MATHEMATICS CLASS- 12
CHAPTER-3
(MATRICES)
Miscellaneous Exercise on Chapter 3
Question 1
If $A$ and $B$ are symmetric matrices, prove that $AB-BA$ is a skew-symmetric matrix.
Solution:
Since $A$ and $B$ are symmetric matrices, we have
$$ A^T=A \quad\text{and}\quad B^T=B. $$Consider the matrix
$$ AB-BA. $$Taking the transpose of both sides,
$$ (AB-BA)^T=(AB)^T-(BA)^T. $$Using the property
$$ (AB)^T=B^TA^T, $$we get
$$ (AB-BA)^T=B^TA^T-A^TB^T. $$Since $A$ and $B$ are symmetric matrices,
$$ A^T=A \quad\text{and}\quad B^T=B, $$therefore,
$$ (AB-BA)^T=BA-AB. $$Now,
$$ BA-AB=-(AB-BA). $$Hence,
$$ (AB-BA)^T=-(AB-BA). $$Since the transpose of the matrix is equal to its negative, $AB-BA$ is a skew-symmetric matrix.
Hence proved.
Question 2
Show that the matrix $B^TAB$ is symmetric or skew-symmetric according as $A$ is symmetric or skew-symmetric.
Solution:
Let
$$ M=B^TAB. $$We shall find the transpose of $M$.
$$ M^T=(B^TAB)^T. $$Using the property
$$ (PQR)^T=R^TQ^TP^T, $$we get
$$ M^T=B^TA^T(B^T)^T. $$Since
$$ (B^T)^T=B, $$therefore,
$$ M^T=B^TA^TB. $$Case 1: When $A$ is a symmetric matrix.
Since $A$ is symmetric,
$$ A^T=A. $$Substituting in the above expression,
$$ M^T=B^TAB. $$But
$$ M=B^TAB. $$Hence,
$$ M^T=M. $$Therefore, $B^TAB$ is a symmetric matrix.
Case 2: When $A$ is a skew-symmetric matrix.
Since $A$ is skew-symmetric,
$$ A^T=-A. $$Substituting this value,
$$ M^T=B^T(-A)B. $$ $$ M^T=-B^TAB. $$But
$$ M=B^TAB. $$Hence,
$$ M^T=-M. $$Therefore, $B^TAB$ is a skew-symmetric matrix.
Hence proved that $B^TAB$ is symmetric or skew-symmetric according as $A$ is symmetric or skew-symmetric.
Question 3
Find the values of $x$, $y$ and $z$ if the matrix
$$ A= \begin{bmatrix} 0&2y&z\\ x&y&-z\\ x&-y&z \end{bmatrix} $$satisfies the equation
$$ A^TA=I. $$Solution:
Since
$$ A^TA=I, $$first find the transpose of matrix $A$.
$$ A^T= \begin{bmatrix} 0&x&x\\ 2y&y&-y\\ z&-z&z \end{bmatrix} $$Now, multiply $A^T$ and $A$.
$$ A^TA= \begin{bmatrix} 0&x&x\\ 2y&y&-y\\ z&-z&z \end{bmatrix} \begin{bmatrix} 0&2y&z\\ x&y&-z\\ x&-y&z \end{bmatrix} $$Multiplying the corresponding rows and columns, we get
$$ A^TA= \begin{bmatrix} 2x^2&0&0\\ 0&6y^2&0\\ 0&0&3z^2 \end{bmatrix} $$But it is given that
$$ A^TA=I= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} $$Equating the corresponding elements of the matrices, we obtain
$$ 2x^2=1, $$ $$ 6y^2=1, $$ $$ 3z^2=1. $$Therefore,
$$ x^2=\frac12, $$ $$ y^2=\frac16, $$ $$ z^2=\frac13. $$Hence,
$$ x=\pm\frac{1}{\sqrt2}, $$ $$ y=\pm\frac{1}{\sqrt6}, $$ $$ z=\pm\frac{1}{\sqrt3}. $$Rationalising the denominators,
$$ x=\pm\frac{\sqrt2}{2}, \qquad y=\pm\frac{\sqrt6}{6}, \qquad z=\pm\frac{\sqrt3}{3}. $$Hence, the required values are
$$ \boxed{ x=\pm\frac{\sqrt2}{2}, \qquad y=\pm\frac{\sqrt6}{6}, \qquad z=\pm\frac{\sqrt3}{3} } $$Question 4
For what values of $x$
$$ \begin{bmatrix} 1&2&1 \end{bmatrix} \begin{bmatrix} 1&2&0\\ 2&0&1\\ 1&0&2 \end{bmatrix} \begin{bmatrix} 0\\ 2\\ x \end{bmatrix} =0? $$Solution:
First, multiply the first two matrices.
