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Chapter 4: Problem 282

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) \(-2\)

Short Answer

Expert verified
The correct value of \(a\) for the system of equations to have infinite solutions is \(a=-1\).

Step by step solution

01

Represent the system of equations in matrix form

The given system of equations can be represented as: \[ \left| \begin{array}{c c c} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{array} \right| \] Next, we need to find the determinant of the matrix.
02

Compute the determinant of the matrix

To find the determinant of the matrix, we first select the first row to find the co-factors and then compute the determinant using cofactor expansion. The determinant of the given matrix is: \[ \begin{vmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix} = 1\begin{vmatrix}1 & a \\ 0 & 1\end{vmatrix} - a\begin{vmatrix}0 & a \\ a & 1\end{vmatrix} \] Simplifying, we get the determinant as: \(Det = 1(1) - a(0 - a^2) = 1 + a^3\)
03

Determine the condition for infinite solutions

To have infinite solutions, the determinant of the matrix must be zero. Thus, \(Det = 1 + a^3 = 0\) Now, we will find the value of 'a' that satisfies this equation.
04

Test the given options for 'a'

Now, we will test the given options for 'a' to find which one satisfies the condition for infinite solutions. (a) 0: \(1 + (0)^3 = 1\) (This doesn't satisfy the condition for infinite solutions) (b) 1: \(1 + (1)^3 = 2\) (This doesn't satisfy the condition for infinite solutions) (c) -1: \(1 + (-1)^3 = 0\) (This satisfies the condition for infinite solutions) (d) -2: \(1 + (-2)^3 = -7\) (This doesn't satisfy the condition for infinite solutions) Since option (c) satisfies the condition for infinite solutions, the correct answer is: \[\boxed{a = -1}\]

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Most popular questions from this chapter

If \(f(x)=\left|\begin{array}{ccc}\sec x & \cos x & \sec ^{2} x+\cos x \operatorname{cosec}^{2} x \\ \cos ^{2} x & \cos ^{2} x & \operatorname{cosec}^{2} x \\ 1 & \cos ^{2} x & \operatorname{cosec}^{2} x\end{array}\right|\) then \({ }^{(\pi / 2)} \int_{0} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\ldots\) (a) \((1 / 3)-(\pi / 3)\) (b) \((1 / 3)-(\pi / 4)\) (c) \((2 / 3)+(\pi / 3)\) (d) \((4 / 3)-(\pi / 4)\)

If $$ \mathrm{A}=\left|\begin{array}{cc} 3 & 1 \\ -9 & -3 \end{array}\right| $$ then \(\mathrm{I}+2 \mathrm{~A}+3 \mathrm{~A}^{2}+\ldots \ldots \infty=\ldots\) (a) \(\mid\)\begin{tabular}{cc|c|cc|c|cc|c|cc} 9 & 1 & (b) & 4 & 1 & (c) & \(\begin{gathered}7 & 2 \\ -18 & -5\end{gathered} \mid\) (d) \(\left|\begin{array}{cc}7 & 2 \\ -5 & -18\end{array}\right|\) \end{tabular}

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) \((7 / 19)\)

\(\left|\begin{array}{ccc}\sqrt{1} 1+\sqrt{3} & \sqrt{2} 0 & \sqrt{5} \\\ \sqrt{15}+\sqrt{2} 2 & \sqrt{25} & \sqrt{10} \\ 3+\sqrt{5} 5 & \sqrt{1} 5 & \sqrt{25}\end{array}\right|=\ldots \ldots\) (a) \(5(5 \sqrt{3}-3 \sqrt{2})\) (b) \(5(3 \sqrt{2}+5 \sqrt{3})\) (c) \(-5(5 \sqrt{3}+3 \sqrt{2})\) (d) \(5(3 \sqrt{2}-5 \sqrt{3})\)

If $$ A=\left|\begin{array}{cc} {[(-1+i \sqrt{3}) /(2 i)]} & {[(-1-i \sqrt{3}) /(2 i)]} \\ {[(1+i \sqrt{3}) /(2 i)]} & {[(1-i \sqrt{3}) /(2 i)]} \end{array}\right| $$ \(\mathrm{i}=\sqrt{(-1)}\) and \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}+2\) then \(\mathrm{f}(\mathrm{A})=\) (a) $$ \left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right| $$ (b) \([(3-\mathrm{i} \sqrt{3}) / 2]\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|\) (c) \([(5-\mathrm{t} \sqrt{3}) / 2]\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|\) (d) \(\quad(2+\mathrm{i} \sqrt{3})\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|\)
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