![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-305.png\")
<\/figure>\n\n\n\nTo construct a 3\u00d73 nonzero matrix ( A ) such that the vector ([-1, 1, -1]^T) is a solution of the homogeneous equation ( A\\mathbf{x} = \\mathbf{0} ), we need to ensure that this vector lies in the null space of ( A ). The null space (or kernel) of a matrix ( A ) is the set of all vectors ( \\mathbf{x} ) for which ( A\\mathbf{x} = \\mathbf{0} ) <\/p>\n\n\n\n
Step-by-Step Construction:<\/strong><\/p>\n\n\n\n Verification:<\/strong><\/p>\n\n\n\n To verify that (\\mathbf{v} = [-1, 1, -1]^T) is in the null space of ( A ): Explanation:<\/strong><\/p>\n\n\n\n The null space of a matrix ( A ) consists of all vectors ( \\mathbf{x} ) such that ( A\\mathbf{x} = \\mathbf{0} ). In this case, we constructed ( A ) so that ([-1, 1, -1]^T) is in its null space. The row space of ( A ) is the orthogonal complement of its null space. By selecting basis vectors for the orthogonal complement and using them as rows of ( A ), we ensure that ( A\\mathbf{v} = \\mathbf{0} ). This method guarantees that the given vector lies in the null space of the constructed matrix.<\/p>\n","protected":false},"excerpt":{"rendered":" Construct a 3×3 nonzero matrix A such that the vector [-1 1 -1] is a solution of Ax = 0. The correct answer and explanation is : To construct a 3\u00d73 nonzero matrix ( A ) such that the vector ([-1, 1, -1]^T) is a solution of the homogeneous equation ( A\\mathbf{x} = \\mathbf{0} ), […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-196828","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196828","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=196828"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196828\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196828"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196828"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196828"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
The given vector (\\mathbf{v} = [-1, 1, -1]^T) should be in the null space of ( A ). To construct ( A ), we can start by considering a basis for its null space that includes (\\mathbf{v}). Since (\\mathbf{v}) is a single vector, it spans a one-dimensional subspace. However, to define a 3\u00d73 matrix with this null space, we need to consider the orthogonal complement of (\\mathbf{v}), which will form the row space of ( A ).<\/li>\n\n\n\n
The orthogonal complement of (\\mathbf{v}) consists of all vectors (\\mathbf{w}) such that (\\mathbf{w} \\cdot \\mathbf{v} = 0). Solving for (\\mathbf{w} = [w_1, w_2, w_3]^T), we get:
[
-w_1 + w_2 – w_3 = 0 \\implies w_2 = w_1 + w_3
]
Therefore, any vector orthogonal to (\\mathbf{v}) can be written as:
[
\\mathbf{w} = [w_1, w_1 + w_3, w_3]^T = w_1[1, 1, 0]^T + w_3[0, 1, 1]^T
]
The vectors ([1, 1, 0]^T) and ([0, 1, 1]^T) form a basis for the orthogonal complement of (\\mathbf{v}).<\/li>\n\n\n\n
Using the basis vectors of the orthogonal complement as rows of ( A ), we have:
[
A = \\begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 1
\\end{bmatrix}
]
To make ( A ) a 3\u00d73 matrix, we can add any nonzero row that is a linear combination of the existing rows or any other vector that maintains the rank of the matrix. For simplicity, we can add the row ([1, 0, -1]), which is also orthogonal to (\\mathbf{v}):
[
A = \\begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 1 \\
1 & 0 & -1
\\end{bmatrix}
]<\/li>\n<\/ol>\n\n\n\n
[
A\\mathbf{v} = \\begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 1 \\
1 & 0 & -1
\\end{bmatrix}
\\begin{bmatrix}
-1 \\
1 \\
-1
\\end{bmatrix}
= \\begin{bmatrix}
1(-1) + 1(1) + 0(-1) \\
0(-1) + 1(1) + 1(-1) \\
1(-1) + 0(1) + (-1)(-1)
\\end{bmatrix}
= \\begin{bmatrix}
-1 + 1 + 0 \\
0 + 1 – 1 \\
-1 + 0 + 1
\\end{bmatrix}
= \\begin{bmatrix}
0 \\
0 \\
0
\\end{bmatrix}
]
Thus, ( A\\mathbf{v} = \\mathbf{0} ), confirming that (\\mathbf{v}) is in the null space of ( A ).<\/p>\n\n\n\n