$$ \begin{aligned} \begin{bmatrix} 1&2&1 \end{bmatrix} & \begin{bmatrix} 1&2&0\\ 2&0&1\\ 1&0&2 \end{bmatrix} \\[4mm] &= \begin{bmatrix} 1(1)+2(2)+1(1) & 1(2)+2(0)+1(0) & 1(0)+2(1)+1(2) \end{bmatrix} \\[4mm] &= \begin{bmatrix} 6&2&4 \end{bmatrix} \end{aligned} $$Now multiply the resulting row matrix by the given column matrix.
$$ \begin{aligned} \begin{bmatrix} 6&2&4 \end{bmatrix} \begin{bmatrix} 0\\ 2\\ x \end{bmatrix} &= 6(0)+2(2)+4x \\[2mm] &=4+4x. \end{aligned} $$It is given that
$$ 4+4x=0. $$Therefore,
$$ 4x=-4 $$ $$ x=-1. $$Hence, the required value of $x$ is
$$ \boxed{x=-1} $$Question 5
If
$$ A= \begin{bmatrix} 3&1\\ -1&2 \end{bmatrix}, $$show that
$$ A^2-5A+7I=O. $$Solution:
First, find $A^2$.
$$ A^2=A\times A $$ $$ = \begin{bmatrix} 3&1\\ -1&2 \end{bmatrix} \begin{bmatrix} 3&1\\ -1&2 \end{bmatrix} $$Multiplying the matrices, we get
$$ \begin{aligned} A^2 &= \begin{bmatrix} 3(3)+1(-1) & 3(1)+1(2)\\ (-1)(3)+2(-1) & (-1)(1)+2(2) \end{bmatrix} \\[2mm] &= \begin{bmatrix} 8&5\\ -5&3 \end{bmatrix} \end{aligned} $$Now, find $5A$.
$$ 5A= 5 \begin{bmatrix} 3&1\\ -1&2 \end{bmatrix} = \begin{bmatrix} 15&5\\ -5&10 \end{bmatrix} $$Also,
$$ 7I= \begin{bmatrix} 7&0\\ 0&7 \end{bmatrix} $$Now calculate $A^2-5A$.
$$ \begin{aligned} A^2-5A &= \begin{bmatrix} 8&5\\ -5&3 \end{bmatrix} - \begin{bmatrix} 15&5\\ -5&10 \end{bmatrix} \\[2mm] &= \begin{bmatrix} -7&0\\ 0&-7 \end{bmatrix} \end{aligned} $$Adding $7I$, we get
$$ \begin{aligned} A^2-5A+7I &= \begin{bmatrix} -7&0\\ 0&-7 \end{bmatrix} + \begin{bmatrix} 7&0\\ 0&7 \end{bmatrix} \\[2mm] &= \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix} \end{aligned} $$Since
$$ \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix} =O, $$therefore,
$$ \boxed{ A^2-5A+7I=O. } $$Hence proved.
Question 6
If
$$ A= \begin{bmatrix} 1&1&1\\ 0&1&2\\ 0&0&1 \end{bmatrix}, $$show that
$$ A^3-3A^2+3A-I=O. $$Solution:
First, find $A^2$.
$$ A^2=A\times A $$ $$ = \begin{bmatrix} 1&1&1\\ 0&1&2\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&1&1\\ 0&1&2\\ 0&0&1 \end{bmatrix} $$Multiplying the matrices, we get
$$ A^2= \begin{bmatrix} 1&2&4\\ 0&1&4\\ 0&0&1 \end{bmatrix} $$Now, find $A^3$.
$$ A^3=A^2\times A $$ $$ = \begin{bmatrix} 1&2&4\\ 0&1&4\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&1&1\\ 0&1&2\\ 0&0&1 \end{bmatrix} $$Multiplying the matrices, we obtain
$$ A^3= \begin{bmatrix} 1&3&9\\ 0&1&6\\ 0&0&1 \end{bmatrix} $$Next, calculate $3A^2$.
$$ 3A^2= 3 \begin{bmatrix} 1&2&4\\ 0&1&4\\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 3&6&12\\ 0&3&12\\ 0&0&3 \end{bmatrix} $$Also,
$$ 3A= 3 \begin{bmatrix} 1&1&1\\ 0&1&2\\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 3&3&3\\ 0&3&6\\ 0&0&3 \end{bmatrix} $$and
$$ I= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}. $$Now, calculate $A^3-3A^2$.
$$ \begin{aligned} A^3-3A^2 &= \begin{bmatrix} 1&3&9\\ 0&1&6\\ 0&0&1 \end{bmatrix} - \begin{bmatrix} 3&6&12\\ 0&3&12\\ 0&0&3 \end{bmatrix} \\[2mm] &= \begin{bmatrix} -2&-3&-3\\ 0&-2&-6\\ 0&0&-2 \end{bmatrix} \end{aligned} $$Now, add $3A$.
