5+6+0& 5-6+0\\ -5+6+0& 6-12+9& -5+6+0\\ 5-6+0& -5+6+0& 6-12+9\end{array}\right]=\left[\begin{array}{ccc}3& 1& -1\\ 1& 3& 1\\ -1& 1& 3\end{array}\right]\mathrm{}$$

Therefore

$$A^{-1}=\frac{1}{4}\left[\begin{array}{ccc}3& 1& -1\\ 1& 3& 1\\ -1& 1& 3\end{array}\right]=\left[\begin{array}{ccc}\frac{3}{4}& \frac{1}{4}& -\frac{1}{4}\\ \frac{1}{4}& \frac{3}{4}& \frac{1}{4}\\ -\frac{1}{4}& \frac{1}{4}& \frac{3}{4}\end{array}\right]\mathrm{}$$

(You can verify $A{A}^{-1}=I$ directly.)

Final answer:

$$\boxed{{\displaystyle A^3-6A^2+9A-4I=O,A^{-1}=\left[\begin{array}{ccc}\frac{3}{4}& \frac{1}{4}& -\frac{1}{4}\\ \frac{1}{4}& \frac{3}{4}& \frac{1}{4}\\ -\frac{1}{4}& \frac{1}{4}& \frac{3}{4}\end{array}\right]\mathrm{}}}$$

 

Question 17 :

Let $A$ be a nonsingular square matrix of order $3\times 3$. Then $\mathrm{\mid }\mathrm{adj}A\mathrm{\mid }$ is equal to

(A) $\mathrm{\mid }A\mathrm{\mid }$
(B) $\mathrm{\mid }A{\mathrm{\mid }}^2$
(C) $\mathrm{\mid }A{\mathrm{\mid }}^3$
(D) $3\mathrm{\mid }A\mathrm{\mid }$

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Solution:

We know the formula for the determinant of the adjugate (adjoint) of a square matrix:

$$\mathrm{\mid }\mathrm{adj}A\mathrm{\mid }=\mathrm{\mid }A{\mathrm{\mid }}^{n-1}$$where $n$ is the order of the square matrix.

Here, $A$ is of order $3\times 3$.
So $n=3$

$$\mathrm{\mid }\mathrm{adj}A\mathrm{\mid }=\mathrm{\mid }A{\mathrm{\mid }}^{3-1}=\mathrm{\mid }A{\mathrm{\mid }}^2\mathrm{}$$

Final Answer:

$$\boxed{{\displaystyle \mathrm{\mid }\mathrm{adj}A\mathrm{\mid }=\mathrm{\mid }A{\mathrm{\mid }}^2}}$$

Hence, the correct option is (B) $\mathrm{\mid }A{\mathrm{\mid }}^2$

Question 18:

If $A$ is an invertible matrix of order $2$, then $\det (A^{-1})$ is equal to

(A) $\det (A)$
(B) ${\displaystyle \frac{1}{\det (A)}}$
(C) $1$
(D) $0$

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Solution:

We know the fundamental property of determinants:

$$\det (A^{-1})=\frac{1}{\det (A)}$$

provided $A$ is invertible (that is, $\det (A)\mathrm{\ne }0$).

Final Answer:

$$\boxed{{\displaystyle \det (A^{-1})=\frac{1}{\det (A)}}}$$

Hence, the correct option is (B) ${\displaystyle \frac{1}{\det (A)}}$.