CBSE Sample Paper 2025-26 · Section A · 1 mark
Q5: Inverse of the diagonal matrix diag ( 3 , 2 , 5 ) \operatorname{diag}(3, 2, 5) diag ( 3 , 2 , 5 )
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The question
The inverse of the matrix [ 3 0 0 0 2 0 0 0 5 ] \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{bmatrix} 3 0 0 0 2 0 0 0 5
(A) [ 0 0 3 0 2 0 5 0 0 ] \begin{bmatrix} 0 & 0 & 3 \\ 0 & 2 & 0 \\ 5 & 0 & 0 \end{bmatrix} 0 0 5 0 2 0 3 0 0
(B) [ 1 3 0 0 0 1 2 0 0 0 1 5 ] \begin{bmatrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{5} \end{bmatrix} 3 1 0 0 0 2 1 0 0 0 5 1
(C) [ − 1 3 0 0 0 − 1 2 0 0 0 − 1 5 ] \begin{bmatrix} -\frac{1}{3} & 0 & 0 \\ 0 & -\frac{1}{2} & 0 \\ 0 & 0 & -\frac{1}{5} \end{bmatrix} − 3 1 0 0 0 − 2 1 0 0 0 − 5 1
(D) [ − 3 0 0 0 − 2 0 0 0 − 5 ] \begin{bmatrix} -3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -5 \end{bmatrix} − 3 0 0 0 − 2 0 0 0 − 5
Exam answer Show all explanations
Let A A A ∣ A ∣ = 3 × 2 × 5 = 30 ≠ 0 |A| = 3 \times 2 \times 5 = 30 \ne 0 ∣ A ∣ = 3 × 2 × 5 = 30 = 0
Why?
A A A
∣ A ∣ ≠ 0 |A| \ne 0 ∣ A ∣ = 0 A A A A − 1 A^{-1} A − 1
The cofactors are A 11 = 10 A_{11} = 10 A 11 = 10 A 22 = 15 A_{22} = 15 A 22 = 15 A 33 = 6 A_{33} = 6 A 33 = 6
Why?
Each diagonal cofactor is the 2 × 2 2 \times 2 2 × 2
A 11 = 2 × 5 − 0 = 10 A 22 = 3 × 5 − 0 = 15 A 33 = 3 × 2 − 0 = 6 \begin{aligned} A_{11} &= 2 \times 5 - 0 = 10 \\ A_{22} &= 3 \times 5 - 0 = 15 \\ A_{33} &= 3 \times 2 - 0 = 6 \end{aligned} A 11 A 22 A 33 = 2 × 5 − 0 = 10 = 3 × 5 − 0 = 15 = 3 × 2 − 0 = 6 In full, each with the sign + + + i + j i + j i + j
A 11 = + ∣ 2 0 0 5 ∣ = ( 2 ) ( 5 ) − ( 0 ) ( 0 ) = 10 − 0 = 10 A 22 = + ∣ 3 0 0 5 ∣ = ( 3 ) ( 5 ) − ( 0 ) ( 0 ) = 15 − 0 = 15 A 33 = + ∣ 3 0 0 2 ∣ = ( 3 ) ( 2 ) − ( 0 ) ( 0 ) = 6 − 0 = 6 \begin{aligned} A_{11} &= +\begin{vmatrix} 2 & 0 \\ 0 & 5 \end{vmatrix} \\ &= (2)(5) - (0)(0) \\ &= 10 - 0 = 10 \\[4pt] A_{22} &= +\begin{vmatrix} 3 & 0 \\ 0 & 5 \end{vmatrix} \\ &= (3)(5) - (0)(0) \\ &= 15 - 0 = 15 \\[4pt] A_{33} &= +\begin{vmatrix} 3 & 0 \\ 0 & 2 \end{vmatrix} \\ &= (3)(2) - (0)(0) \\ &= 6 - 0 = 6 \end{aligned} A 11 A 22 A 33 = + 2 0 0 5 = ( 2 ) ( 5 ) − ( 0 ) ( 0 ) = 10 − 0 = 10 = + 3 0 0 5 = ( 3 ) ( 5 ) − ( 0 ) ( 0 ) = 15 − 0 = 15 = + 3 0 0 2 = ( 3 ) ( 2 ) − ( 0 ) ( 0 ) = 6 − 0 = 6 For example, covering row 1 and column 2 leaves
M 12 = ∣ 0 0 0 5 ∣ = ( 0 ) ( 5 ) − ( 0 ) ( 0 ) = 0 \begin{aligned} M_{12} &= \begin{vmatrix} 0 & 0 \\ 0 & 5 \end{vmatrix} \\ &= (0)(5) - (0)(0) = 0 \end{aligned} M 12 = 0 0 0 5 = ( 0 ) ( 5 ) − ( 0 ) ( 0 ) = 0 Every other minor contains a whole row or column of zeros, so it is 0. The adjoint is the transpose of the cofactor matrix, and a diagonal matrix is its own transpose.
