Using the matrix method, solve the following system of linear equations : 2x+3y+10/z=4,\ 4x-6y+5z=1,\ 6x+9y-20/z=2.

MathsCBSE Class 12 · Sample paper, session 2023-24 · 5 marks · Long answer

CBSE Class 12 Mathematics sample paper 2023-24: question 34

From the official CBSE sample question paper for 2023-24, section D, question 34, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34 · 5 marks · Long answerMatrices

Using the matrix method, solve the following system of linear equations :
$\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2$ .

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

Treat $1/ x$ , $1/ y$ , $1/ z$ as the unknowns so the equations become linear.

Hint 2

Write the system as $A X = B$ , check $∣ A ∣ \neq = 0$ and find adj $A$ from the cofactors.

Hint 3

Compute $X = A^{- 1} B$ , then take reciprocals of the entries.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Let $u = \frac{1}{x}, v = \frac{1}{y}, w = \frac{1}{z}$ . The system becomes $A X = B$ , where $A = 246 3 - 6 9 105 - 20$ , $X = 1/ x 1/ y 1/ z$ and $B = 412$ .
½ mark
$∣ A ∣ = 2 (120 - 45) - 3 (- 80 - 30) + 10 (36 + 36) = 150 + 330 + 720 = 1200 \neq = 0$ , so $A^{- 1}$ exists.
½ mark
Cofactors: $A_{11} = 75, A_{12} = 110, A_{13} = 72$ ; $A_{21} = 150, A_{22} = - 100, A_{23} = 0$ ; $A_{31} = 75, A_{32} = 30, A_{33} = - 24$ .
So $adj A = 7511072 150 - 100 0 7530 - 24$ .
1½ marks
$A^{- 1} = \frac{1}{∣ A ∣} adj A = \frac{1}{1200} 7511072 150 - 100 0 7530 - 24$ .
½ mark
$X = A^{- 1} B = \frac{1}{1200} 7511072 150 - 100 0 7530 - 24 412$ .
½ mark
$= \frac{1}{1200} 300 + 150 + 150 440 - 100 + 60 288 + 0 - 48 = \frac{1}{1200} 600400240 = 1/2 1/3 1/5$ .
½ mark
So $\frac{1}{x} = \frac{1}{2}, \frac{1}{y} = \frac{1}{3}, \frac{1}{z} = \frac{1}{5}$ . Hence $x = 2, y = 3, z = 5$ .
1 mark

Answer: x = 2, y = 3, z = 5

Where the marks are

Writes the system as $A X = B$ with $X$ in terms of $1/ x, 1/ y, 1/ z$ (½ mark)
Finds $∣ A ∣ = 1200 \neq = 0$ (½ mark)
Finds the correct adjoint of $A$ (1½ marks)
Writes $A^{- 1} = \frac{1}{1200} adj A$ (½ mark)
Uses $X = A^{- 1} B$ (½ mark)
Multiplies correctly to get $1/2, 1/3, 1/5$ (½ mark)
States $x = 2, y = 3, z = 5$ (1 mark)

Revise the chapter

Chapter: Matrices (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Matrices, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24: every question, by section.

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CBSE Class 12 Mathematics sample paper 2023-24: question 34

Hints

Step-by-step solution

Revise the chapter

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