NCERT Class 12 Mathematics — Determinants, Exercise 4.5. All 16 questions solved.

Exercise 4.5 is what the chapter has been building towards. Any system of linear equations can be written as a single matrix equation:

$$\mathrm{A}\mathrm{X} = \mathrm{B}$$

where A holds the coefficients, X the unknowns and B the right-hand sides. If A is invertible, multiply on the left by $\mathrm{A}^{-1}$:

$$\mathrm{X} = \mathrm{A}^{-1}\mathrm{B}$$

and the system is solved in one step. The whole exercise is that formula plus the question of when it applies.

Key insight. $|\mathrm{A}|$ decides everything, and there are three cases, not two:

• $|\mathrm{A}| \neq 0$ — consistent, with exactly one solution, given by $\mathrm{X} = \mathrm{A}^{-1}\mathrm{B}$.
• $|\mathrm{A}| = 0$ and $(\mathrm{adj}\,\mathrm{A})\mathrm{B} \neq \mathrm{O}$ — inconsistent, no solution at all.
• $|\mathrm{A}| = 0$ and $(\mathrm{adj}\,\mathrm{A})\mathrm{B} = \mathrm{O}$ — consistent, with infinitely many solutions.

Stopping at $|\mathrm{A}| = 0$ and declaring the system inconsistent is the commonest error in this exercise. That second test has to be done.

Questions 1 to 6 — examine the consistency of the system.

Question 1

$x + 2y = 2$, $2x + 3y = 3$

Solution.

$$\mathrm{A} = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}, \qquad |\mathrm{A}| = 3 – 4 = -1 \neq 0$$

Since the determinant is non-zero, A is invertible and the system has a unique solution.

Question 2

$2x – y = 5$, $x + y = 4$

Solution.

$$|\mathrm{A}| = \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} = 2 + 1 = 3 \neq 0$$

Question 3

$x + 3y = 5$, $2x + 6y = 8$

Solution.

$$|\mathrm{A}| = \begin{vmatrix} 1 & 3 \\ 2 & 6 \end{vmatrix} = 6 – 6 = 0$$

The determinant vanishes, so the second test is needed.

$$\mathrm{adj}\,\mathrm{A} = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}, \qquad (\mathrm{adj}\,\mathrm{A})\mathrm{B} = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix} \neq \mathrm{O}$$

Geometrically the two equations describe parallel lines: the second has twice the coefficients of the first but not twice its right-hand side.

Question 4

$x + y + z = 1$, $2x + 3y + 2z = 2$, $ax + ay + 2az = 4$

Solution.

$$\mathrm{A} = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2a \end{bmatrix}$$

Expanding, and taking the common factor $a$ out of the third row:

$$|\mathrm{A}| = a\begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 1 & 1 & 2 \end{vmatrix} = a\big[1(6-2) – 1(4-2) + 1(2-3)\big] = a(4 – 2 – 1) = a$$

So $|\mathrm{A}| = a$, which is non-zero provided $a \neq 0$, and the system is then consistent with a unique solution.

It is worth being explicit about the excluded case, since the question does not restrict $a$: if $a = 0$ the third equation reads $0 = 4$, which is impossible, so the system is inconsistent there. The textbook’s answer of “consistent” assumes $a \neq 0$.

Question 5

$3x – y – 2z = 2$, $2y – z = -1$, $3x – 5y = 3$

Solution.

$$\mathrm{A} = \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}$$

$$|\mathrm{A}| = 3(0 – 5) + 1(0 + 3) – 2(0 – 6) = -15 + 3 + 12 = 0$$

So compute $(\mathrm{adj}\,\mathrm{A})\mathrm{B}$ with $\mathrm{B} = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$:

$$(\mathrm{adj}\,\mathrm{A})\mathrm{B} = \begin{bmatrix} -5 \\ -3 \\ -6 \end{bmatrix} \neq \mathrm{O}$$

Question 6

$5x – y + 4z = 5$, $2x + 3y + 5z = 2$, $5x – 2y + 6z = -1$

Solution.

$$|\mathrm{A}| = \begin{vmatrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{vmatrix} = 5(18 + 10) + 1(12 – 25) + 4(-4 – 15)$$

$$= 140 – 13 – 76 = 51 \neq 0$$

Questions 7 to 14 — solve by the matrix method.

Question 7

$5x + 2y = 4$, $7x + 3y = 5$

Solution.

$$\mathrm{A} = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}, \qquad |\mathrm{A}| = 15 – 14 = 1$$

$$\mathrm{A}^{-1} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$$

$$\mathrm{X} = \mathrm{A}^{-1}\mathrm{B} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4 \\ 5 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}$$

Question 8

$2x – y = -2$, $3x + 4y = 3$

Solution. $|\mathrm{A}| = 8 + 3 = 11$.

$$\mathrm{X} = \frac{1}{11}\begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} -2 \\ 3 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -5 \\ 12 \end{bmatrix}$$

Question 9

$4x – 3y = 3$, $3x – 5y = 7$

Solution. $|\mathrm{A}| = -20 + 9 = -11$.

$$\mathrm{X} = \frac{1}{-11}\begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}\begin{bmatrix} 3 \\ 7 \end{bmatrix} = \frac{1}{-11}\begin{bmatrix} 6 \\ 19 \end{bmatrix}$$

Question 10

$5x + 2y = 3$, $3x + 2y = 5$

Solution. $|\mathrm{A}| = 10 – 6 = 4$.