$$ \begin{aligned} A^3-3A^2+3A &= \begin{bmatrix} -2&-3&-3\\ 0&-2&-6\\ 0&0&-2 \end{bmatrix} + \begin{bmatrix} 3&3&3\\ 0&3&6\\ 0&0&3 \end{bmatrix} \\[2mm] &= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \end{aligned} $$Finally, subtract the identity matrix.
$$ \begin{aligned} A^3-3A^2+3A-I &= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} - \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \\[2mm] &= \begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&0&0 \end{bmatrix} \end{aligned} $$Since
$$ \begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&0&0 \end{bmatrix} =O, $$therefore,
$$ \boxed{ A^3-3A^2+3A-I=O. } $$Hence proved.
Question 6
Find $x$, if
$$ \begin{bmatrix} x&-5&-1 \end{bmatrix} \begin{bmatrix} 1&0&2\\ 0&2&1\\ 2&0&3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} =0. $$Solution:
First, multiply the second and the third matrices.
$$ \begin{aligned} \begin{bmatrix} 1&0&2\\ 0&2&1\\ 2&0&3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} &= \begin{bmatrix} 1(x)+0(4)+2(1)\\ 0(x)+2(4)+1(1)\\ 2(x)+0(4)+3(1) \end{bmatrix} \\[2mm] &= \begin{bmatrix} x+2\\ 9\\ 2x+3 \end{bmatrix} \end{aligned} $$Now multiply the first matrix by the above column matrix.
$$ \begin{aligned} \begin{bmatrix} x&-5&-1 \end{bmatrix} \begin{bmatrix} x+2\\ 9\\ 2x+3 \end{bmatrix} &= x(x+2)-5(9)-1(2x+3) \\[2mm] &= x^2+2x-45-2x-3 \\[2mm] &= x^2-48. \end{aligned} $$It is given that
$$ x^2-48=0. $$Therefore,
$$ x^2=48 $$ $$ x=\pm\sqrt{48} =\pm4\sqrt3. $$Hence, the required values of $x$ are
$$ \boxed{x=\pm4\sqrt3.} $$Question 7
A manufacturer produces three products $x$, $y$ and $z$ which he sells in two markets. The annual sales are given below.
| Market | $x$ | $y$ | $z$ |
|---|---|---|---|
| I | 10000 | 2000 | 18000 |
| II | 6000 | 20000 | 8000 |
(a)
If the unit sale prices of $x$, $y$ and $z$ are ₹2.50, ₹1.50 and ₹1.00 respectively, find the total revenue in each market with the help of matrix algebra.
Solution:
The sales matrix is
$$ A= \begin{bmatrix} 10000&2000&18000\\ 6000&20000&8000 \end{bmatrix} $$The price matrix is
$$ P= \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix} $$The total revenue in each market is obtained by multiplying the matrices.
$$ R=AP $$ $$ = \begin{bmatrix} 10000&2000&18000\\ 6000&20000&8000 \end{bmatrix} \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix} $$ $$ = \begin{bmatrix} 10000(2.50)+2000(1.50)+18000(1.00)\\ 6000(2.50)+20000(1.50)+8000(1.00) \end{bmatrix} $$ $$ = \begin{bmatrix} 25000+3000+18000\\ 15000+30000+8000 \end{bmatrix} $$ $$ = \begin{bmatrix} 46000\\ 53000 \end{bmatrix} $$Hence, the total revenue is
- Market I = ₹46,000
- Market II = ₹53,000
(b)
If the unit costs of the above three commodities are ₹2.00, ₹1.00 and 50 paise respectively, find the gross profit.
Solution:
The cost matrix is
$$ C= \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix} $$The total cost in each market is
$$ AC = \begin{bmatrix} 10000&2000&18000\\ 6000&20000&8000 \end{bmatrix} \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix} $$ $$ = \begin{bmatrix} 10000(2.00)+2000(1.00)+18000(0.50)\\ 6000(2.00)+20000(1.00)+8000(0.50) \end{bmatrix} $$ $$ = \begin{bmatrix} 20000+2000+9000\\ 12000+20000+4000 \end{bmatrix} $$ $$ = \begin{bmatrix} 31000\\ 36000 \end{bmatrix} $$The gross profit is obtained by subtracting the total cost from the total revenue.