A − 1 = 1 ∣ A ∣ adj A = 1 30 [ 10 0 0 0 15 0 0 0 6 ] = [ 1 3 0 0 0 1 2 0 0 0 1 5 ] \begin{aligned} A^{-1} &= \frac{1}{|A|}\operatorname{adj} A \\ &= \frac{1}{30}\begin{bmatrix} 10 & 0 & 0 \\ 0 & 15 & 0 \\ 0 & 0 & 6 \end{bmatrix} \\ &= \begin{bmatrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{5} \end{bmatrix} \end{aligned} A − 1 = ∣ A ∣ 1 adj A = 30 1 10 0 0 0 15 0 0 0 6 = 3 1 0 0 0 2 1 0 0 0 5 1 Why?
Divide every entry by 30:
10 30 = 1 3 , 15 30 = 1 2 , 6 30 = 1 5 \frac{10}{30} = \frac{1}{3},\ \ \frac{15}{30} = \frac{1}{2},\ \ \frac{6}{30} = \frac{1}{5} 30 10 = 3 1 , 30 15 = 2 1 , 30 6 = 5 1 Shortcut: the inverse of a diagonal matrix has the reciprocals of the diagonal entries on its diagonal.
Check: 3 × 1 3 = 2 × 1 2 = 5 × 1 5 = 1 3 \times \frac{1}{3} = 2 \times \frac{1}{2} = 5 \times \frac{1}{5} = 1 3 × 3 1 = 2 × 2 1 = 5 × 5 1 = 1 A A − 1 = I AA^{-1} = I A A − 1 = I
Why?
Watch out: option (C) is − A − 1 -A^{-1} − A − 1 − I -I − I − A -A − A A A A I I I
Answer: (B) [ 1 3 0 0 0 1 2 0 0 0 1 5 ] \begin{bmatrix} \dfrac{1}{3} & 0 & 0 \\ 0 & \dfrac{1}{2} & 0 \\ 0 & 0 & \dfrac{1}{5} \end{bmatrix} 3 1 0 0 0 2 1 0 0 0 5 1
Where this comes from Chapter 4, Determinants : Inverse of a diagonal matrix.
Question type: Inverse of a matrix
How to spot it: click to see Type 6. Inverse of a matrix: ∣ A ∣ |A| ∣ A ∣
Spot it: "find the inverse" or "find A − 1 A^{-1} A − 1
Here: The inverse is asked: ∣ A ∣ = 3 × 2 × 5 = 30 ≠ 0 |A| = 3 \times 2 \times 5 = 30 \ne 0 ∣ A ∣ = 3 × 2 × 5 = 30 = 0 1 3 , 1 2 , 1 5 \frac13, \frac12, \frac15 3 1 , 2 1 , 5 1
Quick lesson · Worked NCERT solutions · Interactive lesson
Also uses Chapter 3, Matrices .
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