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Matrices and Determinants question

2004 · Shift 0 · Q79
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  5. /2004 · Shift 0 · Q79

Matrices and Determinants question

2004 · Shift 0 · Q79

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A=(1−1121−3111).A = \left( {\begin{matrix} 1 & { - 1} & 1 \\ 2 & 1 & { - 3} \\ 1 & 1 & 1 \\ \end{matrix} } \right).A=​121​−111​1−31​​. and 101010B=(422−50α1−23)B = \left( {\begin{matrix} 4 & 2 & 2 \\ { - 5} & 0 & \alpha \\ 1 & { - 2} & 3 \\ \end{matrix} } \right)B=​4−51​20−2​2α3​​. if BBB is the inverse of matrix AAA, then α\alphaα is
  1. A
    555
  2. B
    −1-1−1
  3. C
    222
  4. D
    −2-2−2
View written solutionFree

Correct answer: A

  1. Interpret the given matrices

We are given

A=(1−1121−3111)A=\begin{pmatrix} 1 & -1 & 1\\ 2 & 1 & -3\\ 1 & 1 & 1 \end{pmatrix}A=​121​−111​1−31​​

and

10B=(422−50α1−23).10B=\begin{pmatrix} 4 & 2 & 2\\ -5 & 0 & \alpha\\ 1 & -2 & 3 \end{pmatrix}.10B=​4−51​20−2​2α3​​.

Since B=A−1B=A^{-1}B=A−1, we use the fact that

A−1=1det⁡A adj⁡(A).A^{-1}=\frac{1}{\det A}\,\operatorname{adj}(A).A−1=detA1​adj(A).

Also, from the form of 10B10B10B, it is convenient to compute A−1A^{-1}A−1 directly.

  1. Find det⁡A\det AdetA

Expand along the first row:

det⁡A=1∣1−311∣−(−1)∣2−311∣+1∣2111∣.\det A= 1\begin{vmatrix}1 & -3\\ 1 & 1\end{vmatrix} -(-1)\begin{vmatrix}2 & -3\\ 1 & 1\end{vmatrix} +1\begin{vmatrix}2 & 1\\ 1 & 1\end{vmatrix}.detA=1​11​−31​​−(−1)​21​−31​​+1​21​11​​.

Now compute each minor:

∣1−311∣=1⋅1−(−3)⋅1=4,\begin{vmatrix}1 & -3\\ 1 & 1\end{vmatrix}=1\cdot 1-(-3)\cdot 1=4,​11​−31​​=1⋅1−(−3)⋅1=4, ∣2−311∣=2⋅1−(−3)⋅1=5,\begin{vmatrix}2 & -3\\ 1 & 1\end{vmatrix}=2\cdot 1-(-3)\cdot 1=5,​21​−31​​=2⋅1−(−3)⋅1=5, ∣2111∣=2⋅1−1⋅1=1.\begin{vmatrix}2 & 1\\ 1 & 1\end{vmatrix}=2\cdot 1-1\cdot 1=1.​21​11​​=2⋅1−1⋅1=1.

Therefore,

det⁡A=1(4)−(−1)(5)+1(1)=4+5+1=10.\det A=1(4)-(-1)(5)+1(1)=4+5+1=10.detA=1(4)−(−1)(5)+1(1)=4+5+1=10.

So,

A−1=110adj⁡(A).A^{-1}=\frac{1}{10}\operatorname{adj}(A).A−1=101​adj(A).

This matches the given form 10B=(some matrix)10B=(\text{some matrix})10B=(some matrix).

  1. Find the cofactor corresponding to the position of α\alphaα

In the given matrix

10B=10A−1=adj⁡(A),10B=10A^{-1}=\operatorname{adj}(A),10B=10A−1=adj(A),

α\alphaα is at position (2,3)(2,3)(2,3) of 10B10B10B.

Now,

adj⁡(A)=(Cji),\operatorname{adj}(A)=(C_{ji}),adj(A)=(Cji​),

where CjiC_{ji}Cji​ are cofactors of AAA.

So entry (2,3)(2,3)(2,3) of adj⁡(A)\operatorname{adj}(A)adj(A) equals cofactor C3,2C_{3,2}C3,2​ of AAA.

Let us compute C3,2C_{3,2}C3,2​.

Delete row 3 and column 2 from AAA:

(112−3)\begin{pmatrix} 1 & 1\\ 2 & -3 \end{pmatrix}(12​1−3​)

Its determinant is

1(−3)−1(2)=−3−2=−5.1(-3)-1(2)=-3-2=-5.1(−3)−1(2)=−3−2=−5.

Therefore,

C3,2=(−1)3+2(−5)=(−1)5(−5)=5.C_{3,2}=(-1)^{3+2}(-5)=(-1)^5(-5)=5.C3,2​=(−1)3+2(−5)=(−1)5(−5)=5.

Hence,

α=5.\alpha=5.α=5.
  1. Check with options

The correct option is:

A: 5\boxed{\text{A: }5}A: 5​
  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So the stored answer is correct.

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