Where Mindful Learning Begins
Where Mindful Learning Begins
This Class 12 Maths NCERT Solutions Chapter 4 Ex 4.5 page covers all 17 questions, solved step-by-step — checking whether a system of linear equations is consistent, and solving systems of two or three variables using the matrix method X=A^{-1}B, exactly the way CBSE Determinants answers are marked.
Any system of linear equations can be written as AX=B, where A is the coefficient matrix, X is the column of unknowns, and B is the column of constants.
If |A|\neq0, A is invertible and the system has a unique solution: X=A^{-1}B=\dfrac{1}{|A|}(\text{adj }A)B.
If |A|=0, check (\text{adj }A)B: if it equals the zero matrix the system is consistent with infinitely many solutions; otherwise the system is inconsistent (no solution).
Writing as AX=B with A=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}: |A| = 1(3)-2(2) = 3-4 = -1 \neq 0
A=\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}: |A| = 2(1)-(-1)(1) = 2+1 = 3 \neq 0
A=\begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}: |A| = 1(6)-3(2) = 6-6 = 0
Since |A|=0, check (\text{adj }A)B, where \text{adj }A = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} and B=\begin{bmatrix} 5 \\ 8 \end{bmatrix}:
(\text{adj }A)B = \begin{bmatrix} 6(5)-3(8) \\ -2(5)+1(8) \end{bmatrix} = \begin{bmatrix} 30-24 \\ -10+8 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix} \neq O
A=\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix},\ B=\begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}. Expanding along R_1: |A| = 1(4)+1(5)+1(1) = 10
Computing all cofactors gives \text{adj }A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}, so A^{-1}=\dfrac{1}{10}\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}
X = A^{-1}B = \dfrac{1}{10}\begin{bmatrix} 16+0+4 \\ -20+0+10 \\ 4+0+6 \end{bmatrix} = \dfrac{1}{10}\begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}
A=\begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{bmatrix},\ B=\begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix}. Expanding along R_1: |A| = 2(5)-3(-5)+3(5) = 10+15+15 = 40
Computing all cofactors gives \text{adj }A = \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}, so A^{-1}=\dfrac{1}{40}\begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}
X = A^{-1}B = \dfrac{1}{40}\begin{bmatrix} 25-12+27 \\ 25+52+3 \\ 25-44-21 \end{bmatrix} = \dfrac{1}{40}\begin{bmatrix} 40 \\ 80 \\ -40 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}
A=\begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix},\ B=\begin{bmatrix} 3 \\ 5 \end{bmatrix}. |A| = 10-6 = 4 \Rightarrow A^{-1}=\dfrac{1}{4}\begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}
X = A^{-1}B = \dfrac{1}{4}\begin{bmatrix} 6-10 \\ -9+25 \end{bmatrix} = \dfrac{1}{4}\begin{bmatrix} -4 \\ 16 \end{bmatrix} = \begin{bmatrix} -1 \\ 4 \end{bmatrix}
A=\begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix},\ B=\begin{bmatrix} -2 \\ 3 \end{bmatrix}. |A| = 8+3 = 11 \Rightarrow A^{-1}=\dfrac{1}{11}\begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}
X = A^{-1}B = \dfrac{1}{11}\begin{bmatrix} -8+3 \\ 6+6 \end{bmatrix} = \dfrac{1}{11}\begin{bmatrix} -5 \\ 12 \end{bmatrix}
A=\begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix},\ B=\begin{bmatrix} 3 \\ 7 \end{bmatrix}. |A| = -20+9 = -11 \Rightarrow A^{-1}=\dfrac{1}{-11}\begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}
X = A^{-1}B = \dfrac{1}{-11}\begin{bmatrix} -15+21 \\ -9+28 \end{bmatrix} = \dfrac{1}{-11}\begin{bmatrix} 6 \\ 19 \end{bmatrix}
A=\begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix},\ B=\begin{bmatrix} 4 \\ 5 \end{bmatrix}. |A| = 15-14 = 1 \Rightarrow A^{-1}=\begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}
X = A^{-1}B = \begin{bmatrix} 12-10 \\ -28+25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}
A=\begin{bmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{bmatrix},\ B=\begin{bmatrix} 1 \\ 3/2 \\ 9 \end{bmatrix}. Expanding along R_1: |A| = 2(13)-1(-5)+1(3) = 26+5+3 = 34
