{"id":19997,"date":"2024-04-15T05:47:00","date_gmt":"2024-04-15T05:47:00","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=19997"},"modified":"2024-04-15T05:47:00","modified_gmt":"2024-04-15T05:47:00","slug":"let-the-eigen-values-of-a-2-a%c2%97-2-matrix-a-be-1-2-with-eigen-vectors-x1-and-x2-respectively-then-the-eigen-values-and-eigen-vectors-of-the-matrix-a2-3a-4i-would-respectively-be-a-2","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/let-the-eigen-values-of-a-2-a%c2%97-2-matrix-a-be-1-2-with-eigen-vectors-x1-and-x2-respectively-then-the-eigen-values-and-eigen-vectors-of-the-matrix-a2-3a-4i-would-respectively-be-a-2\/","title":{"rendered":"Let the eigen values of a 2 \u00c3\u0097 2 matrix A be 1, -2 with eigen vectors x1 and x2 respectively. Then the eigen values and eigen vectors of the matrix A2 – 3A + 4$$I$$ would, respectively, be A. 2, 14; x1, x2 B. 2, 14; x1 + x2, x1 – x2 C. 2, 0; x1, x2 D. 2, 0; x1 + x2, x1 – x2"},"content":{"rendered":"

[amp_mcq option1=”2, 14; x1, x2″ option2=”2, 14; x1 + x2, x1 – x2″ option3=”2, 0; x1, x2″ option4=”2, 0; x1 + x2, x1 – x2″ correct=”option1″]<\/p>\n

The correct answer is $\\boxed{\\text{(B)}}$.<\/p>\n

Let $A$ be a 2×2 matrix with eigenvalues $\\lambda_1=1$ and $\\lambda_2=-2$, and corresponding eigenvectors $x_1$ and $x_2$. Then, we know that $Ax_1=\\lambda_1x_1$ and $Ax_2=\\lambda_2x_2$.<\/p>\n

We can also write this as $A(x_1+x_2)=(\\lambda_1+0)x_1+(\\lambda_2+0)x_2=x_1+x_2$ and $A(x_1-x_2)=(\\lambda_1-0)x_1+(\\lambda_2-0)x_2=x_1-x_2$.<\/p>\n

Therefore, the eigenvalues of $A^2-3A+4I$ are $1+0+4=5$ and $-2+0+4=2$, and the corresponding eigenvectors are $x_1+x_2$ and $x_1-x_2$.<\/p>\n

Here is a more detailed explanation of each option:<\/p>\n