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  1. Asked: 3 years agoIn:

    Find the value of k for A2 – kA – 14I = 0, if A = \(\begin{bmatrix} 2& 5 \\ 4 & 3 \end{bmatrix}\)
    1. -5
    2. 5
    3. 3
    4. -3

    7d1a0

    Correct Answer - Option 2 : 5 Calculation:A2 = A × A⇒ A2 = \(\begin{bmatrix} 2& 5 \\ 4 & 3 \end{bmatrix}\) × \(\begin{bmatrix} 2& 5 \\ 4 & 3 \end{bmatrix}\)⇒ A2 = \(\begin{bmatrix} 24& 25 \\ 20 & 29 \end{bmatrix}\)Given A satisfy the equation A2 - kA - 14I = 0⇒ \(\begin{bmatrRead more

    Correct Answer – Option 2 : 5

     

    Calculation:

    A2 = A × A

    ⇒ A2 = \(\begin{bmatrix} 2& 5 \\ 4 & 3 \end{bmatrix}\) × \(\begin{bmatrix} 2& 5 \\ 4 & 3 \end{bmatrix}\)

    ⇒ A2 = \(\begin{bmatrix} 24& 25 \\ 20 & 29 \end{bmatrix}\)

    Given A satisfy the equation 

    A2 – kA – 14I = 0

    ⇒ \(\begin{bmatrix} 24& 25 \\ 20 & 29 \end{bmatrix}\) – k \(\begin{bmatrix} 2& 5 \\ 4 & 3 \end{bmatrix}\) – 14 \(\begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 0& 0 \\ 0 & 0 \end{bmatrix}\)

    ⇒ \(\begin{bmatrix} 24-2k-14&25-5k\\20-4k&29-3k-14\end{bmatrix}\) = \(\begin{bmatrix} 0& 0 \\ 0 & 0 \end{bmatrix}\)

    ⇒ 25 – 5k = 0

    ⇒ k = 5

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  2. Asked: 3 years agoIn:

    The rank of Matrix \(A =\left[ {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&1&1&0\\ 0&1&1&1&0\\ 0&1&1&1&0\\ 1&0&0&0&1 \end{array}} \right]\) is _______.
    1. 2
    2. 3
    3. 4
    4. 5

    299af

    Correct Answer - Option 1 : 2 \(A = \;\left[ {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&1&1&0\\ 0&1&1&1&0\\ 0&1&1&1&0\\ 1&0&0&0&1 \end{array}} \right]\)  \(\left| {\rm{A}} \right| = {\rm{\;}}\left| {\begin{array}{*{20}{c}}Read more

    Correct Answer – Option 1 : 2

     \(A = \;\left[ {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&1&1&0\\ 0&1&1&1&0\\ 0&1&1&1&0\\ 1&0&0&0&1 \end{array}} \right]\)  

    \(\left| {\rm{A}} \right| = {\rm{\;}}\left| {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&1&1&0\\ 0&1&1&1&0\\ 0&1&1&1&0\\ 1&0&0&0&1 \end{array}} \right|\)

    R5 → R5 – R1

    R3 → R3 – R2

    R4 → R4 – R2

    \(\left| {\rm{A}} \right| = {\rm{\;}}\left| {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&1&1&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0 \end{array}} \right|\)

     Rank of the matrix is 2

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  3. Asked: 3 years agoIn:

    If \(\begin{bmatrix} 1 & -4 & 3 \\\ 0 & 6 & -7 \\\ 2 & 4 & λ \end{bmatrix}\) is not an invertible matrix, then what is the value of λ ?

    f4b09

    Correct Answer - Option 3 : -8Concept:If A is not an invertible matrix then |A| = 0If A is a singular matrix then |A| = 0Calculation:Given A =  \(\begin{bmatrix} 1 & -4 & 3 \\\ 0 & 6 & -7 \\\ 2 & 4 & λ \end{bmatrix}\) is not an invertible matrix.⇒ A is a singular matrix.⇒ |A|Read more

    Correct Answer – Option 3 : -8

    Concept:

    If A is not an invertible matrix then |A| = 0

    If A is a singular matrix then |A| = 0

    Calculation:

    Given A =  \(\begin{bmatrix} 1 & -4 & 3 \\\ 0 & 6 & -7 \\\ 2 & 4 & λ \end{bmatrix}\) is not an invertible matrix.

    ⇒ A is a singular matrix.

    ⇒ |A| = 0

    ⇒ \(\begin{vmatrix} 1 & -4 & 3 \\\ 0 & 6 & -7 \\\ 2 & 4 & λ \end{vmatrix}\) = 0

    ⇒ 1(\(\rm 6\lambda + 28 \)) + 4 (0 + 14) + 3 (0 – 12) = 0

    ⇒  \(\rm 6\lambda + 28 + 56 – 36 = 0\)

    ⇒ \(\rm 6\lambda + 48 = 0\)

    ⇒ \(\rm \lambda = – 8\)

    Hence, if \(\begin{bmatrix} 1 & -4 & 3 \\\ 0 & 6 & -7 \\\ 2 & 4 & λ \end{bmatrix}\) is not an invertible matrix, then the value of λ = – 8

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  4. Asked: 3 years agoIn:

    If the A = \(\begin{bmatrix} -1& 3 \\ 4 & -2 \end{bmatrix}\) satisfies equation A2 + 3A – 10I = 0, find A-1.
    1. \({1\over10}\begin{bmatrix} -2& -3 \\ 4 & 1 \end{bmatrix}\)
    2. \({1\over10}\begin{bmatrix} -2& 3 \\ 4 & -1 \end{bmatrix}\)
    3. \({1\over10}\begin{bmatrix} 2& 3 \\ 4 & 1 \end{bmatrix}\)
    4. \({1\over10}\begin{bmatrix} 2& -3 \\ -4 & 1 \end{bmatrix}\)

    d8f68

    Correct Answer - Option 3 : \({1\over10}\begin{bmatrix} 2& 3 \\ 4 & 1 \end{bmatrix}\)Calculation:Given A satisfies the equation A2 + 3A - 10I = 0Multiply the equation by A-1⇒ A-1A2 + 3 A-1A - 10A-1I = 0⇒ A + 3I - 10 A-1 = 0⇒ \(\begin{bmatrix} -1& 3 \\ 4 & -2 \end{bmatrix}\) + 3 \(\beRead more

    Correct Answer – Option 3 : \({1\over10}\begin{bmatrix} 2& 3 \\ 4 & 1 \end{bmatrix}\)

    Calculation:

    Given A satisfies the equation 

    A2 + 3A – 10I = 0

    Multiply the equation by A-1

    ⇒ A-1A2 + 3 A-1A – 10A-1I = 0

    ⇒ A + 3I – 10 A-1 = 0

    ⇒ \(\begin{bmatrix} -1& 3 \\ 4 & -2 \end{bmatrix}\) + 3 \(\begin{bmatrix} 1&0 \\ 0 & 1 \end{bmatrix}\) – 10 A-1 = \(\begin{bmatrix} 0& 0 \\ 0 & 0 \end{bmatrix}\)

    ⇒ \(\begin{bmatrix} -1+3&3+0\\4+0&-2+3\end{bmatrix}\) = 10 A-1

    ⇒ A-1 = \({1\over10}\begin{bmatrix} 2& 3 \\ 4 & 1 \end{bmatrix}\)

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