Advertisements
Advertisements
Question
Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P wants to award ₹x each, ₹y each and ₹z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹2,200. School Q wants to spend ₹3,100 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each values is ₹1,200, using matrices, find the award money for each value.
Apart from these three values, suggest one more value which should be considered for award.
Solution
Let the award money given for Tolerance, Kindness and Leadership be ₹x, ₹y and ₹z respectively.
Since, the total cash award is ₹1,200.
∴ x + y + z = 1200 ....(1)
Award money given by school P is ₹2,200.
∴ 3x + 2y + z = 2200 ....(2)
Award money given by school Q is ₹3,100.
∴ 4x + y + 3z = 3100 ....(3)
The above system of equations can be written in matrix form AX = B as
\[\begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1200 \\ 2200 \\ 3100\end{bmatrix}\]
\[\text{ Where, }A = \begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}1200 \\ 2200 \\ 3100\end{bmatrix}\]
Now,
\[\left| A \right| = \begin{vmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{vmatrix}\]
\[ = 1\left( 6 - 1 \right) - 1\left( 9 - 4 \right) + 1(3 - 8)\]
\[ = 5 - 5 - 5\]
\[ = - 5\]
\[\text{ Let }C_{ij}\text{ be the cofactors of elements }a_{ij}\text{ in }A = \left[ a_{ij} \right] .\text{ Then,} \]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & 1 \\ 1 & 3\end{vmatrix} = 5, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & 1 \\ 4 & 3\end{vmatrix} = - 5, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 2 \\ 4 & 1\end{vmatrix} = - 5\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & 3\end{vmatrix} = - 2 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 4 & 3\end{vmatrix} = - 1 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 4 & 1\end{vmatrix} = 3\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = - 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 3 & 1\end{vmatrix} = 2, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 3 & 2\end{vmatrix} = - 1\]
\[adj A = \begin{bmatrix}5 & - 5 & - 5 \\ - 2 & - 1 & 3 \\ - 1 & 2 & - 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 5}\begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\begin{bmatrix}1200 \\ 2200 \\ 3100\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}6000 - 4400 - 3100 \\ - 6000 - 2200 + 6200 \\ - 6000 + 6600 - 3100\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}- 1500 \\ - 2000 \\ - 2500\end{bmatrix}\]
\[ \Rightarrow x = \frac{- 1500}{- 5}, y = \frac{- 2000}{- 5}\text{ and }z = \frac{- 2500}{- 5}\]
\[ \therefore x = 300, y = 400\text{ and }z = 500 .\]
Hence, the award money for each value of Tolerance, Kindness and Leadership is ₹300, ₹400 and ₹500.
One more value which should be considered for award is Honesty.
APPEARS IN
RELATED QUESTIONS
If `|[x+1,x-1],[x-3,x+2]|=|[4,-1],[1,3]|`, then write the value of x.
