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Matrices and Determinants question

2021 · 22 Jul · Shift 2 · Q31
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Matrices and Determinants question

2021 · 22 Jul · Shift 2 · Q31

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A = [aij] be a real matrix of order 3 ×\times× 3, such that ai1 + ai2 + ai3 = 1, for i = 1, 2, 3. Then, the sum of all the entries of the matrix A3 is equal to :
  1. A
    2
  2. B
    1
  3. C
    3
  4. D
    9
View written solutionFree

Correct answer: C

  1. Let A=[a11a12a13a21a22a23a31a32a33]A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{bmatrix}A=​a11​a21​a31​​a12​a22​a32​​a13​a23​a33​​​ with the given condition ai1+ai2+ai3=1for i=1,2,3.a_{i1}+a_{i2}+a_{i3}=1 \quad \text{for } i=1,2,3.ai1​+ai2​+ai3​=1for i=1,2,3. This means that the sum of entries in each row of AAA is 111.

  2. Define the column vector e=[111].\mathbf{e}=\begin{bmatrix}1\\1\\1\end{bmatrix}.e=​111​​. Then multiplying AAA by e\mathbf{e}e gives

\begin{bmatrix}1\\1\\1\end{bmatrix}=\mathbf{e}.$$ So, $$A\mathbf{e}=\mathbf{e}.$$ 3. Now apply $A$ repeatedly: $$A^2\mathbf{e}=A(A\mathbf{e})=A\mathbf{e}=\mathbf{e},$$ and similarly, $$A^3\mathbf{e}=\mathbf{e}.$$ Thus each row sum of $A^3$ is also $1$. 4. Since $A^3$ is a $3\times 3$ matrix, and each of its $3$ rows has sum $1$, the sum of all entries of $A^3$ is $$1+1+1=3.$$ 5. Therefore, the correct option is $$\boxed{3}.$$
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