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Chapter 4: Problem 296

Suppose a matrix A satisfies \(\mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}=0 .\) If \(\mathrm{A}^{5}=\mathrm{aA}+\mathrm{bI}\) then the value of \(2 \mathrm{a}-3 \mathrm{~b}\) must be (a) 4135 (b) 1435 (c) 1453 (d) 3145

Short Answer

Expert verified
The correct answer to the problem is 98, which is not among the given options.

Step by step solution

01

Rewrite the equation in terms of A

First, we'll rewrite the given equation \(A^2 - 5A + 7I = 0\) in terms of A: \(A^2 - 5A = -7I\)
02

Express A^3 using given equation for A

Now let's compute the expression for \(A^3\) using the equation we found in Step 1: \(A^3 = A(A^2 - 5A) = -A(7I) = -7A\)
03

Express A^4 using given equation for A

Next, we'll compute the expression for \(A^4\) using the equation in Step 1 and the expression for \(A^3\): \(A^4 = -7A^2\)
04

Express A^5 using given equation for A

Now, let's compute the expression for \(A^5\) using the equation in Step 1 and the expression for \(A^4\): \(A^5 = A(A^4) = A(-7A^2) = -7A^3\) Using the expression for \(A^3\) we found in Step 2: \(A^5 = -7(-7A) = 49A\)
05

Find a and b

Now that we've found the expression for \(A^5\), we can find the values for a and b from the given expression \(A^5 = aA + bI\): Comparing our expression for \(A^5\) to the given expression, we find that: \(a = 49\) and \(b = 0\)
06

Calculate 2a - 3b

Finally, we can substitute the values of a and b we found in Step 5 into the expression \(2a - 3b\): \(2a - 3b = 2(49) - 3(0) = 98\) So, none of the given answer choices (a), (b), (c), or (d) match the correct answer of 98.

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Most popular questions from this chapter

The value of the determinant $$ \left|\begin{array}{ccc} \sin \theta & \cos \theta & \sin 2 \theta \\ \sin [\theta+(2 \pi / 3)] & \cos [\theta+(2 \pi / 3)] & \sin [2 \theta+(4 \pi / 3)] \\ \sin [\theta-(2 \pi / 3)] & \cos [\theta-(2 \pi / 3)] & \sin [2 \theta-(4 \pi / 3)] \end{array}\right| $$ is (a) 0 (b) \(2 \sin \theta\) \(1-\sin 2 \theta \quad\) (d) \(-2 \cos \theta\)

If the system of equations \(x+a y=0, a z+y=0, a x+z=0\) has infinite number of solutions then \(\mathrm{a}=\) (a) 0 (b) 1 (c) \(-1\) (d) \(-2\)

The value of \(\left|\begin{array}{cc}\log _{3} 1024 & \log _{8} 3 \\ \log _{3} 8 & \log _{4} 9\end{array}\right| \times\left|\begin{array}{ll}\log _{2} 3 & \log _{4} 3 \\ \log _{3} 4 & \log _{3} 4\end{array}\right|=\ldots\) (a) 6 (b) 9 (c) 10 (d) 12

If \(z\) is a complex number and \(a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\) are all real, then $$ \left|\begin{array}{lll} a_{1} z+b_{1} z & a_{2} z+b_{2} z & a_{3} z+b_{3} z \\ b_{1} z+a_{1} z & b_{2} z+a_{2} z & b_{3} z+a_{3} z \\ b_{1} z+a_{1} & b_{2} z+a_{2} & b_{3} z+a_{3} \end{array}\right|=\ldots $$ (a) \(|\underline{z}|^{2}\) (b) \(\left(a_{1} a_{2} a_{3}+b_{1} b_{2} b_{3}\right)^{2}|z|^{2}\) (c) c (d) 0

Let $$ A=\left|\begin{array}{ccc} 4 & 6 & 6 \\ 1 & 3 & 2 \\ -1 & -5 & -2 \end{array}\right| $$ If \(\mathrm{q}\) is the angle between two non-zero column vectors \(\mathrm{X}\) such that \(\mathrm{AX}=\lambda \mathrm{X}\) for some scalar \(\lambda\), then \(\tan \theta=\) (a) \([7 /\\{\sqrt{(202)\\}]}\) (b) \((\sqrt{3} / 19)\) (c) \(\sqrt{[} 3 /(202)]\) (d) \((7 / 19)\)
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