NCERT Solutions Class 12 Maths Chapter 4 Ex 4.5 | LearnCBSEHub

Linear Equations

NCERT EXERCISE 4.5 • COMPLETE Q1 — Q16

💡 Matrix Method Workflow

A system of equations $AX = B$ is solved as follows:

Consistent: If $|A| \neq 0$, there is a Unique Solution given by $X = A^{-1}B$.
Inconsistent: If $|A| = 0$ and $(\text{adj } A)B \neq O$.
Consistent/Dependent: If $|A| = 0$ and $(\text{adj } A)B = O$.

Question 01 — 06 • Consistency

Examine the consistency of the systems.

Q1: $x + 2y = 2; \ 2x + 3y = 3$ $A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$.
$|A| = 3 – 4 = -1 \neq 0$.

Result: Consistent (Unique Solution)

Q3: $x + 3y = 5; \ 2x + 6y = 8$ $A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$.
$|A| = 6 – 6 = 0$.
Checking $(\text{adj } A)B$: $\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} 30-24 \\ -10+8 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix} \neq O$.

Result: Inconsistent

Question 07 — 14 • Solving Equations

Solve using Matrix Method ($X = A^{-1}B$).

Example Q7: $5x + 2y = 4; \ 7x + 3y = 5$ $A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}, B = \begin{bmatrix} 4 \\ 5 \end{bmatrix}$.
$|A| = 15 – 14 = 1$.
$A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$.
$X = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \end{bmatrix} = \begin{bmatrix} 12-10 \\ -28+25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}$.

Solution: $x = 2, y = -3$

Example Q13 (3×3): Solving for $x, y, z$

$A = \begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{bmatrix}, \ B = \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix}$.
$|A| = 2(4+1) – 3(-2-3) + 3(-1+6) = 10 + 15 + 15 = 40$.
After calculating cofactors and $A^{-1}$, $X = A^{-1}B = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$.

Solution: $x = 1, y = 2, z = -1$

Question 15

Find $A^{-1}$ and solve the system.

Matrix $A$ is identical to the coefficients of the system.

$|A| = -1$.
$A^{-1} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$ (after sign reversal by $|A|$).
$X = A^{-1} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$.

Solution: $x=1, y=2, z=3$

Question 16 • Application

Find cost per kg of Onion ($x$), Wheat ($y$), and Rice ($z$).

Formulating Equations $4x + 3y + 2z = 60$
$2x + 4y + 6z = 90$
$6x + 2y + 3z = 70$
Solving via Matrix Method

$A = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix}, B = \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}$.
$|A| = 4(12-12) – 3(6-36) + 2(4-24) = 0 + 90 – 40 = 50$.
$X = A^{-1}B = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}$.

Cost per kg: Onion = ₹5, Wheat = ₹8, Rice = ₹8