Tag: Miscellaneous Exercises on Chapter 4

Question 1.
Prove that the determinant

$$\mid \begin{array}{ccc}x& \sin \theta & \cos \theta \\ -\sin \theta & -x& 1\\ \cos \theta & 1& x\end{array}\mid$$

is independent of $\theta$.

Solution.

Compute the determinant by expanding along the first row:

$$\begin{array}{rl}{\displaystyle D}& {\displaystyle =x\mid \begin{array}{cc}-x& 1\\ 1& x\end{array}\mid -\sin \theta \mid \begin{array}{cc}-\sin \theta & 1\\ \cos \theta & x\end{array}\mid +\cos \theta \mid \begin{array}{cc}-\sin \theta & -x\\ \cos \theta & 1\end{array}\mid \mathrm{}}\end{array}$$

Evaluate each $2\times 2$ determinant:

$$\begin{array}{rl}{\displaystyle x\mid \begin{array}{cc}-x& 1\\ 1& x\end{array}\mid }& {\displaystyle =x((-x)(x)-1\cdot 1)=x(-x^2-1)=-x^3-x,}\\ {\displaystyle -\sin \theta \mid \begin{array}{cc}-\sin \theta & 1\\ \cos \theta & x\end{array}\mid }& {\displaystyle =-\sin \theta ((-\sin \theta )x-1\cdot \cos \theta )=-\sin \theta (-x\sin \theta -\cos \theta )}\\ {\displaystyle }& {\displaystyle =x{\sin }^2\theta +\sin \theta \cos \theta ,}\\ {\displaystyle \cos \theta \mid \begin{array}{cc}-\sin \theta & -x\\ \cos \theta & 1\end{array}\mid }& {\displaystyle =\cos \theta ((-\sin \theta )\cdot 1-(-x)\cos \theta )=\cos \theta (-\sin \theta +x\cos \theta )}\\ {\displaystyle }& {\displaystyle =-\sin \theta \cos \theta +x{\cos }^2\theta \mathrm{}}\end{array}$$Add the three parts:

$$\begin{array}{rl}{\displaystyle D}& {\displaystyle =(-x^3-x)+(x{\sin }^2\theta +\sin \theta \cos \theta )+(-\sin \theta \cos \theta +x{\cos }^2\theta )}\\ {\displaystyle }& {\displaystyle =-x^3-x+x({\sin }^2\theta +{\cos }^2\theta )+(\sin \theta \cos \theta -\sin \theta \cos \theta )}\\ {\displaystyle }& {\displaystyle =-x^3-x+x\cdot 1+0=-x^3\mathrm{}}\end{array}$$Thus the determinant equals $-x^3$, which does not depend on $\theta$. 

Question 2.
Evaluate the determinant

$$\mid \begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0\\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\mid \mathrm{}$$

Solution:

Let

$$D=\mid \begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0\\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\mid \mathrm{}$$

We will expand along the second row since it has a zero term.

Step 1: Expansion along second row

$$D=(-\sin \beta )(-1{)}^{2+1}\mid \begin{array}{cc}\cos \alpha \sin \beta & -\sin \alpha \\ \sin \alpha \sin \beta & \cos \alpha \end{array}\mid$$

$$+(\cos \beta )(-1{)}^{2+2}\mid \begin{array}{cc}\cos \alpha \cos \beta & -\sin \alpha \\ \sin \alpha \cos \beta & \cos \alpha \end{array}\mid \mathrm{}$$

Simplify the cofactors:

$$D=\sin \beta \mid \begin{array}{cc}\cos \alpha \sin \beta & -\sin \alpha \\ \sin \alpha \sin \beta & \cos \alpha \end{array}\mid +\cos \beta \mid \begin{array}{cc}\cos \alpha \cos \beta & -\sin \alpha \\ \sin \alpha \cos \beta & \cos \alpha \end{array}\mid \mathrm{}$$

Step 2: Evaluate the 2×2 determinants

For the first:

$$\mid \begin{array}{cc}\cos \alpha \sin \beta & -\sin \alpha \\ \sin \alpha \sin \beta & \cos \alpha \end{array}\mid =(\cos \alpha \sin \beta )(\cos \alpha )-(-\sin \alpha )(\sin \alpha \sin \beta )$$

$$=\sin \beta ({\cos }^2\alpha +{\sin }^2\alpha )=\sin \beta \mathrm{.}$$

For the second:

$$\mid \begin{array}{cc}\cos \alpha \cos \beta & -\sin \alpha \\ \sin \alpha \cos \beta & \cos \alpha \end{array}\mid =(\cos \alpha \cos \beta )(\cos \alpha )-(-\sin \alpha )(\sin \alpha \cos \beta )$$

$$=\cos \beta ({\cos }^2\alpha +{\sin }^2\alpha )=\cos \beta \mathrm{.}$$

Step 3: Substitute back

$$D=\sin \beta (\sin \beta )+\cos \beta (\cos \beta )={\sin }^2\beta +{\cos }^2\beta =1$$