$$ \begin{aligned} \text{Market I Profit} &=46000-31000 \\ &=15000. \end{aligned} $$ $$ \begin{aligned} \text{Market II Profit} &=53000-36000 \\ &=17000. \end{aligned} $$Therefore, the total gross profit is
$$ 15000+17000=32000. $$Hence,
- Gross Profit in Market I = ₹15,000
- Gross Profit in Market II = ₹17,000
- Total Gross Profit = ₹32,000
Question 8
Find the matrix $X$ such that
$$ X \begin{bmatrix} 1&2&3\\ 4&5&6 \end{bmatrix} = \begin{bmatrix} -7&-8&-9\\ 2&4&6 \end{bmatrix}. $$Solution:
Let
$$ X= \begin{bmatrix} a&b\\ c&d \end{bmatrix}. $$Then,
$$ \begin{bmatrix} a&b\\ c&d \end{bmatrix} \begin{bmatrix} 1&2&3\\ 4&5&6 \end{bmatrix} = \begin{bmatrix} -7&-8&-9\\ 2&4&6 \end{bmatrix}. $$Multiplying the matrices, we get
$$ \begin{bmatrix} a+4b&2a+5b&3a+6b\\ c+4d&2c+5d&3c+6d \end{bmatrix} = \begin{bmatrix} -7&-8&-9\\ 2&4&6 \end{bmatrix}. $$Comparing the corresponding elements,
From the first row,
$$ a+4b=-7 $$ $$ 2a+5b=-8 $$Multiplying the first equation by $2$,
$$ 2a+8b=-14. $$Subtracting the second equation,
$$ 3b=-6 $$ $$ b=-2. $$Substituting in the first equation,
$$ a+4(-2)=-7 $$ $$ a=1. $$From the second row,
$$ c+4d=2 $$ $$ 2c+5d=4. $$Multiplying the first equation by $2$,
$$ 2c+8d=4. $$Subtracting the second equation,
$$ 3d=0 $$ $$ d=0. $$Substituting in the first equation,
$$ c=2. $$Hence,
$$ X= \begin{bmatrix} 1&-2\\ 2&0 \end{bmatrix}. $$Verification:
$$ \begin{aligned} \begin{bmatrix} 1&-2\\ 2&0 \end{bmatrix} \begin{bmatrix} 1&2&3\\ 4&5&6 \end{bmatrix} &= \begin{bmatrix} 1-8&2-10&3-12\\ 2&4&6 \end{bmatrix} \\[2mm] &= \begin{bmatrix} -7&-8&-9\\ 2&4&6 \end{bmatrix}. \end{aligned} $$Hence, the required matrix is
$$ \boxed{ X= \begin{bmatrix} 1&-2\\ 2&0 \end{bmatrix} } $$Question 9
If
$$ A= \begin{bmatrix} \alpha&\beta\\ \gamma&-\alpha \end{bmatrix} $$is such that
$$ A^2=I, $$then
(A) $1+\alpha^2+\beta\gamma=0$
(B) $1-\alpha^2+\beta\gamma=0$
(C) $1-\alpha^2-\beta\gamma=0$
(D) $1+\alpha^2-\beta\gamma=0$
Solution:
Calculate $A^2$.
$$ A^2= \begin{bmatrix} \alpha&\beta\\ \gamma&-\alpha \end{bmatrix} \begin{bmatrix} \alpha&\beta\\ \gamma&-\alpha \end{bmatrix} $$ $$ = \begin{bmatrix} \alpha^2+\beta\gamma&0\\ 0&\alpha^2+\beta\gamma \end{bmatrix}. $$Since
$$ A^2=I= \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}, $$therefore,
$$ \alpha^2+\beta\gamma=1. $$Rearranging,
$$ 1-\alpha^2-\beta\gamma=0. $$Correct Answer: (C).
Question 10
If the matrix $A$ is both symmetric and skew-symmetric, then
(A) $A$ is a diagonal matrix
(B) $A$ is a zero matrix
(C) $A$ is a square matrix
(D) None of these
Solution:
If $A$ is symmetric, then
$$ A^T=A. $$If $A$ is skew-symmetric, then
$$ A^T=-A. $$Therefore,
$$ A=-A. $$Hence,
$$ 2A=0 $$ $$ A=0. $$Thus, $A$ is a zero matrix.
Correct Answer: (B).
Question 11
If $A$ is a square matrix such that
$$ A^2=A, $$then $(I+A)^3-7A$ is equal to
(A) $A$
(B) $I-A$
(C) $I$
(D) $3A$
Solution:
Expand $(I+A)^3$.
$$ (I+A)^3 = I+3A+3A^2+A^3. $$Since
$$ A^2=A, $$we have
$$ A^3=A\cdot A^2=A\cdot A=A^2=A. $$Therefore,
$$ (I+A)^3 = I+3A+3A+A = I+7A. $$Hence,
$$ (I+A)^3-7A = I+7A-7A = I. $$Correct Answer: (C).