Computing all cofactors gives \text{adj }A = \begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}, so A^{-1}=\dfrac{1}{34}\begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}
X = A^{-1}B = \dfrac{1}{34}\begin{bmatrix} 13+12+9 \\ 5-15+27 \\ 3-9-45 \end{bmatrix} = \dfrac{1}{34}\begin{bmatrix} 34 \\ 17 \\ -51 \end{bmatrix} = \begin{bmatrix} 1 \\ 1/2 \\ -3/2 \end{bmatrix}
A=\begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix},\ B=\begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}. Expanding along R_1: |A| = 1(7)+1(19)+2(-11) = 7+19-22 = 4
Computing all cofactors gives \text{adj }A = \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}, so A^{-1}=\dfrac{1}{4}\begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}
X = A^{-1}B = \dfrac{1}{4}\begin{bmatrix} 49-5-36 \\ -133+5+132 \\ -77+5+84 \end{bmatrix} = \dfrac{1}{4}\begin{bmatrix} 8 \\ 4 \\ 12 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}
A=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix},\ B=\begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}. Expanding along R_1: |A| = 1(7)-1(-3)+1(-1) = 7+3-1 = 9
Computing all cofactors gives \text{adj }A = \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}, so A^{-1}=\dfrac{1}{9}\begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}
X = A^{-1}B = \dfrac{1}{9}\begin{bmatrix} 42-33+0 \\ 18+0+0 \\ -6+33+0 \end{bmatrix} = \dfrac{1}{9}\begin{bmatrix} 9 \\ 18 \\ 27 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}
This is the same coefficient matrix as in Exercise 4.4, where |A|=-1 and A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}.
With B=\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}: X = A^{-1}B = \begin{bmatrix} -2+0+2 \\ 9+2-6 \\ 6+1-4 \end{bmatrix} = \begin{bmatrix} 0 \\ 5 \\ 3 \end{bmatrix}
A=\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix},\ B=\begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}. Expanding along R_1: |A| = 2(0)+3(-2)+5(1) = 0-6+5 = -1
Computing all cofactors gives \text{adj }A = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}, so A^{-1}=\dfrac{1}{-1}\begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}
X = A^{-1}B = \begin{bmatrix} 0-5+6 \\ -22-45+69 \\ -11-25+39 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}
Let the cost per kg of onion, wheat and rice be ₹x, ₹y and ₹z respectively. The three statements translate to:
4x+3y+2z=60;\quad 2x+4y+6z=90;\quad 6x+2y+3z=70
A=\begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix},\ B=\begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}. Expanding along R_1: |A| = 4(0)-3(-30)+2(-20) = 0+90-40 = 50
Computing all cofactors gives \text{adj }A = \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}, so A^{-1}=\dfrac{1}{50}\begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}
X = A^{-1}B = \dfrac{1}{50}\begin{bmatrix} 0-450+700 \\ 1800+0-1400 \\ -1200+900+700 \end{bmatrix} = \dfrac{1}{50}\begin{bmatrix} 250 \\ 400 \\ 400 \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}
For a diagonal matrix, |A| = xyz, and its cofactor matrix is also diagonal, with each cofactor equal to the product of the other two diagonal entries: A_{11}=yz,\ A_{22}=xz,\ A_{33}=xy.
So \text{adj }A = \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix}, and A^{-1} = \dfrac{1}{xyz}\begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix} = \begin{bmatrix} \tfrac{1}{x} & 0 & 0 \\ 0 & \tfrac{1}{y} & 0 \\ 0 & 0 & \tfrac{1}{z} \end{bmatrix}
Since rows 1 and 2 of A already match the identity matrix, computing A^2 only needs the third row worked out:
Third row of A^2, column 1: a(1)+b(0)+(-1)(a) = a-a = 0; column 2: a(0)+b(1)+(-1)(b) = b-b = 0; column 3: a(0)+b(0)+(-1)(-1) = 1
Rows 1 and 2 of A^2 stay [1,0,0] and [0,1,0] since multiplying by those rows just copies A's own first and second rows.
So A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I, which means A is its own inverse.
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