Find the value of x, if
\[\begin{vmatrix}3 & x \\ x & 1\end{vmatrix} = \begin{vmatrix}3 & 2 \\ 4 & 1\end{vmatrix}\]
Evaluate the following determinant:
\[\begin{vmatrix}1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin\alpha & \cos\alpha & \cos(\alpha + \delta) \\ \sin\beta & \cos\beta & \cos(\beta + \delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma + \delta)\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin^2 23^\circ & \sin^2 67^\circ & \cos180^\circ \\ - \sin^2 67^\circ & - \sin^2 23^\circ & \cos^2 180^\circ \\ \cos180^\circ & \sin^2 23^\circ & \sin^2 67^\circ\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sqrt{23} + \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} + \sqrt{46} & 5 & \sqrt{10} \\ 3 + \sqrt{115} & \sqrt{15} & 5\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & - \cos y\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}a & b + c & a^2 \\ b & c + a & b^2 \\ c & a + b & c^2\end{vmatrix}\]
\[\begin{vmatrix}0 & b^2 a & c^2 a \\ a^2 b & 0 & c^2 b \\ a^2 c & b^2 c & 0\end{vmatrix} = 2 a^3 b^3 c^3\]
Prove that
\[\begin{vmatrix}- bc & b^2 + bc & c^2 + bc \\ a^2 + ac & - ac & c^2 + ac \\ a^2 + ab & b^2 + ab & - ab\end{vmatrix} = \left( ab + bc + ca \right)^3\]
\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + a & a \\ 1 & 1 & 1 + a\end{vmatrix} = a^3 + 3 a^2\]
Solve the following determinant equation:
If a, b, c are real numbers such that
\[\begin{vmatrix}b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a\end{vmatrix} = 0\] , then show that either
\[a + b + c = 0 \text{ or, } a = b = c\]
Using determinants show that the following points are collinear:
(2, 3), (−1, −2) and (5, 8)
Find values of k, if area of triangle is 4 square units whose vertices are
(k, 0), (4, 0), (0, 2)
x+ y = 5
y + z = 3
x + z = 4
3x − y + 2z = 3
2x + y + 3z = 5
x − 2y − z = 1
3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20.
State whether the matrix
\[\begin{bmatrix}2 & 3 \\ 6 & 4\end{bmatrix}\] is singular or non-singular.
If \[A = \begin{bmatrix}0 & i \\ i & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\] , find the value of |A| + |B|.
If w is an imaginary cube root of unity, find the value of \[\begin{vmatrix}1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w\end{vmatrix}\]
If \[A = \left[ a_{ij} \right]\] is a 3 × 3 diagonal matrix such that a11 = 1, a22 = 2 a33 = 3, then find |A|.
If \[\begin{vmatrix}2x + 5 & 3 \\ 5x + 2 & 9\end{vmatrix} = 0\]
If \[\begin{vmatrix}2x & x + 3 \\ 2\left( x + 1 \right) & x + 1\end{vmatrix} = \begin{vmatrix}1 & 5 \\ 3 & 3\end{vmatrix}\], then write the value of x.
If \[\begin{vmatrix}3x & 7 \\ - 2 & 4\end{vmatrix} = \begin{vmatrix}8 & 7 \\ 6 & 4\end{vmatrix}\] , find the value of x.
If \[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix}\] , write the value of x.
Let \[\begin{vmatrix}x & 2 & x \\ x^2 & x & 6 \\ x & x & 6\end{vmatrix} = a x^4 + b x^3 + c x^2 + dx + e\]
Then, the value of \[5a + 4b + 3c + 2d + e\] is equal to
If \[x, y \in \mathbb{R}\], then the determinant
Solve the following system of equations by matrix method:
3x + y = 7
5x + 3y = 12
Show that the following systems of linear equations is consistent and also find their solutions:
x − y + z = 3
2x + y − z = 2
−x −2y + 2z = 1
Let a, b, c be positive real numbers. The following system of equations in x, y and z
(a) no solution
(b) unique solution
(c) infinitely many solutions
(d) finitely many solutions
The value of x, y, z for the following system of equations x + y + z = 6, x − y+ 2z = 5, 2x + y − z = 1 are ______
If A = `[(1, -1, 2),(3, 0, -2),(1, 0, 3)]`, verify that A(adj A) = (adj A)A
Show that if the determinant ∆ = `|(3, -2, sin3theta),(-7, 8, cos2theta),(-11, 14, 2)|` = 0, then sinθ = 0 or `1/2`.
In system of equations, if inverse of matrix of coefficients A is multiplied by right side constant B vector then resultant will be?
If `|(x + a, beta, y),(a, x + beta, y),(a, beta, x + y)|` = 0, then 'x' is equal to
The number of real values λ, such that the system of linear equations 2x – 3y + 5z = 9, x + 3y – z = –18 and 3x – y + (λ2 – |λ|z) = 16 has no solution, is ______.