 Final Answer:

$$\boxed{{\displaystyle D=1}}$$

Question 3.
If

$$A^{-1}=\left(\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right)\text{and}B=\left(\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right),$$

find $(AB{)}^{-1}$

Solution:

We know the property of inverse of a product:

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

Step 1: Find $B^{-1}$

Let

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

To find $B^{-1}$, compute by the adjoint method (or via row operations).
After simplification, we get:

$$B^{-1}=\left(\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right)\mathrm{}$$

Step 2: Compute $(AB{)}^{-1}=B^{-1}{A}^{-1}$

$$A^{-1}=\left(\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right)\mathrm{}$$

Now multiply $B^{-1}$ and $A^{-1}$:

$$(AB{)}^{-1}=\left(\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right)\left(\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right)\mathrm{}$$

Step 3: Matrix multiplication

$$\begin{array}{rl}{\displaystyle (AB{)}_{11}^{-1}}& {\displaystyle =3(3)+2(-15)+6(5)=9-30+30=9,}\\ {\displaystyle (AB{)}_{12}^{-1}}& {\displaystyle =3(-1)+2(6)+6(-2)=-3+12-12=-3,}\\ {\displaystyle (AB{)}_{13}^{-1}}& {\displaystyle =3(1)+2(-5)+6(2)=3-10+12=5,}\\ {\displaystyle (AB{)}_{21}^{-1}}& {\displaystyle =1(3)+1(-15)+2(5)=3-15+10=-2,}\\ {\displaystyle (AB{)}_{22}^{-1}}& {\displaystyle =1(-1)+1(6)+2(-2)=-1+6-4=1,}\\ {\displaystyle (AB{)}_{23}^{-1}}& {\displaystyle =1(1)+1(-5)+2(2)=1-5+4=0,}\\ {\displaystyle (AB{)}_{31}^{-1}}& {\displaystyle =2(3)+2(-15)+5(5)=6-30+25=1,}\\ {\displaystyle (AB{)}_{32}^{-1}}& {\displaystyle =2(-1)+2(6)+5(-2)=-2+12-10=0,}\\ {\displaystyle (AB{)}_{33}^{-1}}& {\displaystyle =2(1)+2(-5)+5(2)=2-10+10=2.}\end{array}$$

Step 4: Write the result

$$(AB{)}^{-1}=\left(\begin{array}{ccc}9& -3& 5\\ -2& 1& 0\\ 1& 0& 2\end{array}\right)\mathrm{}$$

Final Answer:

$$\boxed{{\displaystyle (AB{)}^{-1}=\left(\begin{array}{ccc}9& -3& 5\\ -2& 1& 0\\ 1& 0& 2\end{array}\right)\mathrm{}}}$$

Question 4.
Let

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

Verify that
(i) $[\mathrm{adj}A{]}^{-1}=\mathrm{adj}(A^{-1})$
(ii) $(A^{-1}{)}^{-1}=A$

---

Answer & proof

(i) General identity and verification.

For any invertible square matrix $A$ we have the identity

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

Taking inverses on both sides (and using $(cM{)}^{-1}=c^{-1}{M}^{-1}$ for scalar $c\mathrm{\ne }0$) gives

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

Also for the inverse matrix $A^{-1}$ we have

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

But $\det (A^{-1})=(\det A{)}^{-1}$ and $(A^{-1}{)}^{-1}=A$, so

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

Comparing the two expressions we obtain the required equality for any invertible $A$:

$$(\mathrm{adj}A{)}^{-1}=\mathrm{adj}(A^{-1})\mathrm{}$$

Verification for the given matrix $A$.

Compute $\det A$, $\mathrm{adj}A$ and the two sides explicitly:

$$\det A=-5$$

One finds

$$\mathrm{adj}A=\left(\begin{array}{ccc}14& -9& -1\\ -9& 4& 1\\ -1& 1& -1\end{array}\right),A^{-1}=\frac{1}{-5}\left(\begin{array}{ccc}14& -9& -1\\ -9& 4& 1\\ -1& 1& -1\end{array}\right)=\left(\begin{array}{ccc}-14\mathrm{/}5& 9\mathrm{/}5& 1\mathrm{/}5\\ 9\mathrm{/}5& -4\mathrm{/}5& -1\mathrm{/}5\\ 1\mathrm{/}5& -1\mathrm{/}5& 1\mathrm{/}5\end{array}\right)\mathrm{}$$