$$\mathrm{X} = \frac{1}{4}\begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}\begin{bmatrix} 3 \\ 5 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} -4 \\ 16 \end{bmatrix}$$

Question 11

$2x + y + z = 1$, $x – 2y – z = \dfrac{3}{2}$, $3y – 5z = 9$

Solution.

$$\mathrm{A} = \begin{bmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{bmatrix}$$

$$|\mathrm{A}| = 2(10 + 3) – 1(-5 – 0) + 1(3 – 0) = 26 + 5 + 3 = 34$$

Computing $\mathrm{A}^{-1}$ from the adjoint and multiplying by $\mathrm{B} = \begin{bmatrix} 1 \\ 3/2 \\ 9 \end{bmatrix}$:

$$\mathrm{X} = \begin{bmatrix} 1 \\ 1/2 \\ -3/2 \end{bmatrix}$$

Substituting back into the third equation as a check: $3(\tfrac12) – 5(-\tfrac32) = \tfrac32 + \tfrac{15}{2} = 9$.

Question 12

$x – y + z = 4$, $2x + y – 3z = 0$, $x + y + z = 2$

Solution.

$$|\mathrm{A}| = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{vmatrix} = 1(1 + 3) + 1(2 + 3) + 1(2 – 1) = 4 + 5 + 1 = 10$$

$$\mathrm{X} = \mathrm{A}^{-1}\mathrm{B} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$$

Question 13

$2x + 3y + 3z = 5$, $x – 2y + z = -4$, $3x – y – 2z = 3$

Solution.

$$|\mathrm{A}| = \begin{vmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{vmatrix} = 2(4 + 1) – 3(-2 – 3) + 3(-1 + 6) = 10 + 15 + 15 = 40$$

$$\mathrm{X} = \mathrm{A}^{-1}\mathrm{B} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$$

Question 14

$x – y + 2z = 7$, $3x + 4y – 5z = -5$, $2x – y + 3z = 12$

Solution.

$$|\mathrm{A}| = \begin{vmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{vmatrix} = 1(12 – 5) + 1(9 + 10) + 2(-3 – 8) = 7 + 19 – 22 = 4$$

$$\mathrm{X} = \mathrm{A}^{-1}\mathrm{B} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}$$

Question 15

If $\mathrm{A} = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$, find $\mathrm{A}^{-1}$. Using $\mathrm{A}^{-1}$ solve the system of equations $2x – 3y + 5z = 11$, $3x + 2y – 4z = -5$, $x + y – 2z = -3$.

Solution. Notice first that the coefficient matrix of the system is A — which is the point of the question.

$$|\mathrm{A}| = 2(-4 + 4) + 3(-6 + 4) + 5(3 – 2) = 0 – 6 + 5 = -1$$

Computing the nine cofactors and transposing gives the adjoint; dividing by $|\mathrm{A}| = -1$ negates it:

$$\mathrm{A}^{-1} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$$

Now apply it to $\mathrm{B} = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$:

$$\mathrm{X} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}\begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} = \begin{bmatrix} 0 – 5 + 6 \\ -22 – 45 + 69 \\ -11 – 25 + 39 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$$

Question 16

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method.

Solution. Let $x$, $y$ and $z$ be the cost per kg of onion, wheat and rice. Reading each sentence as an equation:

$$4x + 3y + 2z = 60$$ $$2x + 4y + 6z = 90$$ $$6x + 2y + 3z = 70$$

$$\mathrm{A} = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix}, \qquad |\mathrm{A}| = 4(12 – 12) – 3(6 – 36) + 2(4 – 24) = 0 + 90 – 40 = 50$$

Since $|\mathrm{A}| = 50 \neq 0$ the system has a unique solution, and $\mathrm{X} = \mathrm{A}^{-1}\mathrm{B}$ gives

$$\mathrm{X} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}$$

Checking against the first condition: $4(5) + 3(8) + 2(8) = 20 + 24 + 16 = 60$.

Common mistakes

• Calling a system inconsistent as soon as $|\mathrm{A}| = 0$. That only rules out a unique solution. Compute $(\mathrm{adj}\,\mathrm{A})\mathrm{B}$: zero means infinitely many solutions, non-zero means none. Questions 3 and 5 both need the second step.
• Writing $\mathrm{X} = \mathrm{B}\mathrm{A}^{-1}$. The inverse multiplies from the left: $\mathrm{X} = \mathrm{A}^{-1}\mathrm{B}$. The other order is not even a valid product for a $3\times3$ times a $3\times1$.
• Losing a variable that is missing from an equation. In question 11 the third equation has no $x$, so its coefficient is $0$, not absent. Every row of A must have three entries.
• Forgetting to divide by the determinant. The adjoint is not the inverse. Question 15 is forgiving here because $|\mathrm{A}| = -1$, but the sign still flips.
• Setting up a word problem with the columns transposed. In question 16 each row is one purchase and each column is one commodity. Getting this the wrong way round produces a system that is solvable but wrong.
• Ignoring a parameter’s special value. In question 4, $|\mathrm{A}| = a$ is non-zero only when $a \neq 0$; at $a = 0$ the third equation becomes $0 = 4$.

Practise next

• Miscellaneous Exercise on Chapter 4 — mixed questions using properties of determinants alongside inverses.
• Chapter 3, Matrices — the operations underlying all of this, worth revisiting if the multiplications felt slow.