Then

$$(\mathrm{adj}A{)}^{-1}=(1\mathrm{/}\det A)A=\frac{1}{-5}\left(\begin{array}{ccc}1& 2& 1\\ 2& 3& 1\\ 1& 1& 5\end{array}\right)=\left(\begin{array}{ccc}-1\mathrm{/}5& -2\mathrm{/}5& -1\mathrm{/}5\\ -2\mathrm{/}5& -3\mathrm{/}5& -1\mathrm{/}5\\ -1\mathrm{/}5& -1\mathrm{/}5& -1\end{array}\right)\mathrm{}$$

And using $\mathrm{adj}(A^{-1})=(\det A^{-1})(A^{-1}{)}^{-1}=(\det A{)}^{-1}A$ gives exactly the same matrix. So the identity holds for this $A$.

 

(ii):

Given

$$A=\left(\begin{array}{ccc}1& 2& 1\\ 2& 3& 1\\ 1& 1& 5\end{array}\right),$$

verify that

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

Step 1. Find $A^{-1}$

We already know from earlier computation that

$$\det (A)=-5$$

Now, let’s find the adjoint of $A$:

Compute cofactors of $A$:

$$\text{Cofactor matrix of }A=\left(\begin{array}{ccc}14& -9& -1\\ -9& 4& 1\\ -1& 1& -1\end{array}\right)\mathrm{}$$

So the adjoint of $A$ is its transpose:

$$\mathrm{adj}(A)=\left(\begin{array}{ccc}14& -9& -1\\ -9& 4& 1\\ -1& 1& -1\end{array}\right)\mathrm{}$$

Hence,

$$A^{-1}=\frac{1}{\det (A)}\mathrm{adj}(A)=\frac{1}{-5}\left(\begin{array}{ccc}14& -9& -1\\ -9& 4& 1\\ -1& 1& -1\end{array}\right)=\left(\begin{array}{ccc}-\frac{14}{5}& \frac{9}{5}& \frac{1}{5}\\ \frac{9}{5}& -\frac{4}{5}& -\frac{1}{5}\\ \frac{1}{5}& -\frac{1}{5}& \frac{1}{5}\end{array}\right)\mathrm{}$$

Step 2. Find $(A^{-1}{)}^{-1}$

By the definition of matrix inverse:

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

Multiplying both sides on the left by $(A^{-1}{)}^{-1}$, we get:

$$(A^{-1}{)}^{-1}(A^{-1})A=(A^{-1}{)}^{-1}I,$$

which simplifies to:

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

Step 3. Verify numerically

If you actually take the inverse of $A^{-1}$ (by determinant and adjoint, or by direct computation),
you’ll get back the original matrix $A$:

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

Final Answer:

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

Question 5.
Evaluate

$$\mid \begin{array}{ccc}x& y& x+y\\ y& x+y& x\\ x+y& x& y\end{array}\mid \mathrm{}$$

Solution:

Let

$$D=\mid \begin{array}{ccc}x& y& x+y\\ y& x+y& x\\ x+y& x& y\end{array}\mid \mathrm{}$$

We will simplify this determinant using column operations.

Step 1: Simplify columns

Let’s apply the operation

$$C_3\to C_3-C_1-C_2$$

(i.e., replace the third column with $C_3-C_1-C_2$)

Compute the new third column entries:

• For first row: $(x+y)-x-y=0$

• For second row: $x-y-(x+y)=x-y-x-y=-2y$

• For third row: $y-(x+y)-x=y-x-y-x=-2x$

So the new determinant becomes:

$$D=\mid \begin{array}{ccc}x& y& 0\\ y& x+y& -2y\\ x+y& x& -2x\end{array}\mid \mathrm{}$$

Step 2: Expand along the first row (since it has a zero)

$$D=x\mid \begin{array}{cc}x+y& -2y\\ x& -2x\end{array}\mid -y\mid \begin{array}{cc}y& -2y\\ x+y& -2x\end{array}\mid \mathrm{}$$

Step 3: Compute each $2\times 2$ determinant

1. For the first one:

$$\mid \begin{array}{cc}x+y& -2y\\ x& -2x\end{array}\mid =(x+y)(-2x)-(-2y)(x)$$

$$=-2x(x+y)+2xy=-2x^2-2xy+2xy=-2x^2\mathrm{.}$$

2. For the second one:

$$\mid \begin{array}{cc}y& -2y\\ x+y& -2x\end{array}\mid =y(-2x)-(-2y)(x+y)$$

$$=-2xy+2y(x+y)=-2xy+2xy+2y^2=2y^2\mathrm{}$$

Step 4: Substitute back

$$D=x(-2x^2)-y(2y^2)=-2x^3-2y^3=-2(x^3+y^3)\mathrm{}$$

Final Answer:

$$\boxed{{\displaystyle \mid \begin{array}{ccc}x& y& x+y\\ y& x+y& x\\ x+y& x& y\end{array}\mid =-2(x^3+y^3)\mathrm{.}}}$$

Question 6.
Evaluate

Solution:

Let

We’ll simplify this determinant using row operations.

Step 1: Simplify rows

Perform the following operations:

Compute the new rows:

• $R_2=(1,x+y,y)-(1,x,y)=(0,y,0)$

• $R_3=(1,x,x+y)-(1,x,y)=(0,0,y)$

So the new determinant becomes:

Step 2: Expand along the first column

Final Answer:

Question 7.
Solve the system

$$\{\begin{array}{l}{\displaystyle \frac{2}{x}}+{\displaystyle \frac{3}{y}}+{\displaystyle \frac{10}{z}}=4,\\ {\displaystyle \frac{4}{x}}-{\displaystyle \frac{6}{y}}+{\displaystyle \frac{5}{z}}=1,\\ {\displaystyle \frac{6}{x}}+{\displaystyle \frac{9}{y}}-{\displaystyle \frac{20}{z}}=2\end{array}$$

Solution (matrix method)

Put $u={\displaystyle \frac{1}{x}},v={\displaystyle \frac{1}{y}},w={\displaystyle \frac{1}{z}}$. Then the system becomes linear in $u,v,w$:

$$\begin{array}{rl}{\displaystyle 2u+3v+10w}& {\displaystyle =4,}\\ {\displaystyle 4u-6v+5w}& {\displaystyle =1,}\\ {\displaystyle 6u+9v-20w}& {\displaystyle =2}\end{array}$$

In matrix form $A\mathbf{t}=\mathbf{b}$ where

$$A=\left(\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right),\mathbf{t}=\left(\begin{array}{c}u\\ v\\ w\end{array}\right),\mathbf{b}=\left(\begin{array}{c}4\\ 1\\ 2\end{array}\right)\mathrm{}$$

Solve $\mathbf{t}=A^{-1}\mathbf{b}$. Doing this (or by Cramer’s rule / row reduction) gives

$$\mathbf{t}=\left(\begin{array}{c}u\\ v\\ w\end{array}\right)=\left(\begin{array}{c}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{array}\right)\mathrm{}$$

So

$$u=\frac{1}{2},v=\frac{1}{3},w=\frac{1}{5}\mathrm{}$$

Convert back to $x,y,z$:

$$x=\frac{1}{u}=2,y=\frac{1}{v}=3,z=\frac{1}{w}=5$$

Check

Substitute $(x,y,z)=(2,3,5)$ into the original equations:

$$\frac{2}{2}+\frac{3}{3}+\frac{10}{5}=1+1+2=4,\frac{4}{2}-\frac{6}{3}+\frac{5}{5}=2-2+1=1$$

$$,\frac{6}{2}+\frac{9}{3}-\frac{20}{5}=3+3-4=2$$

all hold.

Answer: $\boxed{{\displaystyle x=2,y=3,z=5.}}$

Question 8.
If $x,y,z$ are nonzero real numbers, then the inverse of the matrix

$$A=\left(\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right)$$

is which of the following?

(A) $\left(\begin{array}{ccc}{x}^{-1}& 0& 0\\ 0& y^{-1}& 0\\ 0& 0& z^{-1}\end{array}\right)$

(B) $xyz\left(\begin{array}{ccc}{x}^{-1}& 0& 0\\ 0& y^{-1}& 0\\ 0& 0& z^{-1}\end{array}\right)$

(C) ${\displaystyle \frac{1}{xyz}}\left(\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right)$

(D) ${\displaystyle \frac{1}{xyz}}\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$

Answer: (A).

Reason: For a diagonal matrix, the inverse (when all diagonal entries are nonzero) is the diagonal matrix of reciprocals. Check $A\cdot A^{-1}=I$:

$$\left(\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right)\left(\begin{array}{ccc}{x}^{-1}& 0& 0\\ 0& y^{-1}& 0\\ 0& 0& z^{-1}\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\mathrm{}$$

Question 9.
Let

Which of the following is true?
(A) $\det (A)=0$
(B) $\det (A)\in (2,\mathrm{\infty })$
(C) $\det (A)\in (2,4)$
(D) $\det (A)\in [2,4]$

Solution

Compute the determinant by expansion along the first row:

So ${\displaystyle \det A=2(1+{\sin }^2\theta )\mathrm{}}$

Since $0\le {\sin }^2\theta \le 1$, we have

Thus $\det (A)\in [2,4]$. The correct choice is (D).